(*) Basic Proportionality Theorem(Thales Theorem)
Theorem:- if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio
Given : In △ABC, DE // BC which intersects sides AB and AC at "d" and "e" respectively
RTP: AD = AE
DB EC
Construction :
Join B,E and C,D and then draw
DM _|_ AC and EN _|_ AB.
(*) Proof :
Area of △ADE = (1/2) * AD * EN
Area of △BDE = (1/2) * BD * EN
So,
ar( △ADE)=(1/2)*AD*EN=AD---(1)
ar( △BDE) (1/2)*BD*EN=BD
Again
Area of △ADE = (1/2)*AE*DM
Area of △CDE = (1/2)*EC*DM
ar(△ADE)=(1/2)*AE*DM=AE ---(2)
ar(△CDE) (1/2)*EC*DM EC
Observe that ⧍BDE and⧍ CDE are on the same base DE and between same parallels BC and DE.
So,
ar(⧍BDE) = ar(⧍CDE) ---(3)
From(1),(2) and (3), we have
AD = AE
DB EC
Hence proved
Is the converse of the above theorem also true? To examine this, let us perform the following activity
(*)Activity:-
Draw an angle XAY on your note book and on ray AX, mark points B1,B2,B3,B4 and B suc that
AB1=B1B2=B2B3=B3B4=B4B=1cm(say)
Similarly on ray AY, mark points C1,C2,C3,C4 and C such that
AC1=C1C2=C2C3=C3C4=C4C=2cm(say)
Join B1, C1 and B, C.
Observe that
AB1 = AC1 = 1 and B1C1 // BC
B1B C1C 4
Similarly, joining
B2C2, B3C3 and B4C4, you see that
AB2 = AC2 = 2 and B2C2 //BC
B2B C2C 3
AB3 = AC3 =3 and B3C3 // BC
B3B C3C 2
AB4 = AC4 = 4 and B4C4 // BC
B4B C4C 1
From this we obtain the following theorem called converse of the Thales theorem
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Theorem-8.2:- if a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side.
Given: In △ABC, a line DE is drawn such that
AD = AE
DB EC
RTP : DE // BC
proof : Assume that DE is not parallel to BC then draw the line DE' parallel to BC
So,
AD = AE'
DB E'C
.^. AE = AE'
EC E'C'
Adding "1" on both sides of the above, you can see that "E" and E' must coincide.
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Now let us solve some examples on Thales theorem and its converse.
Example -1: In △ABC, DE//BC and
AD = 3 , AC = 5.6, Find AE.
DB 5
sol) In △ABC, DE//BC
=> AD = AE (by B.P.T)
DB EC
but
AD = 3
DB 5
So,
AE = 3
EC 5
Given:- AC = 5.6 and AE:EC=3:5
AE = 3
AC - AE 5
AE = 3 (cross multiply)
5.6 - AE 5
5AE = 3*5.6 - 3AE
8AE = 16.8
AE = 16.8
8
AE = 2.1 cm
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Example-2: In the given figure LM//AB AL = x-3, AC=2x, BM=x-2 and BC=2x+3 find the value of "x"
sol) In △ABC, LM // AB
=> AL = BM (by B.P.T)
LC MC
x-3 = x-2
2x-(x-3) (2x+3)-(x-2)
x-3 = x-2 (cross multiply)
x+3 x+5
(x+5)(x-3) = (x-2)(x+3)
x^2 +2x -15 = x^2 +x -6
=> 2x -x = -6 +15
.^. x = 9
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Example-3) The diagonals of a quadrilateral ABCD intersect each other at point 'O' such that
AO = CO. Prove that ABCD is a trapezium
BO DO
Sol) Given : In quadrilateral ABCD,
AO = CO
BO DO
RTP: ABCD is a trapezium
Construction: Through 'O' draw a line parallel to AB which meets DA at X.
Proof : In △DAB, XO//AB (by construction)
DX = DO (by B.P.T)
XA OB
AX = BO
XD OD ------(1)
again
AO = CO (given)
BO DO
AO = BO
CO OD ------(2)
From(1) and (2)
AX = AO
XD CO
In △ADC, XO is a line such that
AX = AO
XD OC
=> XO // DC (by converse of the B.P.T)
=> AB // DC
In quadrilateral ABCD, AB//DC
=> ABCD is a trapezium (by definition)
Hence proved.
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Example 4) In trapezium ABCD, AB//DC, E and F are points on non-parallel sides AD and BC respectively such that EF // AB. Show that
AE = BF
ED FC
Sol) Let us join AC to intersect EF at G.
AB//DC and EF//AB (given)
=> EF// DC (Lines parallel to the same line are parallel to each other)
In △ADC, EG//DC
So
AE = AG (by BPT)
ED GC
Similarly, In △CAB, GF//AB
CG = CF (by BPT)
GA FB
i.e, AG = BF ----(2)
GC FC
From(1) and (2)
AE = BF
ED FC
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Exercise -8.1
(1) In triangle PQR, ST is a line such that
PS = PT and also / PST = /_ PRQ.
SQ TR
Prove that △ PQR is an isosceles triangle.
sol) Given :- PS = PT
SQ TR
2) ST || QR ( By Converse of BPT)
Basic Proportionality Theorem (BPT) :- If a line is drawn parallel to one side of a triangle to intersect the other two side in distinct points, the other two sides are divided in the same ration.
3) /_PST = /_ PQR ( Corresponding angles are always equal)
Given :- /_ PST = /_ PRQ
4) /_PQR = /_PRQ
We know :- Two angles or two sides of an isosceles triangle are equal.
.^. /_\PQR is an isosceles triangle.
(2) In the given figure, LM // CB and LN // CD
prove that AM = AN
AB AD
sol) Consider /_\ ABC
1)Given :- LM || CB
2) AM = AL (by BPT)
MB LC
Consider /_\ ADC
3) Given :- LN ||CD
4) AN = AL (by BPT)
ND LC
equating (2) and (4)
5) AM = AN
MB ND
Taking the reciprocal of (5)
6) MB = ND
AM AN
Adding 1 on both sides
7) MB + 1 = ND + 1
AM AN
8) (MB + AM) = (ND + AN)
AM AN
9) AB = AD
AM AN
10) AM = AN
AB AD
Hence proved
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(3) In the given figure, DE //AC and DF //AE.
Prove that BF = BE
FE EC
sol) Consider /_\ ABE
1) Given :- DF || AE
2) BF = BD (By BPT)
FE DA
Consider /_\ ABC
3) Given :- DE || AC
4) BE = BD ( by BPT)
EC DA
equating (2) and (4)
5) BF = BE
FE EC
Hence proved .
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(4) In the given figure,
AB // CD // EF,
given
AB = 7.5 cm
DC = ycm,
EF = 4.5 cm,
BC = x cm.
Calculate the values of "x" and "y"
sol) Consider △ ACB and △ CEF
Both are similar traingles
.^. AB = BC
EF CF
Given :- AB = 7.5 cm and EF = 4.5 cm
7.5 = x
4.5 3
7.5 *3 = x
4.5
22.5 = x (4.5 * 5 = 22.5)
4.5
5 = x
Consider △ BCD and △ BFE
From BPT theorem, both are similar triangles
x = x+3
y 4.5
y = 5 * 4.5
8
y = 5*45
8*10
y = 225
80
y = 45
16
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(5) Prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side ( using basic proportionality theorem )
sol) Let us assume △ ABC
DE || BC
"D" is the mid-point of AB
RTP :- "E" is the mid-point of AC
We know that :- If a line is drawn parallel to one side of a triangle to intersect the other two side in distinct points, the other two sides are divided in the same ration.
AD = AE
DB EC
DB = AE
DB EC
1 = AE
EC
EC = AE
=> E is the mid-point of AC
Hence proved.
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(6) Prove that a line joining the midpoints of any two sides of a triangle is parallel to the third side ( Using converse of basic proportionality theorem )
sol) Assume :- /_\ ABC
"D" is the mid-point of AB
"E" is the mid-point of AC
RTP :- DE || BC
Proof :- In /_\ ABC,
"D" is the mid-point of AB
=> AD = DB
=> AD = 1 -----(1)
DB
"E" is the mid-point of AC
=> AE = EC
=> AE = 1 -----(2)
EC
eq (1) & (2)
AD = AE
DB EC
.^. DE || BC ( By BPT)
Hence proved.
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(7) In the given figure, DE // OQ and DF // OR . Show that EF // QR.
sol) Consider /_\ POQ
1) Given :- DE || OQ
2) PE = PD (by BPT)
EQ DO
Cnsider /_\ POR :-
3) Given :- DF || OR
4) PF = PD (by BPT)
FR DO
Equating (2) & (4)
5) PE = PF
EQ FR
6) EF || QR ( By BPT)
Hence proved.
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(8) In the adjacent figure A, B and C are points on OP, OQ and OR respectively such that AB // PQ and AC // PR. Show that BC // QR.
sol) Consider /_\ OPQ
1) Given :-AB || PQ
2) OA = OB (By BPT)
AP BQ
Consider /_\ OPR
3) Given :-AC || PR
4) OA = OC (By BPT)
AP CR
equating (2) & (4)
5) OB = OC
BQ CR
6) BC || QR ( By BPT)
Hence proved.
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9) ABCD is a trapezium in which AB || DC and its diagonals intersect each other at point "O". Show that
AO = CO
BO DO
sol) Draw a line EF || CD which is passing through "O".
In /_\ ABC and /_\ EOC
These are similar triangles as per BPT.
AE = BO ------(1)
EC OC
Similarly, in /_\ BOD and /_\ FOD
BF = AO -----(2)
FD OD
in /_\ ABC and /_\ BAD
BO = AO ----(3)
OC OD
Because diagonals of a trapezium divide each other in same ratio
From the above three equations, it is clear
AE = BF
EC FD
Hence , /_\ ABC ~ /_\ BAD
using the third equation
BO = AO
OC OD
or
AO = CO
BO DO
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(10) Draw a line segment of length 7.2 cm and divide it in the ration 5 : 3. Measure the two parts.
sol) Take m = 5 and n = 3
Steps of construction :-
1) Draw line AB of length 7.2
2) Draw a ray AX, making an acute angle(less then 90 degree) with AB.
3) Mark 8 ( 5 +3) points A1, A2. A3.....A8 on AX such that AA1 = A1A2 = so on..
4) join BA8
5) Since we want the ratio 5:3 , Through point A5 (m=5), we draw a line parallel to BA8 (by making an angle equal to /_A A8 B at A5 ) intersecting AB at the point "C".
then, AC : CB = 5 : 3
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AAA Criterion for similarity of Triangles
Theorem-8.3 : In two triangles , if the angles are equal, then the sides opposite to the equal angles are in the same ratio( or proportional) and hence the two triangles are similar.
Given: In triangles ABC and DEF,
∠A = ∠D
∠B = ∠E
∠C = ∠F
R.T.P :-
AB = BC = AC
DE EF DF
Construction: Locate points "P" and "Q" on DE and DF respectively.such that AB=DP and AC=DQ.
Join PQ.
Proof : ABC ~ DPQ (why?)
This gives B = P = E and PQ//EF (how?)
Therefore
DP = DQ (why?)
PE QF
i.e, AB = AC (why?)
DE DF
Similarly
AB = BC
DE EF and So
AB = BC = AC
DE EF DF
Hence proved.
Note : If two angles of a triangle are respectively equal to the two angles of another triangle, then by the angle sum property of a triangle, third angles will also be equal.
So AA similarity criterion is stated as if two angles of one triangle are respectively equal to the two angles triangles, then the two triangles are similar.
What about the converse of the above statement?
If the sides of a triangle are respectively proportional to the sides of another triangle , is it true that their corresponding angles are equal?
Let us exercise it through an activity.
Activity :-
(*) Draw two triangles ABC and DEF such that
AB = 3cm,
BC = 6cm,
CA = 8cm,
DE = 2.5 cm,
EF = 9cm and
FD = 12cm.
sol) we have
AB = 3 = 30 = 10 = 2
DE 4.5 45 15 3
BC = 6 = 2
EF 9 3
CA = 8 = 2
DF 12 3
AB = BC = CA = 2
DE EF FD 3
Now measure the angles of both the triangles. What do you observe? what can you say about the corresponding angles? They are equal, so the triangles are similar. You can verify it for different triangles.
from the above activity, we can give the following criterion for similarity of two traingles.
(*) SSS Criterion for similarity of Triangles
Theorem-8.4: If in two triangles, the sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are equal and hence the triangles are similar.
Given : ∆ABC and ∆DEF are such that
AB = BC = CA
DE EF FD (< 1)
RTP :
∠A = ∠D,
∠B = ∠E,
∠C = ∠F
Construction : Locate points P and Q on DE and DF respectivelysuch that AB = DP and AC = DQ. Join PQ.
Proof :
DP = DQ and PQ // EF (why?)
PE QF
So ∠P = ∠E and ∠Q = ∠F (why ?)
Therefore,
DP = DQ = PQ
DE DF EF
So,
DP = DQ = BC (why?)
DE DF EF
So BC = PQ (Why ?)
∆ABC ≅ ∆DPQ (why ?)
So ∠A = ∠D, ∠B = ∠E and ∠C = ∠F (How ?)
We studied that for similarity of two polygons any one condition is not sufficient. But for the similarity of triangles, there is no need for fulfillment of both the conditions as one automatically implies the other. Now let us look for SAS similarity criterion.
(*) SAS Criterion for similarity of Triangles
Theorem-8.5 :- If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.
Given:- In △ABC and △DEF
AB = AC (<1) and
DE DF
∠A = ∠D
RTP : △ABC ~ △DEF
Construction:- Locate points P and Q on DE and DF respectively such that AB=DP and AC=DQ.
Join PQ.
Proof: PQ//EF and △ABC ~ △DPQ
So
∠A = ∠D
∠B = ∠P
∠C = ∠Q
.^. △ABC ~△ DEF
Construction : To construct a triangle similar to a given triangle as per given scale factor.
a) Construct a triangle similar to a given triangle ABC with its sides equal to (3/4) of corresponding sides of ∆ABC (scale factor (3/4)
Steps :
1). Draw a ray BX, making an acute angle with BC on the side opposite to vertex A.
2). Locate 4 points B1, B2, B3 and B4 on BX so that BB1 = B1B2 =B2B3 = B3B4.3.
3). Join B4C and draw a line through B3 parallel to B4C intersecting BC at C′.
4). Draw a line through C′ parallel to CA to intersect AB at A′.So ∆A′BC′ is the required triangle.
Let us take some examples to illustrate the use of these criteria.
Example-5. A person 1.65m tall casts 1.8m shadow. At the same instance, a lamp-posts casts a shadow of 5.4 m. Find the height of the lamppost.
Solution:In ∆ABC and ∆PQR
∠B = ∠Q= 90 deg.
∠C = ∠R (AC || PR, all sun’s rays are parallel at any instance)
∆ABC ~ ∆PQR ( by AA similarity)
AB = BC
PQ QR (cpst, corresponding parts of Similar triangles)
1.65 = 1.8
PQ 5.4
PQ = 1.65 * 5.4 = 4.95 m.
1.8
The height of the lamp post = 4.95m
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Example-6. A man sees the top of a tower in a mirror which is at a distance of 87.6m from the tower. The mirror is on the ground facing upwards. The man is 0.4m away from the mirror and his height is 1.5m. How tall is the tower?
solution: In ∆ABC & ∆EDC
∠ABC=∠EDC=90°
∠BCA=∠DCE (angle of incidence and angle of reflection are same)
∆ABC ~ ∆EDC (by AA similarity)
AB = BC
ED CD
=> 1.5 = 0.4
h 87.6
h = 1.5 * 87.6 = 328.5m
0.4
Hence , the height of the towers is 328.5 m.
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Example-7). Gopal is worrying that his neighbour can see into his living room from the top floor of his house.
He has decided to build a fence that is high enough to block the view from their top floor window. What should be the height of the fence?
The measurements are given in the figure
solution:- In △ABD & △ACE
∠B = ∠C = 90 deg
∠A = ∠A (common angle)
△ABD ~ △ACE (by AA similarity)
AB = BD
AC CE
2 = BD
3 1.2
BD = 2 *1.2
8
=> 2.4
8
=> 24 = 3 = 0.3m
80 10
Total height of the fence required is 1.5m + 0.3m = 1.8m to block the neighbour's view
Exercise-8.2
1). In the given figure, ∠ADE = ∠B
(!) Show that △ABC ~ △ADE
2) If AD = 3.8 cm, AE = 3.6 cm, BE = 2.1 cm, BC = 4.2 cm find DE.
sol) given:- ∠ADE = ∠B
∠C = 90,
∠A = 90 -0
If ∠A = 90 -0 , ∠B = 0 => ∠AED = 90
Comparing △ABC with △ AED we have
∠A common in both triangles
∠C = ∠AED = 90
∠ADE = ∠B
.^. By AAA property, △ ABC and △ ADE are similar triangles
△ABC ~ △ADE
Hence proved
2sol) Given AD = 3.8 cm, AE = 3.6 cm, BE = 2.1 cm, BC = 4.2
As △ ABC and △ ADE are similar triangles,
We have
AB = BC = AC
AD DE AE
AB = 4.2 = AC
3.8 DE 3.6
But, AE + BE = AB
3.6 + 2.1 = 5.7
Now,
5.7 = 4.2
3.8 DE
DE =4.2 * (2/3)
DE = 1.4 * 2
DE = 2.8 cm
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(2) The perimeters of two similar triangles are 30 cm and 20 cm respectively. If one side of the first triangle is 12 cm, determine the corresponding side of the second triangle.
sol) Since the triangle are similar
perimeter of 1st △ = Any side of 1st △
perimeter of 2nd△ corresponding side of 2nd△
30 = 12
20 x
x = 20 *12
30
x = 8 cm
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(3) A girl of height 90 cm is walking away from the base of a lamp post at a speed of 1.2 m/sec. If the lamp post is 3.6m above the ground . Find the length of her shadow after 4 seconds.
sol) Lamp post (PQ) = 3.6 m
Height of girl(ST) = 90 cm
(As lamp post is in "m" convert "cm" into "m")
1m = 100 cm
Height of girl(ST) = 90 m = 0.9 m
100
Speed = 1.2 m/sec,
RTP :- Length of her shadow i.e. TR
Solution :- The girl walks "QT" distance in 4 seconds
We have formula :-
Speed = Distance
Time
1.2 = QT
4
1.2 * 4 = QT
4.8 = QT
QT = 4.8 m
Now,
In /_\ PQR and /_\ STR
/_R = /_R [Common]
/_Q = /_T ( Both 90 deg because lamp post as well as the girl are standing to the ground)
.^. , using AA similarity criterion
ST = RT (by cpst)
PQ QR
0.9 = RT
3.6 RT + TQ
1 = x
4 x + 4.8
x + 4.8 = 4x
3x = 4.8
x = 1.6m
RT = 1.6 m
.^. The length of her shadow after 4 sec = 1.6m
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(4) CM and RN are respectively the medians of
Prove that
sol) Given :- /_\ ABC ~ /_\PQR
So, AB = BC = CA ----(1)
PQ QR RP
and /_A = /_P, /_B = /_Q , /_C = /_R ----(2)
But AB = 2 AM and PQ = 2 PN ( As "CM" and "RN" are medians)
So, from (1), 2AM = CA
2PN RP
i.e, AM = CA -------(3)
PN RP
Also , /_\ MAC = /_\NPR (from 2) -------------(4)
So, from (3) and (4),
/_\ AMC ~ /_\PNR (SAS similarity) -------(5)
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(2) CM = AB
RN PQ
sol) /_\ AMC ~ /_\PNR (SAS similarity) ----(5)
from (5), CM = CA -----(6)
RN RP
---------------------------------------
AB = BC = CA ----(1)
PQ QR RP
But, CA = AB
RP PQ ( From 1 )----------( 7 )
.^. CM = AB (From (6) & (7) ------(8)
RN PQ
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sol) From (1)
AB = BC
PQ QR
.^. CM = BC ( from (8)] -----(9)
RN PQ
Also, CM = AB = 2BM
RN PQ 2QN
i.e, CM = BM ------(10)
RN QN
from (9) and (10)
i.e, CM = BC = BM
RN QR QN
.^. /_\ CMB ~ /_\RNQ (SSS similarity)
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(5) Diagonals AC and BD of a trapezium ABCD with AB //DC intersect each other at the point "O". Using the criterion of similarly for two triangles, show that
OA = OB
OC OD
sol)
Consider /_\ AOB and /_\ DOC
/_AOB = /_DOC (vertically opposite angles)
/_OAB = /_OCD ( alternative interior angle)
/_OBA = /_ODC ( alternative interior angle)
.^. /_\ AOB ~ /_\ DOC ( AAA test of similarity)
OA = OB
OC OD (corresponding parts of similar triangle)
Hence proved.
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(6) AB, CD , PQ are perpendicular to BD. AB = x, Cd =y and PQ = Z.
Prove that
1 + 1 = 1
x y z
sol) In /_\BCD. PQ || CD
BQ = PQ -------(1)
BD CD
In /_\ABD, PQ || AB
QD = PQ
BD AB
1 - BQ = PQ
BD AB
1 - PQ = PQ (from 1)
CD AB
1 = PQ ( 1 + 1 )
CD AB
1 = 1 + 1
PQ CD AB
But , given AB = x, CD = y and PQ = z
1 = 1 + 1
Z Y X
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(7) A flag pole 4m tall casts a 6m, shadow. At the same time, a nearby building casts a shadow of 24m. How tall is the building?
sol)
In /_\ ABC and /_\ PQR
/_B = /_Q = 90 deg
/_C = /_R (AC || PR, all sun's rays are parallel at any instance)
/_\ABC ~ /_\PQR ( by AA similarity)
AB = BC
PQ = QR (cpst)
4 = 6
PQ 24
PQ = 4 * 24
6
PQ = 16 m. the height of the building
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(8) CD and GH are respectively the bisectors of /_ACB and /_EGF such that D and H lie on sides AB and FE of /_\ABC and /_\ FEG respectively.
then show that
(1) CD = AC
GH FG
sol)
1) Given :- /_\ ABC ~ /_\FEG
2) /_BAC = /_EFG (Corresponding angle of similar triangle)
3) /_ABC = /_FEG (corresponding angle of similar triangle)
4) /_ACB = /_FGE
5) 1/2 /_ACB = 1/2 /_FGE
6) /_ACD = /_FGH and /_BCD = /_EFGH
consider /_\ ACD and /_\ FGH
7) /_DAC = /_HFG (from 2)
8) /_ACD = /_EGH (from 6)
9) Also, /_ADC = /_FGH
If the two angle of triangle are equal to the two angle of another triangle, then by angle sum property of /_\, 3rd angle will also be equal.
10) .^. /_\ADC ~ /_\FHG (By AAA test of similarity)
11) CD = AC (CPST)
GH FG
Consider /_\DCB amd /_|HGE
12) /_DBC = /_HEG (from 3)
13) /_BCD = /_FGH (from 6)
14) also, /_BDC = /_EHG
15) /_\DCB ~ /_\HGE
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(9) AX and DY are altitudes of two similar triangles
Prove that AX : DY = AB : DE.
sol)
Given :- /_\ ABC ~ /_\DEF
2) /_ABC = /_DEF ( CAST)
Consider /_\ABX and /_\DEY
3) /_ABX = /_DEY (from 2)
4) /_AXB = /_DYE = 90 deg
5) /_BAX = /_EDY
6) /_\ABX ~ /_\DEY (by AAA test of similarity)
7) AX = AB (CPST)
DY DE
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(10) Construct a triangle shadow similar to the given /_\ ABC, with its sides equal to 5/3 of the corresponding sides of the triangle ABC.
sol) Given a triangle ABC, we are required to construct a triangle whose sides are of the corresponding sides of /_\ABC
Steps of Construction :-
1) Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.
2) Locate 5 points ( the greater of 5 and 3 in 5/3) B1,B2,B3,B4 and B5 so that BB1=B1B2=B2B3=B3B4=B4B5.
3) Join B3( the 3rd point , 3 being smaller of 3 and 5 in) to C and draw a line through B5 parallel to B3C, intersecting the extended line segment BC at C'.
4) Draw a line through C' parallel to CA intersecting the extended line segment BA at A'.
Then A'BC' is the required triangle.
For justification of the construction , note that /_\ABC ~ /_\A'BC'.
.^., AB = AC = BC
A'B A'C' BC'
So, BC = 5 and
BC' 3
.^. A'B = A'C' = BC' = 5
AB AC BC 3
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(11) Construct a triangle of sides 4cm, 5cm, and 6 cm. Then, construct a triangle similar to it, whose sides are 2/3 of the corresponding sides of the first triangle.
sol) Construct a triangle ABC with given sides , AB = 4cm, BC = 5cm, CA = 6cm.
Given a triangle ABC, we are required to construct a triangle whose sides are 2/3 of the corresponding sides of /_\ABC.
Steps of construction :-
1) Draw any ray BX making an angle 30 deg with the base BC of /_\ ABC on the opposite side of the vertex A.
2)Locate seven points B1,B2,B3 on BX so that BB1=b1B2=B2B3
Note:- The number of points should be greater of "m" and 'N" in the scale factor m/n)
3) Join B2- C' and draw a line through B3 || B2C, intersecting the line segment BC at C,
4) Draw a line through "C" parallel to C'A intersecting the line segment BA at A'B'C is the required triangle.
*********************************************************************************
(12) Construct an Isosceles triangle whose base is 8cm and altitude is 4cm. Then, draw another triangle whose sides are 1(1/2) times the corresponding sides of the isosceles triangle.
sol) An isosceles tranagle whose base is 8cm and altitude is 4cm. Scale factor is 1(1/2) = 3/2.
Steps of construction:
1) Draw a line segment BC= 8cm.
2) Draw a perpendicular bisector AD of BC
3) Join AB and AC we get an isosceles triangle /_\ABC
4) Construct an acute angle /_CBX downwards.
5) On BX make three equal parts.
6) Join C to B2 and draw a line through B3 parallel to B2C intersecting the line extended line segment BC at C'
7) Again draw a parallel line C'A' to AC cutting BP at A'
8) /_\A'BC' is the required triangle.
************************************************
Areas of Similar Triangles
For two similar triangles , ratio of their corresponding sides is the same.
Theorem-8.6: The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.
Given : △ABC ~ △PQR
RTP :
ar(△ABC) = (AB )^2 = (BC)^2 = (CA)^2
ar(△PQR) = (PQ)^2 = (QR)^2 = (RP)^2
Construction: Draw AM _|_ BC and PN_|_QR
Proof:
ar(△ABC) = (1/2)*BC*AM = BC*AM ---(1)
ar(△PQR) (1/2)*QR*PN QR*PN
In △ABM & △PQN
∠B = ∠Q (△ABC~ △PQR)
∠M = ∠N = 90 deg
.^. △ABM ~ △PQN (by AA similarity)
AM = AB------(2)
PN PQ
Also △ABC ~ △PQR(given)
AB = BC = AC
PQ QR PR -------(3)
Therefore,
ar(△ABC) = AB * AB = (AB)^2
ar(△PQR) PQ PQ (PQ)^2
Now by using (3), we get
ar(△ABC) = (AB)^2 = (BC)^2 = (AC)^2
ar(△PQR) (PQ)^2 (QR)^2 (PR)^2
Hence proved.
*****************************************************
Example-8. prove that if the area of two similar traingles are equal, then they are congruent.
sol) △ABC ~ △PQR
ar(ABC)=(AB)^2= (BC)^2 = (AC)^2
ar(PQR) (PQ)^2 (QR)^2 (PR)^2
But
ar(△ABC) = 1 --->(Areas are equal)
ar(△PQR)
(AB)^2 = (BC)^2 = (AC)^2 = 1
(PQ)^2 (QR)^2 (PR)^2
So
AB^2 = PQ^2
BC^2 = QR^2
AC^2 = PR^2
From which we get
AB = PQ
BC = QR
AC = PR
.^. △ABC ~ △PQR ( by SSS congruency)
*****************************************************
Example-9. ABC ~ DEF and their areas are respectively 64cm^2 and 121cm^2. If EF = 15.4 cm, then find BC.
sol) ar(△ABC) = (BC)^2
ar(△DEF) (EF)^2
64 = (BC)^2
121= (15.4)^2
8 = (BC)
11 = 15.4
BC = 8 * 15.4
11
BC = 11.2 cm.
******************************************************
Example-10. Diagonals of a trapezium ABCD with AB//DC, intersect each other at the point 'O'. If AB = 2CD, find the ratio of triangles AOB and COD
sol) In trapezium ABCD, AB//DC also AB=2CD
In △AOB and △COD
∠AOB = ∠COD (vertically opposite angles)
∠OAB = ∠OCD (alternate interior angles)
△AOB ~ △COD ( by AA similarity)
ar(△AOB) = AB^2
ar(△COD) DC^2
(2DC)^2 = 4
(DC)^2 1
.^. ar(△AOB) : ar(△COD) = 4:1
Exercise- 8.3
(1) Equilateral triangles are drawn on the three sides of a right angled triangle. Show that the area of the triangle on the hypotenuse is equal to the sum of the areas of triangles on the other two sides.
sol) Given right angled triangle is ABC with AC as hypotenuse
let
AB = a
BC = b
AC = c
we have a^2 + b^2 = c^2 --------(1)
We know that area of equilateral triangle
= _/3 * side^2
4
Area of ACD = _/3 c^2
4
Area of BCF = _/3 b^2
4
Area of AEB = _/3 a^2
4
Area of AEB + Area of BCF
= _/3 (a^2 + b^2)
4
from 1
a^2 + b^2 = c^2
Area of AEB + Area of BCF
= _/3 c^2 = Area of ACD
4
**********************************************
(2) Prove that the area of the equilateral triangle described on the side of a square is half the area of the equilateral triangles described on its diagonal.
sol) Let us take a square with side 'a'
Then the diagonal of square will be a_/2
Area of equilateral traingle with side '
a'= -/3 a^2
4
Area of equilateral triangle with side
a_/2 = _/3 (a_/2)^2
4
=> _/3 2a^2
4
Ratio of two areas can be given as follows :
_/3 a^2
4
----------- = 1 /2
_/3 2a^2
4
*************************************
(3) D, E, F are mid points of sides BC, CA, AB of /_\ ABC . Find the ratio of areas of /_\ DEF and /_\ ABC.
sol)
Since "D" and "F" are midpoints of AB and AC,
DE || AF or DF || BE
similarly EF || AB or EF || DB
AFED is a parallelogram as both pair of opposite side are parallel
By the property of parallelogram
/_DBE = /_DFE
or
/_DEF = /_ABC ----(1)
similarly
/_FEB = /_ACB ----(2)
In /_\DEF and /_\ABC
equation (1) and (2)
/_\DEF ~ /_\CAB
ar (/_\DEF)
ar (/_\ABC)
=> = DE^2
2DE^2
=> 1
4
ar (/_\DEF) : ar (/_\ABC) = 1:4
*****************************
(4) In /_\ ABC, XY // AC and XY divides the traingle into two parts of equal area. Find the
ratio of AX .
XB
sol)
Given :- XY || AC
=> /_1 = /_3 and /_2 = /_4 [ Corresponding angles]
=> /_\BXY ~ /_\BAC [ AA similarity]
=> ar (/_\BXY)
ar (/_\BAC)
=> BX^2 [By theorem] ----(1)
BA^2
Also, we are given that
ar (/_\BXY) = 1/2 ar (/_\BAC)
=> ar (/_\BXY)
ar (/_\BAC)
=> 1/2 -------(2)
From (1) and (2)
BX^2 = 1
BA^2 2
BX = 1
BA _/2
Now, AX
XB
=> AB -BX
XB
=> AB - 1
BX
=> _/2 -1
1
AX = _/2-1
XB 1
**************
(5) Prove that the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
sol)
In case of two similar triangles ABC and PQR
ar (ABC) = AB^2 = AC^2
ar (PQR) PQ^2 PR^2
Let us assume "AD" adn "PM" are the medians of these two triangles.
Then;
AB^2 = AC^2 = AD^2
PQ^2 PR^2 PM^2
Hence ,
ar (ABC) = AD^2
ar (PQR) PM^2
*******************************************************
(6) /_\ABC ~ /_\ DEF , BC = 3cm, EF = 4cm and area of /_\ ABC = 54 cm^2. Determine the area of /_\ DEF.
sol) Given BC = 3cm and EF = 4 cm
ar (/_\ABC) = [ BC ]^2
ar (/_\DEF) [ EF ]^2
54 = 9
ar (/_\DEF) 16
ar (/_\DEF) = 54 * 16
9
ar (/_\DEF) = 96 cm^2
************************
(7) ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP =1cm, and BP = 3cm, AQ = 1.5 cm, CQ = 4.5 cm.
Prove that ( area of /_\APQ = 1/16 area of /_\ABC)
sol)
From above figure its evident that /_\ABC ~ /_\APQ
We know that
Area of /_\APQ
Area of /_\ABC
=> [ AP ]^2
AP +PB
=> [ AQ ]^2
AQ + QC
=> 1
16
=> Area of /_\APQ = 1/16 Area of /_\ABC
******************************************
(8) The area of two similar triangles are 81 cm^2 and 49 cm^2 respectively. If the attitude of the bigger triangle is 4.5cm. Find the corresponding attitude for the smaller triangle.
sol)
Let AD and PM are corresponding altitude of triangle.
ar (/_\ ABC ) = [ AD ]^2
ar (/_\DEF) [ PM]^2
81 = (4.5)^2
49 DQ
DQ = 7 * 4.5 = 3.5 cm
9
*******************************************************
Theorem-8.7 : If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, then the triangles on both sides of the perpendicular are similar to the whole triangle and to each other.
Proof: ABC is a right triangle, right angled at "B". Let "BD" be the perpendicular to hypotenuse "AC"
In △ADB and △ABC
∠A = ∠A
and ∠ADB = ∠ABC
So △ADB ~ △ABC -----(1)
Similarly,△ADB ~ △ABC ----(2)
So from (1) and (2), triangles on both sides of the perpendicular BD are similar to the whole triangle ABC.
Also since △ADB ~ △ABC
△BDC ~ △ABC
So △ADB ~ △BDC
*******************************************************
Pythagoras Theorem (Baudhayan Theorem)
Theorem-8.8 : In a right triangle, the square of hypotenuse is equal to the sum of the squares of the other two sides.
Given : △ABC is a right triangle right angled at "B"
RTP : AC^2 = AB^2 + BC^2
Construction : Draw BD _|_ AC
proof :△ ADB ~ △ABC
AD = AB (sides are proportional)
AB AC
AD . AC = (AB)^2 ------(1)
Also, △BDC ~ △ABC
CD = BC
BC AC
CD . AC = BC^2 ----(2)
On adding (1) & (2)
AD.AC + CD. AC = AB^2 + BC^2
AC(AD + CD) = AB^2 + BC^2
AC.AC = AB^2 + BC^2
AC^2 = AB^2 + BC^2
The above theorem was earlier given by an ancient Indian mathematician Baudhayan(about 800 BC) in the following form.
" The diagonal of a rectangle produces by itself the same area as produced by its both sides (i.e, length and breadth)." So sometimes, this theorem is also referred to as the Baudhayan theorem.
Theorem-8.9: In a triangle if square of one side is equal to the sum of squares of the other two sides, then the angle opposite to the first side is a right angle and the triangle is a right angled triangle.
Given : In △ABC ,
AC^2 = AB^2 + BC^2
RTP: ∠B = 90 deg
Construction : Construct a right angled triangle △PQR right angled at "Q" such that PQ = AB and QR = BC.
Proof : In △PQR ,
PR^2 = PQ^2 + QR^2 (Phy theorem as ∠Q = 90 deg)
PR^2 = AB^2 + BC^2 ( by construction) ----(1)
but AC^2 = AB^2 + BC^2 ( given) ----(2)
.^. AC = PR from (1) and (2)
Now in △ABC and △PQR
AB = PQ ( by construction)
BC = QE ( by construction)
AC = PR ( proved)
.^. △ABC ~ △PQR ( by SSS congruency)
.^. ∠B = ∠Q (by cpct)
but ∠Q = 90 deg ( by constrcution)
.^. ∠B = 90 deg
Hence proved.
**********************************************************
**
Example-11. A ladder 25m long reaches a window of building 20m above the ground, determine the distance of the foot of the ladder from the building.
sol) In △ABC , ∠C = 90 deg
AB^2 = AC^2 + BC^2 ( By pythagorous theorem)
(25)^2 = (20)^2 + BC^2
BC^2 = 625 - 400 = 225
BC = √ 225 = 15m
Hence, the foot of the ladder is at a distance of 15m from the building.
*************************************************
Example-12. BL and CM are medians of a triangle ABC right angled at "A".
Prove that 4(BL^2 + CM^2) = 5BC^2.
sol) BL and CM are medians of △ABC in which ∠A = 90deg
In △ ABC
BC^2 = AB^2 + AC^2 (Phythagorous theorem) --(1)
In △ ABL
BL^2 = AL^2 + AB^2
So
BL^2 = (AC/2)^2 + AB^2 ( L is the midpoint of AC)
BL^2 = (AC^2/4) + AB^2
4BL^2 = AC^2 + 4AB^2
In △CMA ,
CM^2 = AC^2 + AM^2
CM^2 = AC^2 + (AB)^2 (M is the midpoint of AB)
(2)^2
CM^2 = AC^2 + (AB^2)
4
4CM^2 = 4AC^2 + AB^2 ----(3)
On adding (2) and (3), we get
4(BL^2 + CM^2) = 5(AC^2 + AB^2)
.^. 4(BL^2 + CM^2) = 5BC^2 (from 1)
*******************************************************
Example-13. 'O' is any point inside a rectangle ABCD.
Prove that OB^2 + OD^2 = OA^2 + OC^2
sol) Through 'O' draw PQ//BC so that 'P' lies on 'AB' and 'Q' lies on 'DC'
Now PQ//BC
.^. PQ_|_AB & PQ_|_DC (∠B = ∠C = 90deg)
So,
∠BPQ = 90 deg &
∠CQP = 90 deg
.^. BPQC and APQD are both rectangles.
Now from △OPB.
OB^2 = BP^2 + OP^2 ------(1)
Similarly from △OQD,
we have OD^2 = OQ^2 + DQ^2 ---(2)
from △OQC
we have OC^2 = OQ^2 + CQ^2 ----(3)
and from △OAP,
OA^2 = AP^2 + OP^2
Adding (1) and (2)
OB^2 + OD^2 = BP^2 + OP^2 + OQ^2 + DQ^2
=> CQ^2 + OP^2 + OQ^2 + AP^2 (BP=CQ and DQ=AP)
=> CQ^2 + OQ^2 + OP^2 + AP^2
=> OC^2 + OA^2 ( from (3) & (4))
******************************************************
Example-14. The hypotenuse of a right triangle is 6m more than twice of the shortest side. If the third side is 2m, less than the hypotenuse, find the sides of the triangle.
sol) Let the shortest side be 'X' m
Then hypotenuse = (2x+6)m and third side=(2x+4)m
by Pythagores theorem, we have
(2x+6)^2 = x^2 + (2x+4)^2
4x^2 + 24x + 36 = x^2 + 4x^2 + 16x + 16
x^2 - 8x - 20 = 0
(x-10)(x+2) = 0
x = 10 or x = -2
but "x" can't be negative as side of a traingle.
.^. x = 10
Hence, the sides of the triangle are 10m, 26m, and 24m.
********************************************************
Example -15. ABC is a right triangle right angled at "C". Let BC = a , CA= b, AB = c and let 'P' be the length of perpendicular from "C" on "AB" .
prove that
(!) pc = ab
(!!) (1/p^2) = (1/a^2) + (1/b^2)
sol) (!) CD_|_AB and CD = p.
Area of △ABC = (1/2) * AB * CD
=> (1/2)*cp.
also
Area of △ABC = (1/2) * BC * AC
=> (1/2)*ab
(1/2)*cp = (1/2)*ab
cp = ab ------(1)
(!!) Since △ABC is a right triangle right angled at "C"
AB^2 = BC^2 + AC^2
c^2 = a^2 + b^2
(ab)^2 = a^2 + b^2
(p)^2
1 = a^2 + b^2
p^2 a^2b^2
=> 1 + 1
a^2 b^2
******************************************************
Exercise - 8.4
(1) prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
sol)
ABCD is a rhombus in which diagonals AC and BD intersect at point "O".
RTP :- AB^2 + BC^2 + CD^2 + DA^2 = AC^2 + DB^2
In /_\AOB : AB^2 = AO^2 + BO^2
In /_\BOC : BC^2 = CO^2 + BO^2
In /_\COD : CD^2 = DO^2 + CO^2
In /_\AOD : AD^2 = DO^2 + AO^2
Adding the above four equations, we get;
AB^2 + BC^2 + CD^2 + DA^2 = AO^2 + BO^2 + CO^2 + BO^2 + DO^2 + CO^2 + DO^2 + AO^2
OR
AB^2 + BC^2 + CD^2 + DA^2 = 2(AO^2 + BO^2 + CO^2 + DO^2)
AB^2 + BC^2 + CD^2 + DA^2 = 2(2AO^2 + 2BO^2)
***************************************
Since AO^2 = CO^2 and BO^2 = DO^2
**************************************
AB^2 + BC^2 + CD^2 + DA^2 = 4(AO^2 + BO^2) -------(1)
Now, let us take the sum of squares of diagonals ;
AC^2 + DB^2 = (AO + CO)^2 + (DO + BO)^2
=> (2AO)^2 + (2DO)^2
=> 4AO^2 + 4BO^2 ------(2)
from equations (1) and (2), it is clear ;
AB^2 + BC^2 + CD^2 + DA^2 = AC^2 + DB^2
Hence proved.
********************************************
(2) ABC is a right triangle right angled at "B". Let "D" and "E" be any points on AB and BC respectively.
Prove that AE^2 + CD^2 = AC^2 + DE ^2
sol) In right angled /_\ABE and /_\DBC, we have
AE^2 = AB^2 + BE^2 --------(1)
DC^2 = DB^2 + BC^2 -------(2)
Adding (1) and (2)
AE^2 + DC^2 = AB^2 + BE^2 + DB^2 + BC^2
= (AB^2 + BC^2)_ + ( BE^2 + DB^2)
**************************************************
Since AB^2 + BC^2 = AC^2 in right angled triangle ABC
***************************************************
= ( AC^2 + DE^2)
Hence proved.
********************************************
(3) Prove that three times the square of any side of an equilateral triangle is equal to four times the square of the altitude.
sol)
Given :- An equilateral triangle ABC, in which AD _|_BC
RTP :- 3AB^2 = 4AD^2
Proof :- Let AB= BC = CA = a
In /_\ ABD and /_\ACD
AB=AC, AD = AD
and /_ADB = /_ADC ( each 90 deg)
.^. /_\ ABD = /_\ ACD
.^. BD = CD = a (CPCTE)
2
Now,
In /_\ ABD, /_D = 90 deg
.^. AB^2 = BD^2 + AD^2
or
AB^2 = [CD]^2 + AD^2
[2}^2
=> [ AB ]^2 + AD^2
2
AB^2 = [AB]^2 + AD^2
[ 2 }^2
4AB^2 = AB^2 + 4AD^2
3AB^2 = 4AD^2
*************************
(4) PQR is a triangle right angled at "P" and "M" is a point on "QR" such that PM 1 QR.
Show that PM^2 = QM . MR.
sol)
Let /_MPR = x
In /_\ MPR
/_MRP = 180 - 90 -x
/_MRP = 90 - x
Similarly in /_\ MPQ
/_MPQ = 90 - /_MPR
=> 90 - x
/_MQP = 180 - 90 - (90-x)
/_MQP = x
In /_\QMP and /_\PMR
/_MPQ = /_MRP
/_PMQ = /_RMP
/_MQP = /_MPR
/_\QMP ~ /_\PMR [by AAA criteria]
QM = MP
PM MR
PM^2 = MR * QM
Hence proved.
**********************************
(5). ABD is a triangle right angled at "A" and AC 1 BD
1) AB^2 = BC . BD.
sol) consider two triangles ACB and DAB
we have /_ABC = /_DBC (common)
/_ACB = /_DAB (Each is right angle)
/_CAB = /_ADB (Third angle)
.^. Triangle are similar and their corresponding sides must be proportional.
i.e, /_ADC = /_ADB
AC = CB = AB
DA AB DB
.^. AB^2 = BC * BD
********************
2) AC^2 = BC .DC
sol) /_BDA = /_BDC = 90
/_3 = /_2 = 90 /_1
[.^. /_1 + /_2 = 90, /_1 + /_3 = 90 ]
/_2 + /_4 = 90 /_2
[.^. /_1 + /_2 = 90, /_2 + /_4 = 90]
/_\ ADB ~ /_\BDC [AAA criterion of similarity].
Their corresponding sides must be proportional.
DC = CA = DA
AC CB AB
=> DC = CA
AC CB
=> CA^2 = BC * DC
******************************************
3) AD^2 = BD.CD.
sol) In two triangles ADB and ABC, we have
/_ADC = /_ADB ( Common)
/_DCA = /_DAB ( Each is right angle)
/_DAC = /_DBA (Third angle)
/_DCA = /_DAB ( AAA similarity)
Triangle ADB and ABC are similar and so their corresponding sides must be proportion.
DC = CA = DA
DA CB DB
DC = DA
DA DB
AD^2 = DB * DC
Hence proved.
************************
(6) ABC is an isosceles triangle right angled at "C". Prove that AB^2 = 2AC^2.
sol) Since the triangle is right angled at "C" therefore the side "AB" is the hypotenuse. Let the base of the triangle be "AC" and the altitude be "BC".
Applying the Pythagorean theorem
AB^2 = AC^2 + BC^2
Since the triangle is isosceles triangle two of the sides shall be equal.
Therefore AC = BC
Thus AB^2 = AC^2 + AC^2
AB^2 = 2AC^2
Hence proved.
********************************
(7) "O" is any point in the interior triangle ABC. OD 1 BC, OE 1 AC and OF 1 AB , show that
1) OA^2 + OB^2 + OC^2 - OD^2 - OE^2 - OF^2 = Af^2 + BD^2 + CE^2
sol) Given :- /_\ ABC and "O" is a point in the interior of a triangle ABC where, OD _|_ BC, OE _|_ AC, OF _|| AB
Proof :-
1) Let us join the point "O" from A,B, and C.
2) Using pythagoras theorem,
(Hyp)^2 = (Height)^2 + (Base)^2
3) Ina right angle triangle OAF.
(OA)^2 = AF^2 + OF^2 ------(1)
4) In right angle triangle ODB
(OB)^2 = OD^2 + BD^2 -----(2)
5) In a right angle triangle OEC
(OC)^2 = (OE)^2 + (EC)^2 -----(3)
Adding (1) + (2) + (3)
(OA)^2 + (OB)^2 + (OC)^2 = AF^2 + OF^2 + OD^2 + BD^2 + (OE)^2 + (EC)^2
OA^2 + OB^2 + OC^2 - OD^2 - OE^2 - OF^2 = Af^2 + BD^2 + CE^2
Hence proved.
*************************************
2) AF^2 + BD^2 + CE^2 = AE^2 + CD^2 + BF^2.
sol) Using Pythagoras theorem.
(Hyp)^2 = (height)^2 + (Base)^2
1) In /_\ ODB,
OB^2 = OD^2 +BD^2
2) In /_\ OFB,
OB^2 = OF^2 +FB^2
3) In /_\ OFA,
OA^2 = OF^2 + AF^2
4) In /_\ OEA,
OA^2 = OE^2 + AE^2
5) In /_\ OEC,
OC^2 = OE^2 + CE^2
6) In /_\ ODC,
OC^2 = OD^2 + CD^2
L.H.S :- AF^2 + BD^2 + CE^2
=> ( OA^2 - OF^2) + ( OB^2 - OD^2) + ( OC^2 - OE^2)
=> OA^2 + OB^2 + OC^2 - OF^2 - OD^2 - OE^2
R.H.S :- AE^2 + CD^2 + BF^2
=> (OA^2 - OE^2) + ( OC^2 - OD^2) + (OB^2 - OF^2)
=> OA^2 + OB^2 + OC^2 - OF^2 - OD^2 - OE^2
Since L.H.S = R.H.S
Hence proved
**************************************
(8) A wire attached to vertical pole of height 18m is 24m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut.
sol) let AB = 24 m be a wire attached to a vertical pole
BC = 18m
To keep the wire taut, let it be fixed at "A"
AB^2 = AC^2 + BC^2
(24)^2 = AC^2 + (18)^2
AC^2 = 576 - 324
AC = 6√7
Hence, the stake may be placed at a distance 6_/7m the base of pole
*********************************************************
(9) Two poles of heights 6m and 11m stand on a plane ground . If the distance between the feet of the poles is 12m find the distance between their tops.
sol )
Given :- BC = 6 cm and AD = 11m, BC = ED
So, AE = AD - ED
= 11 - 6
= 5m
BE = CD = 12m
Find, AB = ?
Solution :- Now , In /_\ ABE, /_E = 90
AB^2 = AE^2 + BE^2
AB^2 = (5)^2 + (12)^2
AB^2 = 169
AB = 13m
The distance between their tops is 13m.
*****************************************
(10) In a equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC. Prove that 9AD^2 = 7Ab^2.
sol)
ABC is a equilateral triangle,
AB = BC = AC and BD = 1/3 BC
Construction :-
Draw AE _|_ BC
Proof :-
/_\ABE ~ /_\ ACE
.^. BE = EC = BC / 2
Now in /_\ABE,
AB^2 = BE^2 + AE^2
also, AD^2 = AE^2 + DE^2
.^. AB^2 - AD^2 = BE^2 - DE^2
=> BE^2 - ( BE - BD)^2
=> ( BC)^2 - ( BC - BC ) ^2
2 2 3
=> (AB)^2 - ( AB - AB)^2
2 2 3
AB^2 - AD^2 = 2AB^2
9
or
7AB^2 = 9AD^2
******************************
11) In the given figure, ABC is a triangle right angled at "B, D and E are points on BC trisect it.
prove that 8AE^2 = 3AC^2 + 5AD^2.
sol) In /_\ABD, /_B = 90 deg
.^. AC^2 = AB^2 +BC^2 -----(1)
similarly , AE^2 = AB^2 + BE^2 -----(2)
and AD^2 = AB^2 + BD^2 -----(3)
From (1)
3AC^2 = 3AB^2 + 3BC^2 -----(4)
From (2)
5AD^2 = 5AB^2 + 5BD^2 ----(5)
Adding equation (4) and (5)
3AC^2 + 5AD^2 = 8AB^2 + 3BC^2 + 5BD^2
=> 8AB^2 + 3(3 BE )^2 + 5 ( BE)^2
2 2
=> 8 (AB^2 + BE^2)
=> 8AE^2
**************************
12) ABC is an isosceles triangle right angled at "B". Similar traingles ACD and ABE are constructed on sides AC and AB. Find the ratio between the areas of /_\ABE and /_\ ACD.
sol) Given /_\ABC is an isosceles triangle in which /_B = 90 deg
=> AB = BC
By Pythagoras theorem, we have
AC^2 = AB^2 + BC^2
=> AC^2 = AB^2 + AB^2
[ Since AB = BC ]
=> AC^2 = 2AB^2 --------(1)
It is also given that /_\ABE ~ /_\ACD
ratio of areas of similar triangles is equal to ratio of squares of their corresponding sides.
ar (/_\ ABC) = AB^2
ar(/_\ ACD) AC^2
ar(/_\ABC) = AB^2
ar(/_\ACD) 2AB^2 --------(1)
ar(/_\ABC) = 1
ar(/_\ACD) 2
.^. ar(/_\ABC) : ar(/_\ACD) = 1:2
******************************
Optional exercise:-
1). In the given figure
QT = QR
PR QS
and ∠1 and ∠2
prove that △PQS ~ △ TQR
sol)Given :- ∠1 = ∠ 2
QT = QR
PR QS
R.T.P : △PQS ~ △TQR
Proof : Since ∠1 = ∠2
Step 1) PR = QP [ sides opposite to equal angles are equal]
Step 2) Putting PR=QP in
QR = QT
QS PR
QR = QT -------(2)
QS QP
In △PQS and △TQR
∠PQS = ∠TQR [common angles]
QR = QT -----(from(2))
QS QP
Therefore,
△PQS~ △TQR [ SAS similarity]
Hence Proved
*****************************************
2) Ravi is 1.82m tall. He wants to find the height of a tree in his backyard. from the tree's base he walked 12.20m. along the tree's shadow to a position where the end of his shadow exactly overlaps the end of the tree's shadow. He is now 6.10m from the end of the shadow. How tall is the tree?
sol)
In △EDA and ⧍CBA
∠B = ∠D = 90 deg
∠A = ∠A ( common angle)
AA Traingle similarity :- if two traingles are similar it means that all the corresponding angle pairs are congruent and all corresponding sides are proportional
.^. by AA similarity
△EDA ~ △CBA
Now,
ED = DA
CB BA
ED = DB + BA
1.82 6.10
ED = 12.20 + 6.10
1.82 6.10
ED = 18.30
1.82 6.10
ED = 3 * (1.82)
ED = X = 5.46 is the height of the tree
*******************************************
3) The diagonal AC of a parallelogram ABCD intersects DP at the point "Q" where "P" is any point on side "AB". Prove that
CQ * PQ = QA * QD
sol) Given :- The diagonal "AC of a parallelogram "ABCD" intersects "DP" at the point "Q"
Where "p" is any point on side "AB"
R.T.P:- CQ * PQ = QA * QD
Proof :- In △APQ & △CDQ
∠AQP = ∠CQD [ vertically opposite angles]
∠PAQ = ∠DCQ [ alternate angles]
.^. △APQ ~ △CQD [ A.A criterion]
PQ = QA
QD CQ (cross multiply)
CQ * PQ = QA * QD
Hence proved.
*****************************************
4) △ABC and △AMP are two right triangles right angled at "B" and "M" respectively.
(*) Prove that
(!) △ABC ~ △ AMP
sol) Given :-
∠ABC = 90deg ,∠AMP = 90deg
RTP :- △ABC ~ △AMP
Proof :- In △ABC and △AMP
∠A = ∠A (common angle)
∠ABC = ∠AMP = 90 (Both 90 deg)
Using AA similarity
△ABC ~ △AMP
***************
(!!) CA = BC
PA MP
sol) We know when two traingles are similar, the ratio of their corresponding sides is proportional.
As we already proved in (!)
△ABC ~ △AMP
By using AAA criterion
CA = BC
PA MP
Hence proved
*******************************************
(5) An aeroplane leaves an airport and flies due north at a speed of 1000 kmph. At the same time another aeroplane leaves the same airport and flies due west at a speed of 1200 kmph. How far apart will the two planes be after 1(1/2) hour?
sol) condition (1) : 1st plane flies due north at a speed of 1000 kmph
We have
Distance = Speed * Time
Distance = 1000 * 1(1/2)
Distance = 1000 * (3/2)
Distance = 1000 * 1.5 => 1500 km
(*) condition (2) :- 2nd plane due west
Distance = 1200 * 1.5 => 1800 km
In △ABC
BC is distance travelled by first aeroplane and BA is the distance travelled by second aeroplane.
Applying pythagoars Theorem.
AC^2 = AB^2 + BC^2
AC^2 = (1500)^2 + (1800)^2
AC^2 = 2,25,0000 + 3,24,0000
AC = √5,49,0000
AC = √ 549 * √10,000 3 549
3 183
AC = √549 * √(100)^2 61 61
1
AC = 100 √(3)^2 * √61
AC = 3 *100 √61
AC = 300√61 km
Hence, the two planes would be 300√61 km from each other.
******************************************
(6) In a right triangle ABC right angled at C, P, and Q are points on sides AC and CB respectively which divide these sides in the ratio of 2:1.
Prove that
(1) 9AQ^2 = 9 AC^2 + 4BC^2
sol) Given :- "P" divides "AC" in the ratio 2:1 and "Q" divides "BC" in the ratio 2:1
CQ = (2/3) BC
PC = (2/3) AC
(!) In ACQ, we have
AQ^2 = AC^2 + CQ^2
AQ^2 = AC^2 + (2/3)BC)^2
AQ^2 = AC^2 + (4/9) BC^2
9AQ^2 = AC^2 +4BC^2
Hence proved.
********************************
(2) 9BP^2 = 9BC^2 + 4AC^2
sol) In PCB, we have
BP^2 = BC^2 + PC^2
BP^2 = BC^2 + (2/3)AC)^2
BP^2 = BC^2 + (4/9)AC^2
BP^2 = 9BC^2 + 4AC^2
9
9BP^2 = 9BC^2 + 4AC^2
Hence proved
*************************************
(3) 9(AQ^2 + BP^2) = 13AB^2
sol) Adding (1) and (2) we get
9AQ^2 + 9BP^2 = 9 AC^2 +9BC^2 + 4BC^2 + 4AC^2
9(AQ^2 +BP^2) = 13AC^2 + 13BC^2
9(AQ^2 + BP^2) = 13(AC^2 + BC^2)
.^. 9(AQ^2 + BP^2) = 13AB^2
Hence proved.
***************************************
Theorem:- if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio
Given : In △ABC, DE // BC which intersects sides AB and AC at "d" and "e" respectively
RTP: AD = AE
DB EC
Construction :
Join B,E and C,D and then draw
DM _|_ AC and EN _|_ AB.
(*) Proof :
Area of △ADE = (1/2) * AD * EN
Area of △BDE = (1/2) * BD * EN
So,
ar( △ADE)=(1/2)*AD*EN=AD---(1)
ar( △BDE) (1/2)*BD*EN=BD
Again
Area of △ADE = (1/2)*AE*DM
Area of △CDE = (1/2)*EC*DM
ar(△ADE)=(1/2)*AE*DM=AE ---(2)
ar(△CDE) (1/2)*EC*DM EC
Observe that ⧍BDE and⧍ CDE are on the same base DE and between same parallels BC and DE.
So,
ar(⧍BDE) = ar(⧍CDE) ---(3)
From(1),(2) and (3), we have
AD = AE
DB EC
Hence proved
Is the converse of the above theorem also true? To examine this, let us perform the following activity
(*)Activity:-
Draw an angle XAY on your note book and on ray AX, mark points B1,B2,B3,B4 and B suc that
AB1=B1B2=B2B3=B3B4=B4B=1cm(say)
Similarly on ray AY, mark points C1,C2,C3,C4 and C such that
AC1=C1C2=C2C3=C3C4=C4C=2cm(say)
Join B1, C1 and B, C.
Observe that
AB1 = AC1 = 1 and B1C1 // BC
B1B C1C 4
Similarly, joining
B2C2, B3C3 and B4C4, you see that
AB2 = AC2 = 2 and B2C2 //BC
B2B C2C 3
AB3 = AC3 =3 and B3C3 // BC
B3B C3C 2
AB4 = AC4 = 4 and B4C4 // BC
B4B C4C 1
From this we obtain the following theorem called converse of the Thales theorem
*******************************************
Theorem-8.2:- if a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side.
Given: In △ABC, a line DE is drawn such that
AD = AE
DB EC
RTP : DE // BC
proof : Assume that DE is not parallel to BC then draw the line DE' parallel to BC
So,
AD = AE'
DB E'C
.^. AE = AE'
EC E'C'
Adding "1" on both sides of the above, you can see that "E" and E' must coincide.
*********************************************
Now let us solve some examples on Thales theorem and its converse.
Example -1: In △ABC, DE//BC and
AD = 3 , AC = 5.6, Find AE.
DB 5
sol) In △ABC, DE//BC
=> AD = AE (by B.P.T)
DB EC
but
AD = 3
DB 5
So,
AE = 3
EC 5
Given:- AC = 5.6 and AE:EC=3:5
AE = 3
AC - AE 5
AE = 3 (cross multiply)
5.6 - AE 5
5AE = 3*5.6 - 3AE
8AE = 16.8
AE = 16.8
8
AE = 2.1 cm
***********************************************************************
Example-2: In the given figure LM//AB AL = x-3, AC=2x, BM=x-2 and BC=2x+3 find the value of "x"
sol) In △ABC, LM // AB
=> AL = BM (by B.P.T)
LC MC
x-3 = x-2
2x-(x-3) (2x+3)-(x-2)
x-3 = x-2 (cross multiply)
x+3 x+5
(x+5)(x-3) = (x-2)(x+3)
x^2 +2x -15 = x^2 +x -6
=> 2x -x = -6 +15
.^. x = 9
*************************************************
Example-3) The diagonals of a quadrilateral ABCD intersect each other at point 'O' such that
AO = CO. Prove that ABCD is a trapezium
BO DO
Sol) Given : In quadrilateral ABCD,
AO = CO
BO DO
RTP: ABCD is a trapezium
Construction: Through 'O' draw a line parallel to AB which meets DA at X.
Proof : In △DAB, XO//AB (by construction)
DX = DO (by B.P.T)
XA OB
AX = BO
XD OD ------(1)
again
AO = CO (given)
BO DO
AO = BO
CO OD ------(2)
From(1) and (2)
AX = AO
XD CO
In △ADC, XO is a line such that
AX = AO
XD OC
=> XO // DC (by converse of the B.P.T)
=> AB // DC
In quadrilateral ABCD, AB//DC
=> ABCD is a trapezium (by definition)
Hence proved.
*******************************************
Example 4) In trapezium ABCD, AB//DC, E and F are points on non-parallel sides AD and BC respectively such that EF // AB. Show that
AE = BF
ED FC
Sol) Let us join AC to intersect EF at G.
AB//DC and EF//AB (given)
=> EF// DC (Lines parallel to the same line are parallel to each other)
In △ADC, EG//DC
So
AE = AG (by BPT)
ED GC
Similarly, In △CAB, GF//AB
CG = CF (by BPT)
GA FB
i.e, AG = BF ----(2)
GC FC
From(1) and (2)
AE = BF
ED FC
******************************************
Exercise -8.1
(1) In triangle PQR, ST is a line such that
PS = PT and also / PST = /_ PRQ.
SQ TR
Prove that △ PQR is an isosceles triangle.
sol) Given :- PS = PT
SQ TR
2) ST || QR ( By Converse of BPT)
Basic Proportionality Theorem (BPT) :- If a line is drawn parallel to one side of a triangle to intersect the other two side in distinct points, the other two sides are divided in the same ration.
3) /_PST = /_ PQR ( Corresponding angles are always equal)
Given :- /_ PST = /_ PRQ
4) /_PQR = /_PRQ
We know :- Two angles or two sides of an isosceles triangle are equal.
.^. /_\PQR is an isosceles triangle.
prove that AM = AN
AB AD
sol) Consider /_\ ABC
1)Given :- LM || CB
2) AM = AL (by BPT)
MB LC
Consider /_\ ADC
3) Given :- LN ||CD
4) AN = AL (by BPT)
ND LC
equating (2) and (4)
5) AM = AN
MB ND
Taking the reciprocal of (5)
6) MB = ND
AM AN
Adding 1 on both sides
7) MB + 1 = ND + 1
AM AN
8) (MB + AM) = (ND + AN)
AM AN
9) AB = AD
AM AN
10) AM = AN
AB AD
Hence proved
****************************
(3) In the given figure, DE //AC and DF //AE.
Prove that BF = BE
FE EC
sol) Consider /_\ ABE
1) Given :- DF || AE
2) BF = BD (By BPT)
FE DA
Consider /_\ ABC
3) Given :- DE || AC
4) BE = BD ( by BPT)
EC DA
equating (2) and (4)
5) BF = BE
FE EC
Hence proved .
*******************************************************
(4) In the given figure,
AB // CD // EF,
given
AB = 7.5 cm
DC = ycm,
EF = 4.5 cm,
BC = x cm.
Calculate the values of "x" and "y"
sol) Consider △ ACB and △ CEF
Both are similar traingles
.^. AB = BC
EF CF
Given :- AB = 7.5 cm and EF = 4.5 cm
7.5 = x
4.5 3
7.5 *3 = x
4.5
22.5 = x (4.5 * 5 = 22.5)
4.5
5 = x
Consider △ BCD and △ BFE
From BPT theorem, both are similar triangles
x = x+3
y 4.5
y = 5 * 4.5
8
y = 5*45
8*10
y = 225
80
y = 45
16
*******************************************************
(5) Prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side ( using basic proportionality theorem )
sol) Let us assume △ ABC
DE || BC
"D" is the mid-point of AB
RTP :- "E" is the mid-point of AC
We know that :- If a line is drawn parallel to one side of a triangle to intersect the other two side in distinct points, the other two sides are divided in the same ration.
AD = AE
DB EC
DB = AE
DB EC
1 = AE
EC
EC = AE
=> E is the mid-point of AC
Hence proved.
***********************
(6) Prove that a line joining the midpoints of any two sides of a triangle is parallel to the third side ( Using converse of basic proportionality theorem )
sol) Assume :- /_\ ABC
"D" is the mid-point of AB
"E" is the mid-point of AC
RTP :- DE || BC
Proof :- In /_\ ABC,
"D" is the mid-point of AB
=> AD = DB
=> AD = 1 -----(1)
DB
"E" is the mid-point of AC
=> AE = EC
=> AE = 1 -----(2)
EC
eq (1) & (2)
AD = AE
DB EC
.^. DE || BC ( By BPT)
Hence proved.
*******************************
(7) In the given figure, DE // OQ and DF // OR . Show that EF // QR.
sol) Consider /_\ POQ
1) Given :- DE || OQ
2) PE = PD (by BPT)
EQ DO
Cnsider /_\ POR :-
3) Given :- DF || OR
4) PF = PD (by BPT)
FR DO
Equating (2) & (4)
5) PE = PF
EQ FR
6) EF || QR ( By BPT)
Hence proved.
******************************************************
(8) In the adjacent figure A, B and C are points on OP, OQ and OR respectively such that AB // PQ and AC // PR. Show that BC // QR.
sol) Consider /_\ OPQ
1) Given :-AB || PQ
2) OA = OB (By BPT)
AP BQ
Consider /_\ OPR
3) Given :-AC || PR
4) OA = OC (By BPT)
AP CR
equating (2) & (4)
5) OB = OC
BQ CR
6) BC || QR ( By BPT)
Hence proved.
****************************
9) ABCD is a trapezium in which AB || DC and its diagonals intersect each other at point "O". Show that
AO = CO
BO DO
sol) Draw a line EF || CD which is passing through "O".
In /_\ ABC and /_\ EOC
These are similar triangles as per BPT.
AE = BO ------(1)
EC OC
Similarly, in /_\ BOD and /_\ FOD
BF = AO -----(2)
FD OD
in /_\ ABC and /_\ BAD
BO = AO ----(3)
OC OD
Because diagonals of a trapezium divide each other in same ratio
From the above three equations, it is clear
AE = BF
EC FD
Hence , /_\ ABC ~ /_\ BAD
using the third equation
BO = AO
OC OD
or
AO = CO
BO DO
*********************************
(10) Draw a line segment of length 7.2 cm and divide it in the ration 5 : 3. Measure the two parts.
sol) Take m = 5 and n = 3
Steps of construction :-
1) Draw line AB of length 7.2
2) Draw a ray AX, making an acute angle(less then 90 degree) with AB.
3) Mark 8 ( 5 +3) points A1, A2. A3.....A8 on AX such that AA1 = A1A2 = so on..
4) join BA8
5) Since we want the ratio 5:3 , Through point A5 (m=5), we draw a line parallel to BA8 (by making an angle equal to /_A A8 B at A5 ) intersecting AB at the point "C".
then, AC : CB = 5 : 3
*********************************************************************************
AAA Criterion for similarity of Triangles
Theorem-8.3 : In two triangles , if the angles are equal, then the sides opposite to the equal angles are in the same ratio( or proportional) and hence the two triangles are similar.
Given: In triangles ABC and DEF,
∠A = ∠D
∠B = ∠E
∠C = ∠F
R.T.P :-
AB = BC = AC
DE EF DF
Construction: Locate points "P" and "Q" on DE and DF respectively.such that AB=DP and AC=DQ.
Join PQ.
Proof : ABC ~ DPQ (why?)
This gives B = P = E and PQ//EF (how?)
Therefore
DP = DQ (why?)
PE QF
i.e, AB = AC (why?)
DE DF
Similarly
AB = BC
DE EF and So
AB = BC = AC
DE EF DF
Hence proved.
Note : If two angles of a triangle are respectively equal to the two angles of another triangle, then by the angle sum property of a triangle, third angles will also be equal.
So AA similarity criterion is stated as if two angles of one triangle are respectively equal to the two angles triangles, then the two triangles are similar.
What about the converse of the above statement?
If the sides of a triangle are respectively proportional to the sides of another triangle , is it true that their corresponding angles are equal?
Let us exercise it through an activity.
Activity :-
(*) Draw two triangles ABC and DEF such that
AB = 3cm,
BC = 6cm,
CA = 8cm,
DE = 2.5 cm,
EF = 9cm and
FD = 12cm.
sol) we have
AB = 3 = 30 = 10 = 2
DE 4.5 45 15 3
BC = 6 = 2
EF 9 3
CA = 8 = 2
DF 12 3
AB = BC = CA = 2
DE EF FD 3
Now measure the angles of both the triangles. What do you observe? what can you say about the corresponding angles? They are equal, so the triangles are similar. You can verify it for different triangles.
from the above activity, we can give the following criterion for similarity of two traingles.
(*) SSS Criterion for similarity of Triangles
Theorem-8.4: If in two triangles, the sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are equal and hence the triangles are similar.
Given : ∆ABC and ∆DEF are such that
AB = BC = CA
DE EF FD (< 1)
RTP :
∠A = ∠D,
∠B = ∠E,
∠C = ∠F
Construction : Locate points P and Q on DE and DF respectivelysuch that AB = DP and AC = DQ. Join PQ.
Proof :
DP = DQ and PQ // EF (why?)
PE QF
So ∠P = ∠E and ∠Q = ∠F (why ?)
Therefore,
DP = DQ = PQ
DE DF EF
So,
DP = DQ = BC (why?)
DE DF EF
So BC = PQ (Why ?)
∆ABC ≅ ∆DPQ (why ?)
So ∠A = ∠D, ∠B = ∠E and ∠C = ∠F (How ?)
We studied that for similarity of two polygons any one condition is not sufficient. But for the similarity of triangles, there is no need for fulfillment of both the conditions as one automatically implies the other. Now let us look for SAS similarity criterion.
(*) SAS Criterion for similarity of Triangles
Theorem-8.5 :- If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.
Given:- In △ABC and △DEF
AB = AC (<1) and
DE DF
∠A = ∠D
RTP : △ABC ~ △DEF
Construction:- Locate points P and Q on DE and DF respectively such that AB=DP and AC=DQ.
Join PQ.
Proof: PQ//EF and △ABC ~ △DPQ
So
∠A = ∠D
∠B = ∠P
∠C = ∠Q
.^. △ABC ~△ DEF
Construction : To construct a triangle similar to a given triangle as per given scale factor.
a) Construct a triangle similar to a given triangle ABC with its sides equal to (3/4) of corresponding sides of ∆ABC (scale factor (3/4)
Steps :
1). Draw a ray BX, making an acute angle with BC on the side opposite to vertex A.
2). Locate 4 points B1, B2, B3 and B4 on BX so that BB1 = B1B2 =B2B3 = B3B4.3.
3). Join B4C and draw a line through B3 parallel to B4C intersecting BC at C′.
4). Draw a line through C′ parallel to CA to intersect AB at A′.So ∆A′BC′ is the required triangle.
Let us take some examples to illustrate the use of these criteria.
Example-5. A person 1.65m tall casts 1.8m shadow. At the same instance, a lamp-posts casts a shadow of 5.4 m. Find the height of the lamppost.
Solution:In ∆ABC and ∆PQR
∠B = ∠Q= 90 deg.
∠C = ∠R (AC || PR, all sun’s rays are parallel at any instance)
∆ABC ~ ∆PQR ( by AA similarity)
AB = BC
PQ QR (cpst, corresponding parts of Similar triangles)
1.65 = 1.8
PQ 5.4
PQ = 1.65 * 5.4 = 4.95 m.
1.8
The height of the lamp post = 4.95m
********************************************************
Example-6. A man sees the top of a tower in a mirror which is at a distance of 87.6m from the tower. The mirror is on the ground facing upwards. The man is 0.4m away from the mirror and his height is 1.5m. How tall is the tower?
solution: In ∆ABC & ∆EDC
∠ABC=∠EDC=90°
∠BCA=∠DCE (angle of incidence and angle of reflection are same)
∆ABC ~ ∆EDC (by AA similarity)
AB = BC
ED CD
=> 1.5 = 0.4
h 87.6
h = 1.5 * 87.6 = 328.5m
0.4
Hence , the height of the towers is 328.5 m.
********************************************************
Example-7). Gopal is worrying that his neighbour can see into his living room from the top floor of his house.
He has decided to build a fence that is high enough to block the view from their top floor window. What should be the height of the fence?
The measurements are given in the figure
solution:- In △ABD & △ACE
∠B = ∠C = 90 deg
∠A = ∠A (common angle)
△ABD ~ △ACE (by AA similarity)
AB = BD
AC CE
2 = BD
3 1.2
BD = 2 *1.2
8
=> 2.4
8
=> 24 = 3 = 0.3m
80 10
Total height of the fence required is 1.5m + 0.3m = 1.8m to block the neighbour's view
Exercise-8.2
1). In the given figure, ∠ADE = ∠B
(!) Show that △ABC ~ △ADE
2) If AD = 3.8 cm, AE = 3.6 cm, BE = 2.1 cm, BC = 4.2 cm find DE.
sol) given:- ∠ADE = ∠B
∠C = 90,
∠A = 90 -
If ∠A = 90 -
Comparing △ABC with △ AED we have
∠A common in both triangles
∠C = ∠AED = 90
∠ADE = ∠B
.^. By AAA property, △ ABC and △ ADE are similar triangles
△ABC ~ △ADE
Hence proved
2sol) Given AD = 3.8 cm, AE = 3.6 cm, BE = 2.1 cm, BC = 4.2
As △ ABC and △ ADE are similar triangles,
We have
AB = BC = AC
AD DE AE
AB = 4.2 = AC
3.8 DE 3.6
But, AE + BE = AB
3.6 + 2.1 = 5.7
Now,
5.7 = 4.2
3.8 DE
DE =
DE = 1.4 * 2
DE = 2.8 cm
**********************************************************
(2) The perimeters of two similar triangles are 30 cm and 20 cm respectively. If one side of the first triangle is 12 cm, determine the corresponding side of the second triangle.
sol) Since the triangle are similar
perimeter of 1st △ = Any side of 1st △
perimeter of 2nd△ corresponding side of 2nd△
30 = 12
20 x
x = 20 *12
30
x = 8 cm
***********************************************************
(3) A girl of height 90 cm is walking away from the base of a lamp post at a speed of 1.2 m/sec. If the lamp post is 3.6m above the ground . Find the length of her shadow after 4 seconds.
sol) Lamp post (PQ) = 3.6 m
Height of girl(ST) = 90 cm
(As lamp post is in "m" convert "cm" into "m")
1m = 100 cm
Height of girl(ST) = 90 m = 0.9 m
100
Speed = 1.2 m/sec,
RTP :- Length of her shadow i.e. TR
Solution :- The girl walks "QT" distance in 4 seconds
We have formula :-
Speed = Distance
Time
1.2 = QT
4
1.2 * 4 = QT
4.8 = QT
QT = 4.8 m
Now,
In /_\ PQR and /_\ STR
/_R = /_R [Common]
/_Q = /_T ( Both 90 deg because lamp post as well as the girl are standing to the ground)
.^. , using AA similarity criterion
ST = RT (by cpst)
PQ QR
0.9 = RT
3.6 RT + TQ
1 = x
4 x + 4.8
x + 4.8 = 4x
3x = 4.8
x = 1.6m
RT = 1.6 m
.^. The length of her shadow after 4 sec = 1.6m
****************************************
(4) CM and RN are respectively the medians of
Prove that
sol) Given :- /_\ ABC ~ /_\PQR
So, AB = BC = CA ----(1)
PQ QR RP
and /_A = /_P, /_B = /_Q , /_C = /_R ----(2)
But AB = 2 AM and PQ = 2 PN ( As "CM" and "RN" are medians)
So, from (1), 2AM = CA
2PN RP
i.e, AM = CA -------(3)
PN RP
Also , /_\ MAC = /_\NPR (from 2) -------------(4)
So, from (3) and (4),
/_\ AMC ~ /_\PNR (SAS similarity) -------(5)
*****************************************
(2) CM = AB
RN PQ
sol) /_\ AMC ~ /_\PNR (SAS similarity) ----(5)
from (5), CM = CA -----(6)
RN RP
---------------------------------------
AB = BC = CA ----(1)
PQ QR RP
=======================
RP PQ ( From 1 )----------( 7 )
.^. CM = AB (From (6) & (7) ------(8)
RN PQ
*************************************
sol) From (1)
AB = BC
PQ QR
.^. CM = BC ( from (8)] -----(9)
RN PQ
Also, CM = AB = 2BM
RN PQ 2QN
i.e, CM = BM ------(10)
RN QN
from (9) and (10)
i.e, CM = BC = BM
RN QR QN
.^. /_\ CMB ~ /_\RNQ (SSS similarity)
************************************
(5) Diagonals AC and BD of a trapezium ABCD with AB //DC intersect each other at the point "O". Using the criterion of similarly for two triangles, show that
OA = OB
OC OD
sol)
Consider /_\ AOB and /_\ DOC
/_AOB = /_DOC (vertically opposite angles)
/_OAB = /_OCD ( alternative interior angle)
/_OBA = /_ODC ( alternative interior angle)
.^. /_\ AOB ~ /_\ DOC ( AAA test of similarity)
OA = OB
OC OD (corresponding parts of similar triangle)
Hence proved.
*********************************************
(6) AB, CD , PQ are perpendicular to BD. AB = x, Cd =y and PQ = Z.
Prove that
1 + 1 = 1
x y z
sol) In /_\BCD. PQ || CD
BQ = PQ -------(1)
BD CD
In /_\ABD, PQ || AB
QD = PQ
BD AB
1 - BQ = PQ
BD AB
1 - PQ = PQ (from 1)
CD AB
1 = PQ ( 1 + 1 )
CD AB
1 = 1 + 1
PQ CD AB
But , given AB = x, CD = y and PQ = z
1 = 1 + 1
Z Y X
************************************
(7) A flag pole 4m tall casts a 6m, shadow. At the same time, a nearby building casts a shadow of 24m. How tall is the building?
sol)
In /_\ ABC and /_\ PQR
/_B = /_Q = 90 deg
/_C = /_R (AC || PR, all sun's rays are parallel at any instance)
/_\ABC ~ /_\PQR ( by AA similarity)
AB = BC
PQ = QR (cpst)
4 = 6
PQ 24
PQ = 4 * 24
6
PQ = 16 m. the height of the building
*************************************
(8) CD and GH are respectively the bisectors of /_ACB and /_EGF such that D and H lie on sides AB and FE of /_\ABC and /_\ FEG respectively.
then show that
(1) CD = AC
GH FG
sol)
1) Given :- /_\ ABC ~ /_\FEG
2) /_BAC = /_EFG (Corresponding angle of similar triangle)
3) /_ABC = /_FEG (corresponding angle of similar triangle)
4) /_ACB = /_FGE
5) 1/2 /_ACB = 1/2 /_FGE
6) /_ACD = /_FGH and /_BCD = /_EFGH
consider /_\ ACD and /_\ FGH
7) /_DAC = /_HFG (from 2)
8) /_ACD = /_EGH (from 6)
9) Also, /_ADC = /_FGH
If the two angle of triangle are equal to the two angle of another triangle, then by angle sum property of /_\, 3rd angle will also be equal.
10) .^. /_\ADC ~ /_\FHG (By AAA test of similarity)
11) CD = AC (CPST)
GH FG
Consider /_\DCB amd /_|HGE
12) /_DBC = /_HEG (from 3)
13) /_BCD = /_FGH (from 6)
14) also, /_BDC = /_EHG
15) /_\DCB ~ /_\HGE
**********************************************
(9) AX and DY are altitudes of two similar triangles
Prove that AX : DY = AB : DE.
sol)
Given :- /_\ ABC ~ /_\DEF
2) /_ABC = /_DEF ( CAST)
Consider /_\ABX and /_\DEY
3) /_ABX = /_DEY (from 2)
4) /_AXB = /_DYE = 90 deg
5) /_BAX = /_EDY
6) /_\ABX ~ /_\DEY (by AAA test of similarity)
7) AX = AB (CPST)
DY DE
******************************************
(10) Construct a triangle shadow similar to the given /_\ ABC, with its sides equal to 5/3 of the corresponding sides of the triangle ABC.
sol) Given a triangle ABC, we are required to construct a triangle whose sides are of the corresponding sides of /_\ABC
Steps of Construction :-
1) Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.
2) Locate 5 points ( the greater of 5 and 3 in 5/3) B1,B2,B3,B4 and B5 so that BB1=B1B2=B2B3=B3B4=B4B5.
3) Join B3( the 3rd point , 3 being smaller of 3 and 5 in) to C and draw a line through B5 parallel to B3C, intersecting the extended line segment BC at C'.
4) Draw a line through C' parallel to CA intersecting the extended line segment BA at A'.
Then A'BC' is the required triangle.
For justification of the construction , note that /_\ABC ~ /_\A'BC'.
.^., AB = AC = BC
A'B A'C' BC'
So, BC = 5 and
BC' 3
.^. A'B = A'C' = BC' = 5
AB AC BC 3
**********************************
(11) Construct a triangle of sides 4cm, 5cm, and 6 cm. Then, construct a triangle similar to it, whose sides are 2/3 of the corresponding sides of the first triangle.
sol) Construct a triangle ABC with given sides , AB = 4cm, BC = 5cm, CA = 6cm.
Given a triangle ABC, we are required to construct a triangle whose sides are 2/3 of the corresponding sides of /_\ABC.
Steps of construction :-
1) Draw any ray BX making an angle 30 deg with the base BC of /_\ ABC on the opposite side of the vertex A.
2)Locate seven points B1,B2,B3 on BX so that BB1=b1B2=B2B3
Note:- The number of points should be greater of "m" and 'N" in the scale factor m/n)
3) Join B2- C' and draw a line through B3 || B2C, intersecting the line segment BC at C,
4) Draw a line through "C" parallel to C'A intersecting the line segment BA at A'B'C is the required triangle.
*********************************************************************************
(12) Construct an Isosceles triangle whose base is 8cm and altitude is 4cm. Then, draw another triangle whose sides are 1(1/2) times the corresponding sides of the isosceles triangle.
sol) An isosceles tranagle whose base is 8cm and altitude is 4cm. Scale factor is 1(1/2) = 3/2.
Steps of construction:
1) Draw a line segment BC= 8cm.
2) Draw a perpendicular bisector AD of BC
3) Join AB and AC we get an isosceles triangle /_\ABC
4) Construct an acute angle /_CBX downwards.
5) On BX make three equal parts.
6) Join C to B2 and draw a line through B3 parallel to B2C intersecting the line extended line segment BC at C'
7) Again draw a parallel line C'A' to AC cutting BP at A'
8) /_\A'BC' is the required triangle.
************************************************
Areas of Similar Triangles
For two similar triangles , ratio of their corresponding sides is the same.
Theorem-8.6: The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.
Given : △ABC ~ △PQR
RTP :
ar(△ABC) = (AB )^2 = (BC)^2 = (CA)^2
ar(△PQR) = (PQ)^2 = (QR)^2 = (RP)^2
Construction: Draw AM _|_ BC and PN_|_QR
Proof:
ar(△ABC) = (1/2)*BC*AM = BC*AM ---(1)
ar(△PQR) (1/2)*QR*PN QR*PN
In △ABM & △PQN
∠B = ∠Q (△ABC~ △PQR)
∠M = ∠N = 90 deg
.^. △ABM ~ △PQN (by AA similarity)
AM = AB------(2)
PN PQ
Also △ABC ~ △PQR(given)
AB = BC = AC
PQ QR PR -------(3)
Therefore,
ar(△ABC) = AB * AB = (AB)^2
ar(△PQR) PQ PQ (PQ)^2
Now by using (3), we get
ar(△ABC) = (AB)^2 = (BC)^2 = (AC)^2
ar(△PQR) (PQ)^2 (QR)^2 (PR)^2
Hence proved.
*****************************************************
Example-8. prove that if the area of two similar traingles are equal, then they are congruent.
sol) △ABC ~ △PQR
ar(ABC)=(AB)^2= (BC)^2 = (AC)^2
ar(PQR) (PQ)^2 (QR)^2 (PR)^2
But
ar(△ABC) = 1 --->(Areas are equal)
ar(△PQR)
(AB)^2 = (BC)^2 = (AC)^2 = 1
(PQ)^2 (QR)^2 (PR)^2
So
AB^2 = PQ^2
BC^2 = QR^2
AC^2 = PR^2
From which we get
AB = PQ
BC = QR
AC = PR
.^. △ABC ~ △PQR ( by SSS congruency)
*****************************************************
Example-9. ABC ~ DEF and their areas are respectively 64cm^2 and 121cm^2. If EF = 15.4 cm, then find BC.
sol) ar(△ABC) = (BC)^2
ar(△DEF) (EF)^2
64 = (BC)^2
121= (15.4)^2
8 = (BC)
11 = 15.4
BC = 8 * 15.4
11
BC = 11.2 cm.
******************************************************
Example-10. Diagonals of a trapezium ABCD with AB//DC, intersect each other at the point 'O'. If AB = 2CD, find the ratio of triangles AOB and COD
sol) In trapezium ABCD, AB//DC also AB=2CD
In △AOB and △COD
∠AOB = ∠COD (vertically opposite angles)
∠OAB = ∠OCD (alternate interior angles)
△AOB ~ △COD ( by AA similarity)
ar(△AOB) = AB^2
ar(△COD) DC^2
(2DC)^2 = 4
(DC)^2 1
.^. ar(△AOB) : ar(△COD) = 4:1
Exercise- 8.3
(1) Equilateral triangles are drawn on the three sides of a right angled triangle. Show that the area of the triangle on the hypotenuse is equal to the sum of the areas of triangles on the other two sides.
sol) Given right angled triangle is ABC with AC as hypotenuse
let
AB = a
BC = b
AC = c
we have a^2 + b^2 = c^2 --------(1)
We know that area of equilateral triangle
= _/3 * side^2
4
Area of ACD = _/3 c^2
4
Area of BCF = _/3 b^2
4
Area of AEB = _/3 a^2
4
Area of AEB + Area of BCF
= _/3 (a^2 + b^2)
4
from 1
a^2 + b^2 = c^2
Area of AEB + Area of BCF
= _/3 c^2 = Area of ACD
4
**********************************************
(2) Prove that the area of the equilateral triangle described on the side of a square is half the area of the equilateral triangles described on its diagonal.
sol) Let us take a square with side 'a'
Then the diagonal of square will be a_/2
Area of equilateral traingle with side '
a'= -/3 a^2
4
Area of equilateral triangle with side
a_/2 = _/3 (a_/2)^2
4
=> _/3 2a^2
4
Ratio of two areas can be given as follows :
_/3
4
----------- = 1 /2
_/3 2
4
*************************************
(3) D, E, F are mid points of sides BC, CA, AB of /_\ ABC . Find the ratio of areas of /_\ DEF and /_\ ABC.
sol)
Since "D" and "F" are midpoints of AB and AC,
DE || AF or DF || BE
similarly EF || AB or EF || DB
AFED is a parallelogram as both pair of opposite side are parallel
By the property of parallelogram
/_DBE = /_DFE
or
/_DEF = /_ABC ----(1)
similarly
/_FEB = /_ACB ----(2)
In /_\DEF and /_\ABC
equation (1) and (2)
/_\DEF ~ /_\CAB
ar (/_\DEF)
ar (/_\ABC)
=> = DE^2
2DE^2
=> 1
4
ar (/_\DEF) : ar (/_\ABC) = 1:4
*****************************
(4) In /_\ ABC, XY // AC and XY divides the traingle into two parts of equal area. Find the
ratio of AX .
XB
sol)
Given :- XY || AC
=> /_1 = /_3 and /_2 = /_4 [ Corresponding angles]
=> /_\BXY ~ /_\BAC [ AA similarity]
=> ar (/_\BXY)
ar (/_\BAC)
=> BX^2 [By theorem] ----(1)
BA^2
Also, we are given that
ar (/_\BXY) = 1/2 ar (/_\BAC)
=> ar (/_\BXY)
ar (/_\BAC)
=> 1/2 -------(2)
From (1) and (2)
BX^2 = 1
BA^2 2
BX = 1
BA _/2
Now, AX
XB
=> AB -BX
XB
=> AB - 1
BX
=> _/2 -1
1
AX = _/2-1
XB 1
**************
(5) Prove that the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
sol)
In case of two similar triangles ABC and PQR
ar (ABC) = AB^2 = AC^2
ar (PQR) PQ^2 PR^2
Let us assume "AD" adn "PM" are the medians of these two triangles.
Then;
AB^2 = AC^2 = AD^2
PQ^2 PR^2 PM^2
Hence ,
ar (ABC) = AD^2
ar (PQR) PM^2
*******************************************************
(6) /_\ABC ~ /_\ DEF , BC = 3cm, EF = 4cm and area of /_\ ABC = 54 cm^2. Determine the area of /_\ DEF.
sol) Given BC = 3cm and EF = 4 cm
ar (/_\ABC) = [ BC ]^2
ar (/_\DEF) [ EF ]^2
54 = 9
ar (/_\DEF) 16
ar (/_\DEF) = 54 * 16
9
ar (/_\DEF) = 96 cm^2
************************
(7) ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP =1cm, and BP = 3cm, AQ = 1.5 cm, CQ = 4.5 cm.
Prove that ( area of /_\APQ = 1/16 area of /_\ABC)
sol)
From above figure its evident that /_\ABC ~ /_\APQ
We know that
Area of /_\APQ
Area of /_\ABC
=> [ AP ]^2
AP +PB
=> [ AQ ]^2
AQ + QC
=> 1
16
=> Area of /_\APQ = 1/16 Area of /_\ABC
******************************************
(8) The area of two similar triangles are 81 cm^2 and 49 cm^2 respectively. If the attitude of the bigger triangle is 4.5cm. Find the corresponding attitude for the smaller triangle.
sol)
Let AD and PM are corresponding altitude of triangle.
ar (/_\ ABC ) = [ AD ]^2
ar (/_\DEF) [ PM]^2
81 = (4.5)^2
49 DQ
DQ = 7 * 4.5 = 3.5 cm
9
*******************************************************
Theorem-8.7 : If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, then the triangles on both sides of the perpendicular are similar to the whole triangle and to each other.
Proof: ABC is a right triangle, right angled at "B". Let "BD" be the perpendicular to hypotenuse "AC"
In △ADB and △ABC
∠A = ∠A
and ∠ADB = ∠ABC
So △ADB ~ △ABC -----(1)
Similarly,△ADB ~ △ABC ----(2)
So from (1) and (2), triangles on both sides of the perpendicular BD are similar to the whole triangle ABC.
Also since △ADB ~ △ABC
△BDC ~ △ABC
So △ADB ~ △BDC
*******************************************************
Pythagoras Theorem (Baudhayan Theorem)
Theorem-8.8 : In a right triangle, the square of hypotenuse is equal to the sum of the squares of the other two sides.
Given : △ABC is a right triangle right angled at "B"
RTP : AC^2 = AB^2 + BC^2
Construction : Draw BD _|_ AC
proof :△ ADB ~ △ABC
AD = AB (sides are proportional)
AB AC
AD . AC = (AB)^2 ------(1)
Also, △BDC ~ △ABC
CD = BC
BC AC
CD . AC = BC^2 ----(2)
On adding (1) & (2)
AD.AC + CD. AC = AB^2 + BC^2
AC(AD + CD) = AB^2 + BC^2
AC.AC = AB^2 + BC^2
AC^2 = AB^2 + BC^2
The above theorem was earlier given by an ancient Indian mathematician Baudhayan(about 800 BC) in the following form.
" The diagonal of a rectangle produces by itself the same area as produced by its both sides (i.e, length and breadth)." So sometimes, this theorem is also referred to as the Baudhayan theorem.
Theorem-8.9: In a triangle if square of one side is equal to the sum of squares of the other two sides, then the angle opposite to the first side is a right angle and the triangle is a right angled triangle.
Given : In △ABC ,
AC^2 = AB^2 + BC^2
RTP: ∠B = 90 deg
Construction : Construct a right angled triangle △PQR right angled at "Q" such that PQ = AB and QR = BC.
Proof : In △PQR ,
PR^2 = PQ^2 + QR^2 (Phy theorem as ∠Q = 90 deg)
PR^2 = AB^2 + BC^2 ( by construction) ----(1)
but AC^2 = AB^2 + BC^2 ( given) ----(2)
.^. AC = PR from (1) and (2)
Now in △ABC and △PQR
AB = PQ ( by construction)
BC = QE ( by construction)
AC = PR ( proved)
.^. △ABC ~ △PQR ( by SSS congruency)
.^. ∠B = ∠Q (by cpct)
but ∠Q = 90 deg ( by constrcution)
.^. ∠B = 90 deg
Hence proved.
**********************************************************
**
Example-11. A ladder 25m long reaches a window of building 20m above the ground, determine the distance of the foot of the ladder from the building.
sol) In △ABC , ∠C = 90 deg
AB^2 = AC^2 + BC^2 ( By pythagorous theorem)
(25)^2 = (20)^2 + BC^2
BC^2 = 625 - 400 = 225
BC = √ 225 = 15m
Hence, the foot of the ladder is at a distance of 15m from the building.
*************************************************
Example-12. BL and CM are medians of a triangle ABC right angled at "A".
Prove that 4(BL^2 + CM^2) = 5BC^2.
sol) BL and CM are medians of △ABC in which ∠A = 90deg
In △ ABC
BC^2 = AB^2 + AC^2 (Phythagorous theorem) --(1)
In △ ABL
BL^2 = AL^2 + AB^2
So
BL^2 = (AC/2)^2 + AB^2 ( L is the midpoint of AC)
BL^2 = (AC^2/4) + AB^2
4BL^2 = AC^2 + 4AB^2
In △CMA ,
CM^2 = AC^2 + AM^2
CM^2 = AC^2 + (AB)^2 (M is the midpoint of AB)
(2)^2
CM^2 = AC^2 + (AB^2)
4
4CM^2 = 4AC^2 + AB^2 ----(3)
On adding (2) and (3), we get
4(BL^2 + CM^2) = 5(AC^2 + AB^2)
.^. 4(BL^2 + CM^2) = 5BC^2 (from 1)
*******************************************************
Example-13. 'O' is any point inside a rectangle ABCD.
Prove that OB^2 + OD^2 = OA^2 + OC^2
sol) Through 'O' draw PQ//BC so that 'P' lies on 'AB' and 'Q' lies on 'DC'
Now PQ//BC
.^. PQ_|_AB & PQ_|_DC (∠B = ∠C = 90deg)
So,
∠BPQ = 90 deg &
∠CQP = 90 deg
.^. BPQC and APQD are both rectangles.
Now from △OPB.
OB^2 = BP^2 + OP^2 ------(1)
Similarly from △OQD,
we have OD^2 = OQ^2 + DQ^2 ---(2)
from △OQC
we have OC^2 = OQ^2 + CQ^2 ----(3)
and from △OAP,
OA^2 = AP^2 + OP^2
Adding (1) and (2)
OB^2 + OD^2 = BP^2 + OP^2 + OQ^2 + DQ^2
=> CQ^2 + OP^2 + OQ^2 + AP^2 (BP=CQ and DQ=AP)
=> CQ^2 + OQ^2 + OP^2 + AP^2
=> OC^2 + OA^2 ( from (3) & (4))
******************************************************
Example-14. The hypotenuse of a right triangle is 6m more than twice of the shortest side. If the third side is 2m, less than the hypotenuse, find the sides of the triangle.
sol) Let the shortest side be 'X' m
Then hypotenuse = (2x+6)m and third side=(2x+4)m
by Pythagores theorem, we have
(2x+6)^2 = x^2 + (2x+4)^2
4x^2 + 24x + 36 = x^2 + 4x^2 + 16x + 16
x^2 - 8x - 20 = 0
(x-10)(x+2) = 0
x = 10 or x = -2
but "x" can't be negative as side of a traingle.
.^. x = 10
Hence, the sides of the triangle are 10m, 26m, and 24m.
********************************************************
Example -15. ABC is a right triangle right angled at "C". Let BC = a , CA= b, AB = c and let 'P' be the length of perpendicular from "C" on "AB" .
prove that
(!) pc = ab
(!!) (1/p^2) = (1/a^2) + (1/b^2)
sol) (!) CD_|_AB and CD = p.
Area of △ABC = (1/2) * AB * CD
=> (1/2)*cp.
also
Area of △ABC = (1/2) * BC * AC
=> (1/2)*ab
(1/2)*cp = (1/2)*ab
cp = ab ------(1)
(!!) Since △ABC is a right triangle right angled at "C"
AB^2 = BC^2 + AC^2
c^2 = a^2 + b^2
(ab)^2 = a^2 + b^2
(p)^2
1 = a^2 + b^2
p^2 a^2b^2
=> 1 + 1
a^2 b^2
******************************************************
Exercise - 8.4
(1) prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
sol)
ABCD is a rhombus in which diagonals AC and BD intersect at point "O".
RTP :- AB^2 + BC^2 + CD^2 + DA^2 = AC^2 + DB^2
In /_\AOB : AB^2 = AO^2 + BO^2
In /_\BOC : BC^2 = CO^2 + BO^2
In /_\COD : CD^2 = DO^2 + CO^2
In /_\AOD : AD^2 = DO^2 + AO^2
Adding the above four equations, we get;
AB^2 + BC^2 + CD^2 + DA^2 = AO^2 + BO^2 + CO^2 + BO^2 + DO^2 + CO^2 + DO^2 + AO^2
OR
AB^2 + BC^2 + CD^2 + DA^2 = 2(AO^2 + BO^2 + CO^2 + DO^2)
AB^2 + BC^2 + CD^2 + DA^2 = 2(2AO^2 + 2BO^2)
***************************************
Since AO^2 = CO^2 and BO^2 = DO^2
**************************************
AB^2 + BC^2 + CD^2 + DA^2 = 4(AO^2 + BO^2) -------(1)
Now, let us take the sum of squares of diagonals ;
AC^2 + DB^2 = (AO + CO)^2 + (DO + BO)^2
=> (2AO)^2 + (2DO)^2
=> 4AO^2 + 4BO^2 ------(2)
from equations (1) and (2), it is clear ;
AB^2 + BC^2 + CD^2 + DA^2 = AC^2 + DB^2
Hence proved.
********************************************
(2) ABC is a right triangle right angled at "B". Let "D" and "E" be any points on AB and BC respectively.
Prove that AE^2 + CD^2 = AC^2 + DE ^2
sol) In right angled /_\ABE and /_\DBC, we have
AE^2 = AB^2 + BE^2 --------(1)
DC^2 = DB^2 + BC^2 -------(2)
Adding (1) and (2)
AE^2 + DC^2 = AB^2 + BE^2 + DB^2 + BC^2
= (AB^2 + BC^2)_ + ( BE^2 + DB^2)
**************************************************
Since AB^2 + BC^2 = AC^2 in right angled triangle ABC
***************************************************
= ( AC^2 + DE^2)
Hence proved.
********************************************
(3) Prove that three times the square of any side of an equilateral triangle is equal to four times the square of the altitude.
sol)
Given :- An equilateral triangle ABC, in which AD _|_BC
RTP :- 3AB^2 = 4AD^2
Proof :- Let AB= BC = CA = a
In /_\ ABD and /_\ACD
AB=AC, AD = AD
and /_ADB = /_ADC ( each 90 deg)
.^. /_\ ABD = /_\ ACD
.^. BD = CD = a (CPCTE)
2
Now,
In /_\ ABD, /_D = 90 deg
.^. AB^2 = BD^2 + AD^2
or
AB^2 = [CD]^2 + AD^2
[2}^2
=> [ AB ]^2 + AD^2
2
AB^2 = [AB]^2 + AD^2
[ 2 }^2
4AB^2 = AB^2 + 4AD^2
3AB^2 = 4AD^2
*************************
(4) PQR is a triangle right angled at "P" and "M" is a point on "QR" such that PM 1 QR.
Show that PM^2 = QM . MR.
sol)
Let /_MPR = x
In /_\ MPR
/_MRP = 180 - 90 -x
/_MRP = 90 - x
Similarly in /_\ MPQ
/_MPQ = 90 - /_MPR
=> 90 - x
/_MQP = 180 - 90 - (90-x)
/_MQP = x
In /_\QMP and /_\PMR
/_MPQ = /_MRP
/_PMQ = /_RMP
/_MQP = /_MPR
/_\QMP ~ /_\PMR [by AAA criteria]
QM = MP
PM MR
PM^2 = MR * QM
Hence proved.
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(5). ABD is a triangle right angled at "A" and AC 1 BD
1) AB^2 = BC . BD.
sol) consider two triangles ACB and DAB
we have /_ABC = /_DBC (common)
/_ACB = /_DAB (Each is right angle)
/_CAB = /_ADB (Third angle)
.^. Triangle are similar and their corresponding sides must be proportional.
i.e, /_ADC = /_ADB
AC = CB = AB
DA AB DB
.^. AB^2 = BC * BD
********************
2) AC^2 = BC .DC
sol) /_BDA = /_BDC = 90
/_3 = /_2 = 90 /_1
[.^. /_1 + /_2 = 90, /_1 + /_3 = 90 ]
/_2 + /_4 = 90 /_2
[.^. /_1 + /_2 = 90, /_2 + /_4 = 90]
/_\ ADB ~ /_\BDC [AAA criterion of similarity].
Their corresponding sides must be proportional.
DC = CA = DA
AC CB AB
=> DC = CA
AC CB
=> CA^2 = BC * DC
******************************************
3) AD^2 = BD.CD.
sol) In two triangles ADB and ABC, we have
/_ADC = /_ADB ( Common)
/_DCA = /_DAB ( Each is right angle)
/_DAC = /_DBA (Third angle)
/_DCA = /_DAB ( AAA similarity)
Triangle ADB and ABC are similar and so their corresponding sides must be proportion.
DC = CA = DA
DA CB DB
DC = DA
DA DB
AD^2 = DB * DC
Hence proved.
************************
(6) ABC is an isosceles triangle right angled at "C". Prove that AB^2 = 2AC^2.
sol) Since the triangle is right angled at "C" therefore the side "AB" is the hypotenuse. Let the base of the triangle be "AC" and the altitude be "BC".
Applying the Pythagorean theorem
AB^2 = AC^2 + BC^2
Since the triangle is isosceles triangle two of the sides shall be equal.
Therefore AC = BC
Thus AB^2 = AC^2 + AC^2
AB^2 = 2AC^2
Hence proved.
********************************
(7) "O" is any point in the interior triangle ABC. OD 1 BC, OE 1 AC and OF 1 AB , show that
1) OA^2 + OB^2 + OC^2 - OD^2 - OE^2 - OF^2 = Af^2 + BD^2 + CE^2
sol) Given :- /_\ ABC and "O" is a point in the interior of a triangle ABC where, OD _|_ BC, OE _|_ AC, OF _|| AB
Proof :-
1) Let us join the point "O" from A,B, and C.
2) Using pythagoras theorem,
(Hyp)^2 = (Height)^2 + (Base)^2
3) Ina right angle triangle OAF.
(OA)^2 = AF^2 + OF^2 ------(1)
4) In right angle triangle ODB
(OB)^2 = OD^2 + BD^2 -----(2)
5) In a right angle triangle OEC
(OC)^2 = (OE)^2 + (EC)^2 -----(3)
Adding (1) + (2) + (3)
(OA)^2 + (OB)^2 + (OC)^2 = AF^2 + OF^2 + OD^2 + BD^2 + (OE)^2 + (EC)^2
OA^2 + OB^2 + OC^2 - OD^2 - OE^2 - OF^2 = Af^2 + BD^2 + CE^2
Hence proved.
*************************************
2) AF^2 + BD^2 + CE^2 = AE^2 + CD^2 + BF^2.
sol) Using Pythagoras theorem.
(Hyp)^2 = (height)^2 + (Base)^2
1) In /_\ ODB,
OB^2 = OD^2 +BD^2
2) In /_\ OFB,
OB^2 = OF^2 +FB^2
3) In /_\ OFA,
OA^2 = OF^2 + AF^2
4) In /_\ OEA,
OA^2 = OE^2 + AE^2
5) In /_\ OEC,
OC^2 = OE^2 + CE^2
6) In /_\ ODC,
OC^2 = OD^2 + CD^2
L.H.S :- AF^2 + BD^2 + CE^2
=> ( OA^2 - OF^2) + ( OB^2 - OD^2) + ( OC^2 - OE^2)
=> OA^2 + OB^2 + OC^2 - OF^2 - OD^2 - OE^2
R.H.S :- AE^2 + CD^2 + BF^2
=> (OA^2 - OE^2) + ( OC^2 - OD^2) + (OB^2 - OF^2)
=> OA^2 + OB^2 + OC^2 - OF^2 - OD^2 - OE^2
Since L.H.S = R.H.S
Hence proved
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(8) A wire attached to vertical pole of height 18m is 24m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut.
sol) let AB = 24 m be a wire attached to a vertical pole
BC = 18m
To keep the wire taut, let it be fixed at "A"
AB^2 = AC^2 + BC^2
(24)^2 = AC^2 + (18)^2
AC^2 = 576 - 324
AC = 6√7
Hence, the stake may be placed at a distance 6_/7m the base of pole
*********************************************************
(9) Two poles of heights 6m and 11m stand on a plane ground . If the distance between the feet of the poles is 12m find the distance between their tops.
sol )
Given :- BC = 6 cm and AD = 11m, BC = ED
So, AE = AD - ED
= 11 - 6
= 5m
BE = CD = 12m
Find, AB = ?
Solution :- Now , In /_\ ABE, /_E = 90
AB^2 = AE^2 + BE^2
AB^2 = (5)^2 + (12)^2
AB^2 = 169
AB = 13m
The distance between their tops is 13m.
*****************************************
(10) In a equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC. Prove that 9AD^2 = 7Ab^2.
sol)
ABC is a equilateral triangle,
AB = BC = AC and BD = 1/3 BC
Construction :-
Draw AE _|_ BC
Proof :-
/_\ABE ~ /_\ ACE
.^. BE = EC = BC / 2
Now in /_\ABE,
AB^2 = BE^2 + AE^2
also, AD^2 = AE^2 + DE^2
.^. AB^2 - AD^2 = BE^2 - DE^2
=> BE^2 - ( BE - BD)^2
=> ( BC)^2 - ( BC - BC ) ^2
2 2 3
=> (AB)^2 - ( AB - AB)^2
2 2 3
AB^2 - AD^2 = 2AB^2
9
or
7AB^2 = 9AD^2
******************************
11) In the given figure, ABC is a triangle right angled at "B, D and E are points on BC trisect it.
prove that 8AE^2 = 3AC^2 + 5AD^2.
sol) In /_\ABD, /_B = 90 deg
.^. AC^2 = AB^2 +BC^2 -----(1)
similarly , AE^2 = AB^2 + BE^2 -----(2)
and AD^2 = AB^2 + BD^2 -----(3)
From (1)
3AC^2 = 3AB^2 + 3BC^2 -----(4)
From (2)
5AD^2 = 5AB^2 + 5BD^2 ----(5)
Adding equation (4) and (5)
3AC^2 + 5AD^2 = 8AB^2 + 3BC^2 + 5BD^2
=> 8AB^2 + 3(3 BE )^2 + 5 ( BE)^2
2 2
=> 8 (AB^2 + BE^2)
=> 8AE^2
**************************
12) ABC is an isosceles triangle right angled at "B". Similar traingles ACD and ABE are constructed on sides AC and AB. Find the ratio between the areas of /_\ABE and /_\ ACD.
sol) Given /_\ABC is an isosceles triangle in which /_B = 90 deg
=> AB = BC
By Pythagoras theorem, we have
AC^2 = AB^2 + BC^2
=> AC^2 = AB^2 + AB^2
[ Since AB = BC ]
=> AC^2 = 2AB^2 --------(1)
It is also given that /_\ABE ~ /_\ACD
ratio of areas of similar triangles is equal to ratio of squares of their corresponding sides.
ar (/_\ ABC) = AB^2
ar(/_\ ACD) AC^2
ar(/_\ABC) = AB^2
ar(/_\ACD) 2AB^2 --------(1)
ar(/_\ABC) = 1
ar(/_\ACD) 2
.^. ar(/_\ABC) : ar(/_\ACD) = 1:2
******************************
Optional exercise:-
1). In the given figure
QT = QR
PR QS
and ∠1 and ∠2
prove that △PQS ~ △ TQR
sol)Given :- ∠1 = ∠ 2
QT = QR
PR QS
R.T.P : △PQS ~ △TQR
Proof : Since ∠1 = ∠2
Step 1) PR = QP [ sides opposite to equal angles are equal]
Step 2) Putting PR=QP in
QR = QT
QS PR
QR = QT -------(2)
QS QP
In △PQS and △TQR
∠PQS = ∠TQR [common angles]
QR = QT -----(from(2))
QS QP
Therefore,
△PQS~ △TQR [ SAS similarity]
Hence Proved
*****************************************
2) Ravi is 1.82m tall. He wants to find the height of a tree in his backyard. from the tree's base he walked 12.20m. along the tree's shadow to a position where the end of his shadow exactly overlaps the end of the tree's shadow. He is now 6.10m from the end of the shadow. How tall is the tree?
sol)
In △EDA and ⧍CBA
∠B = ∠D = 90 deg
∠A = ∠A ( common angle)
AA Traingle similarity :- if two traingles are similar it means that all the corresponding angle pairs are congruent and all corresponding sides are proportional
.^. by AA similarity
△EDA ~ △CBA
Now,
ED = DA
CB BA
ED = DB + BA
1.82 6.10
ED = 12.20 + 6.10
1.82 6.10
ED = 18.30
1.82 6.10
ED = 3 * (1.82)
ED = X = 5.46 is the height of the tree
*******************************************
3) The diagonal AC of a parallelogram ABCD intersects DP at the point "Q" where "P" is any point on side "AB". Prove that
CQ * PQ = QA * QD
sol) Given :- The diagonal "AC of a parallelogram "ABCD" intersects "DP" at the point "Q"
Where "p" is any point on side "AB"
R.T.P:- CQ * PQ = QA * QD
Proof :- In △APQ & △CDQ
∠AQP = ∠CQD [ vertically opposite angles]
∠PAQ = ∠DCQ [ alternate angles]
.^. △APQ ~ △CQD [ A.A criterion]
PQ = QA
QD CQ (cross multiply)
CQ * PQ = QA * QD
Hence proved.
*****************************************
4) △ABC and △AMP are two right triangles right angled at "B" and "M" respectively.
(*) Prove that
(!) △ABC ~ △ AMP
sol) Given :-
∠ABC = 90deg ,∠AMP = 90deg
RTP :- △ABC ~ △AMP
Proof :- In △ABC and △AMP
∠A = ∠A (common angle)
∠ABC = ∠AMP = 90 (Both 90 deg)
Using AA similarity
△ABC ~ △AMP
***************
(!!) CA = BC
PA MP
sol) We know when two traingles are similar, the ratio of their corresponding sides is proportional.
As we already proved in (!)
△ABC ~ △AMP
By using AAA criterion
CA = BC
PA MP
Hence proved
*******************************************
(5) An aeroplane leaves an airport and flies due north at a speed of 1000 kmph. At the same time another aeroplane leaves the same airport and flies due west at a speed of 1200 kmph. How far apart will the two planes be after 1(1/2) hour?
sol) condition (1) : 1st plane flies due north at a speed of 1000 kmph
We have
Distance = Speed * Time
Distance = 1000 * 1(1/2)
Distance = 1000 * (3/2)
Distance = 1000 * 1.5 => 1500 km
(*) condition (2) :- 2nd plane due west
Distance = 1200 * 1.5 => 1800 km
In △ABC
BC is distance travelled by first aeroplane and BA is the distance travelled by second aeroplane.
Applying pythagoars Theorem.
AC^2 = AB^2 + BC^2
AC^2 = (1500)^2 + (1800)^2
AC^2 = 2,25,0000 + 3,24,0000
AC = √5,49,0000
AC = √ 549 * √10,000 3 549
3 183
AC = √549 * √(100)^2 61 61
1
AC = 100 √(3)^2 * √61
AC = 3 *100 √61
AC = 300√61 km
Hence, the two planes would be 300√61 km from each other.
******************************************
(6) In a right triangle ABC right angled at C, P, and Q are points on sides AC and CB respectively which divide these sides in the ratio of 2:1.
Prove that
(1) 9AQ^2 = 9 AC^2 + 4BC^2
sol) Given :- "P" divides "AC" in the ratio 2:1 and "Q" divides "BC" in the ratio 2:1
CQ = (2/3) BC
PC = (2/3) AC
(!) In ACQ, we have
AQ^2 = AC^2 + CQ^2
AQ^2 = AC^2 + (2/3)BC)^2
AQ^2 = AC^2 + (4/9) BC^2
9AQ^2 = AC^2 +4BC^2
Hence proved.
********************************
(2) 9BP^2 = 9BC^2 + 4AC^2
sol) In PCB, we have
BP^2 = BC^2 + PC^2
BP^2 = BC^2 + (2/3)AC)^2
BP^2 = BC^2 + (4/9)AC^2
BP^2 = 9BC^2 + 4AC^2
9
9BP^2 = 9BC^2 + 4AC^2
Hence proved
*************************************
(3) 9(AQ^2 + BP^2) = 13AB^2
sol) Adding (1) and (2) we get
9AQ^2 + 9BP^2 = 9 AC^2 +9BC^2 + 4BC^2 + 4AC^2
9(AQ^2 +BP^2) = 13AC^2 + 13BC^2
9(AQ^2 + BP^2) = 13(AC^2 + BC^2)
.^. 9(AQ^2 + BP^2) = 13AB^2
Hence proved.
***************************************
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