PRAYAS Education

SIMILAR TRIANGLES

 INTRODUCTION

In earlier classes, you have learnt about congruence of two geometric figures, and also some basic theorems and
results on the congruence of triangle. Two geometric figures having same shape and size are congruent to each
other but two geometric figures having same shape are called similar. Two congruent geometric figures are always
similar but the converse may or may not be true.
All regular polygons of same number of sides such as equilateral triangle, squares, etc, are similar. All circles are
similar.
In some cases, we can easily notice that two geometric figures are not similar. For example, a triangle and a
rectangle can never be similar. In case, we are given two triangles, they may appear to be similar but actually they
may not be similar. So, we need some criteria to determine the similarity of two geometric figures. In particular,
we shall discuss similar triangles.

 HISTORICAL FACTS
EUCLID was a very great Greek mathematician born about 2400
years ago. He is called the father of geometry
because he was the first to establish a school of mathematics in
Alexandria. He wrote a book on geometry called “The Elements”
which has 13 volumes and has been used as a text book for over
2000 years. This book was further systematized by the great
mathematician of Greece tike Thales, Pythagoras. Pluto and
Aristotle.
Abraham Lincoln, as a young lawyer was of the view that this
greek book was a splendid sharpner of human mind
and improver his power of logic and language.
A king once asked Euclid, “Isn’t there an easier way to
understand geometry”
Euclid replied : “There is no royal-road way to geometry.
Every one has to think for himself when studying.”
THALES (640-546 B.C.) a Greek mathematician was the first
who initiated and formulated the theoretical study of geometry to
make astronomy a more exact science. He is said to have
introduced geometry in Greece. He is believed to have found the
heights of the pyramids in Egypt, using shadows and the principle
of similar triangles. The use of similar triangles has made possible
the measurements of heights and distances. He proved the well-
known and very useful theorem credited after his name : Thales
Theorem.

 CONGRUENT FIGURES
Two geometrical figures are said to be congruent, provided they must have same shape and same size. Congruent
figures are alike in every respect.
EX. 1. Two squares of the same length.
2. Two circle of the same radii.
3. Two rectangles of the same dimensions.
4. Two wings of a fan.
5. Two equilateral triangles of same length.
 PRAYAS Education

 SIMILAR FIGURES
Two figures are said to be similar, if they have the same shape. Similar figures may differ in size. Thus, two
congruent figures are always similar, but two similar figures need not be congruent.
EX. 1. Any two line segments are similar.
2. Any two equilateral triangles are similar
3. Any two squares are similar.
4. Any two circles are similar.

We use the symbol ‘~’ to indicate similarity of figures.

 SIMILAR TRIANGLES

∆ ABC and ∆ DEF are said to be similar, if their corresponding angles are equal and the corresponding sides are

proportional.

i.e., when ∠ A = ∠ D, ∠ B = ∠ E, ∠ C = ∠ F

AB BC AC
and = = .
DE EF DF

And, we write ∆ ABC ~ ∆ DEF.

The sign ‘~’ is read as ‘is similar to’.

THEOREM-1 (Thales Theorem or Basic Proportionality Theorem ) : If a line is drawn parallel to one side
of a triangle intersecting the other two sides, then the other two sides are divided in the same ratio.

Given : A ∆ ABC in which line  parallel to BC (DE║BC) intersecting AB at D and AC at E.

AD AE
To prove : =
DB EC

Construction : Join D to C and E to B. Through E drawn EF perpendicular to AB i.e., EF ⊥ AB and through D

draw DG ⊥ AC.
 PRAYAS Education

Proof :
STATEMENT REASON
1 1
1. Area of ( ∆ ADE) = ( AD × EF ) Area of ∆ = base × altitude
2 2
1
Area of ( ∆ BDE) = (BD × EF)
2
1
AD × EF
Area ( ∆ADE ) AD By 1.
2. =2 =
Area (∆BDE ) 1 BD × EF DB
2
1
AE × DG
Area (∆ADE ) 2 AE
3. = = Similarly
Area (∆BDE ) 1
EC × DG EC
2 ( ∆ CDE)
4. Area ( ∆ BDE) = Area ∆ s BDE and CDE are on the same base BC and between the
same parallel lines DE and BC.
Area (∆ADE ) AE
5. =
Area (∆BDE ) EC By 3. & 4.
6. AD AE
= By 1. & 5.
DB EC
Hence proved.
THEOREM-2 (Converse of Basic Proportionality Theorem) : If a line divided any two sides of a triangle
proportionally, the line is parallel to the third side.
AD AE
Given : A ∆ ABC and DE is a line meeting AB and AC at D and E respectively such that =
DB EC
To prove : DE║BC

Proof :
STATEMENT REASON
1. If possible, let DE be not parallel to BC. Then,
draw DF║BC
AD AE
2. = By Basic Proportionality Theorem.
DB FC
AD AE
3. = Given
DB EC
AF AE
4. ∴ = From 2 and 3.
FC EC
AF AE Adding 1 on both sides.
⇒ +1 = +1
FC EC
AF + FC AE + EC
⇒ = By adding.
FC EC
AC AC
⇒ = AF + FC = AC and AE + EC = AC.
FC EC
⇒ FC = EC ⇒ E and F coincide.
But, DF║BC. Hence DE║BC.
Hence, proved.
 PRAYAS Education

Ex.1 In the adjoining figure. DE║BC.

(i) If AD = 3.4 cm, AB = 8.5 cm and AC = 13.5 cm, find AE
AD 3
(ii) If = and AC = 9.6 cm, find AE.
DB 5

AD AE
Sol. (i) Since DE║BC, we have =
AB AC
3.4 AE 3.4 × 13.5
∴ = ⇒ = 5.4
8.5 13.5 8.5
Hence, AE = 5.4 cm.

AD AE
(i) =
Since DE║BC, we have
DB EC
AE 3  AD 3 
∴ =  = (Given)
AC 5  DB 5 
Let AE = x cm. Then, EC = (AC – AE) = (9.6 – x) cm.
x 3
∴ = ⇒ 5 x = 3(9.6 − x)
9.6 − x 5
⇒ 5x = 28.8 – 3x ⇒ 8x = 28.8 ⇒ x = 3.6.
∴ AE = 3.6 cm.

Ex.2 In the adjoining figure, AD = 5.6 cm, AB = 8.4 cm, AE = 3.8 cm and AC = 5.7 cm. Show that DE║BC.
Sol. We have, AD = 5.6 cm, DB = (AB – AD) = (8.4 – 5.6) cm = 2.8 cm.
AE = 3.8 cm, EC = (AC – AE) = (5.7 – 3.8) cm = 1.9 cm.

AD 5.6 2 AE 3.8 2
∴ = = and =
DB 2.8 1 EC 1.9 1

AD AE
Thus, =
DB EC

∴ DE divides AB and AC proportionally.

Hence, DE║BC
PS PT
Ex.3 In fig, = and ∠ PST = ∠ PRQ. Prove that PQR is an isosceles triangle. [NCERT]
SQ TR
PS PT
Sol. It is given that =
SQ TR
So, ST║QR [Theorem]
Therefore, ∠ PST = ∠ PQR [Corresponding angles] – (1)

Also, it is given that

∠ PST = ∠ PRQ (2)
So, ∠ PRQ = ∠ PQR [Form 1 and 2]

Therefore PQ = PR [Sides opposite the equal angles]

i.e., PQR is an isosceles triangle.
 PRAYAS Education

Ex.4 Prove that any line parallel to parallel sides of a trapezium divides the non-parallel sides proportionally (i.e., In the
same ratio).
OR
ABCD is a trapezium with DE║AB. E and F are points on AD and BC respectively such that EF║AB. Show that
AE BF
= [NCERT]
ED FC
Sol. We are given trapezium ABCD.
CD ║BA
EF║AB and CD both
We join AC.
It mets EF at O.
In ∆ ACD, OE║CD
AO CF
⇒ = …(i)
OC ED
(Basic Proportionality Theorem)
In ∆ CAB, OF║AB
CO CF
⇒ = [B.P.T.]
OA FB
AO BF
⇒ = …(ii)
OC FC
From (i) and (ii)
AE BF
=
ED FC
Hence, proved.

Ex.5 Prove that the internal bisector of an angle of a triangle divides the
opposite side in the ratio of the sides containing the angle.
(Internal Angle Bisector Theorem)
Sol. Given : A ∆ ABC in which AD is the internal bisector of ∠ A.
BD AB
To Prove : =
DC AC
Construction : Draw CE║DA, meeting BA produced at E.

Proof :
STATEMENT REASON
1. ∠ 1 = ∠ 2 AD is the bisector of ∠ A
2. ∠ 2 = ∠ 3 Alt. ∠ s are equal, as CE║DA and AC is the transversal
3. ∠ 1 = ∠ 4 Corres. ∠ s are equal, as CE║DA and BE is the transversal
4. ∠ 3 = ∠ 4 From 1, 2 and 3.
5. AE = AC Sides opposite to equal angles are equal
6. In ∆ BCE, DA║CE
BD BA By B. P. T.
⇒ =
DC AE
BD AB
⇒ = Using 5
DC AC
Hence, proved.
 PRAYAS Education

Remark : The external bisector of an angle divides the opposite side externally in the ratio of the sides containing
the angle. i.e., if in a ∆ ABC, AD is the bisector of the exterior of angle ∠ A and intersect BC produced in
BD AB
D, = .
CD AC

 AXIOMS OF SIMILARITY OF TRIANGLES

1. AA (Angle-Angle) Axiom of Similarity :
If two triangles have two pairs of corresponding angles equal, then the triangles are similar. In the given figure,
∆ ABC and ∆ DEF are such that
∠ A = ∠ D and ∠ B = ∠ E.
∴ ∆ ABC ~ ∆ DEF

2. SAS (Side-Angle-Side) Axiom of Similarity :

If two triangles have a pair of corresponding angles equal and the sides including them proportional, then the
triangles are similar.
In the given fig, ∆ ABC and ∆ DEF are such that
AB AC
∠ A = ∠ D and =
DE DF
∴ ∆ ABC ~ ∆ DEF

3. SSS (Side- Side- Side) Axiom of Similarity :

If two triangles have three pair of corresponding sides proportional, then the triangles are similar.
If in ∆ ABC and ∆ DEF we have :
AB AC BC
= = , then ∆ ABC ~ ∆ DEF .
DE DF EF
Ex.6. In figure, find ∠ L.
Sol. In ∆ ABC and ∆ LMN,
AB 4.4 2
= =
LM 11 5
BC 4 2 CA 3.6 2
= = and = =
MN 10 5 NL 9 5
AB BC CA
⇒ = =
LM MN NL
⇒ ∆ ABC ~ ∆ LMN (SSS Similarity)
⇒ ∠ L = ∠ A = 180 – ∠ B – ∠ C
0

= 1800 – 500 – 700 = 600

∴ ∠ L = 600
 PRAYAS Education

Ex.7 In the figure, AB ⊥ BC, DE ⊥ AC, and GF ⊥ BC, Prove that ∆ ADE ~ ∆ GCF.
Sol. ∠ 1 + ∠ 4 = ∠ 1 + ∠ 2 (each side = 900)
⇒ ∠ 4 =∠ 2
⇒ ∠A =∠G …(i)
Also ∠E =∠F …(ii) (each equal to 900)
From (i) and (ii), we get AA similarity for triangle ADE and GCF.
⇒ ∆ ADE ~ ∆ GCF
QT QR
Ex.8 In fig, = and ∠ 1 = ∠ 2. Prove that ∆ PQS ~ ∆ TQR.
PR QS
Sol. ∠1 = ∠2 (Given)
⇒ PR = PQ …(i)
(Sides opposite to equal angles in ∆ QRP)
QT QR
Also = (Given) …(i)
PR QS
From (i) and (ii), we have
QT QR QP QS
= ⇒ = …(iii)
PR QS QT QR
Now, in triangles PQR and TQR, we have
∠ PQS = ∠ TQR (each = ∠ 1)
PQ QS
and = (from (3))
TQ QR
⇒ ∆ PQS ~ ∆ TQR (SAS Similarity)
Ex.9 In fig, CD and GH are respectively, the medians of ∆ ABC and ∆ FEG, If ∆ ABC ~ ∆ FEG, prove that
(i) ∆ ADC ~ ∆ FHG
CD AB
(ii) = (NCERT)
GH FE
Sol. ∆ ABC ~ ∆ FEG (given)
⇒ ∠ A = ∠ F, …(i) ( the corresponding angles of the similar triangles are equal)
AC AB
Also, = (Corresponding sides are proportional)
FG FE
AC 2 AB  D is mid − po int of AB 
⇒ =  
FG 2 FH  H is mid − po int of FE 
AC FG
⇒ = …(ii)
AD FH

Now, in triangles ADC and FHG, we have

AC FG
∠ A = ∠ F and = (By (i) and (ii))
AD FH
⇒ ∆ ADC~ ∆ FHG (SAS similarity)
(ii) ∆ ADC ~ ∆ FHG
CD AD
⇒ = (Corresponding sides proportional)
GH FH
CD 2 × AD CD AB
⇒ = ⇒ =
GH 2 × FH GH FE
 PRAYAS Education

Ex,10 ABC is a right triangle, right angled at B. If BD is the length of perpendicular drawn from B to AC. Prove that :
(i) ∆ ADB ~ ∆ ABC and hence AB2 = AD × AC (ii) ∆ BDC ~ ∆ ABC and hence BC2 = D × AC
1 1 1
(iii) ∆ ADB ~ ∆ BDC and hence BD2 = AD × DC (iv) 2
+ 2
+
AB BC BD 2
Sol. Given : ABC is right angled triangle at B and BD ⊥ AC
To prove :
(i) ∆ ADB ~ ∆ ABC and hence AB2 = AD × AC
(ii) ∆ BDC ~ ∆ ABC and hence BC2 = CD × AC
(iii) ∆ ADB ~ ∆ BDC and hence BD2 = AD × DC
1 1 1
(iv) 2
+ 2
+
AB BC BD 2
Proof : (i) In two triangles ADB and ABC, we have :
∠ BAD = ∠ BAC (Common)
∠ ADB = ∠ ABC (Each is right angle)
∠ ABD = ∠ ACB (Third angle)
∠ ADB = ∠ ABC (AAA Similarity)
Triangle ADB and ABC are similar and so their corresponding sides must be proportion.
AD DB AB AD AB
= = ⇒ = ⇒ AB × AB = AC × AD ⇒ AB 2 = AD × AC This proves
AB DC AC AB AC
(a).
(ii) Again consider two triangles BDC and ABC, we have
∠ BCD = ∠ ACB (Common)
∠ BDC = ∠ ABC (Each is right angle)
∠ DBC = ∠ BAC (Third angle
∴ Triangle are similar and their corresponding sides must be proportional.
BD DC BC
i.e., ∠ BAD = ∠ BAC = =
AB BC AC
(iii) In two triangle ADB and BDC, we have :
DC BC
⇒ = ⇒ BC × BC = DC × AC ⇒ BC 2 = CD × AC This proves (ii)
BC AC
∠ BDA = ∠ BDC = 900
∠ 3 = ∠ 2 = 900 ∠ 1 [ ∠ 1 + ∠ 2 = 900, ∠ 1 + ∠ 3 = 900]
∠ 1 = ∠ 4 = 90 ∠ 2
0
[ ∠ 1 + ∠ 2 = 900, ∠ 2 + ∠ 4 = 900]
∆ ADB ~ ∆ BDC (AAA criterion of similarity)
⇒ Their corresponding sides must be proportional.
AD DB AB AD DB
= = ⇒ = ⇒ BD × BD = AD × AC
BD DC BC BD DC
∴ BD is the mean proportional of AD and DC
(iv) From (i), we have : AB2 = AD × AC
(ii), we have : BC2 = CD × AC
(iii), we have : BD2 = ADDC
1 1 1 1 1  1 1 
Consider + = + + +
AB 2
BC 2
AD × AC CD × AC AC  AD DC 

1 1 1  DC AD  1  AD + DC  1  AC 
+ = + = =
AB 2
BC 2
AC  AD DC  AC  AD × DC  AC  AD × DC 
   
1 1
= = (from (iii))
AD × DC BD 2
1 1 1
2
+ 2
=
AB BC BD 2
 PRAYAS Education

Thus we have proved the following :

If a perpendicular is drawn from the vertex containing the right angle of a right triangle to the hypotenuse
then:
(a) Thu triangle on each side of the perpendicular are similar to each other and also similar to the
original triangle.
i.e., ∆ ADB ~ ∆ BDC, ∆ ADB ~ ∆ ABC, ∆ BDC ~ ∆ ABC
(b) The square of the perpendicular is equal to the product of the length of two parts into which the
hypotenuse is divided by the perpendicular i.e., BD2 = AD × 0DC.

 RESULTS ON AREA OF SIMILAR TRIANGLES

Theorem-3 : The areas of two similar triangles are proportional to the squares on their corresponding
sides.
Given : ∆ ABC ~ ∆ DEF

Area of ∆ABC AB 2 BC 2 AC 2
To prove : = = =
Area of ∆DEF DE 2 EF 2 DF 2

Construction : Draw AL ⊥ BC and DM ⊥ EF.

Proof:

STATEMENT REASON
1
× BC × AL
Area ∆ABC 2
1. = 1
Area ∆DEF 1 × EF × DM Area of ∆ = × Base × Height
2
2
Area ∆ABC BC AL
⇒ = ×
Area ∆DEF EF DM
2. In ∆ ALB and ∆ DME, we have
(i) ∠ ALB = ∠ DME
Each equal to 900
(ii) ∠ ABL = ∠ DEM
∴ ∆ ALB ~ ∆ DME ∆ ABC ~ ∆ DEF ⇒ ∠ B = ∠ E
AA=axiom
AL AB
⇒ =
DM DE Corresponding sides of similar ∆ s are proportional.
3. ∆ ABC ~ ∆ DEF
AB BC AC Given.
⇒ = =
DE EF DF Corresponding sides of similar ∆ s are proportional.
AL BC
= From 2 and 3.
4. DM EF

AL BC
5. Substituting= in 1, we get :
DM EF
Area ∆ABC BC 2
=
Area ∆DEF EF 2

6. Combining 3 and 5, we get :

Area ∆ABC AB 2 BC 2 AC 2
= = =
Area ∆DEF DE 2 EF 2 DF 2
 PRAYAS Education

Conrollary-1 : The areas of two similar triangles are proportional to the squares on their corresponding altitude.

Given : ∆ ABC ~ ∆ DEF, AL ⊥ BC and DM ⊥ EF.

Area of ∆ABC AL2

To prove : =
Area of ∆DEF DM 2
Proof :

STATEMENT REASON
1
× BC × AL
Area ∆ABC 1
1. = 2 Area of ∆ = × Base × Height
Area ∆DEF 1 × EF × DM 2
2
Area ∆ABC BC AL
⇒ = =
Area ∆DEF EF DM
2.
In ∆ ALB and ∆ DME, we have Each equal to 900
(i) ∠ ALB = ∠ DME ∆ ABC ~ ∆ DEF ⇒ ∠ B = ∠ E
(ii) ∠ ABL = ∠ DEM AA=axiom
⇒ ∆ ALB ~ ∆ DME
AL AB Corresponding sides of similar ∆ s are proportional.
⇒ =
DM DE
3. Given.
∆ ABC ~ ∆ DEF
Corresponding sides of similar ∆ s are proportional.
AB BC AC
⇒ = =
DE EF DF
4. From 2 and 3.
BC AL
=
EF DM
5.
BC AL
Substituting= in 1, we get :
EF DM
Area ∆ABC AL2
=
Area ∆DEF DM 2

Hence, proved

Corollary-2 : The areas of two similar triangles proportional to the squares on their corresponding medians.

Given : ∆ ABC ~ ∆ DEF and AP, PQ are their medians.

Area of ∆ABC AP 2
To prove : =
Area of ∆DEF DQ 2
Proof :
 PRAYAS Education

STATEMENT REASON
1. ∆ ABC ~ ∆ DEF Given
Area ∆ABC AB 2
⇒ = ……I. Area of two similar ∆ s are proportional to the squares on
Area ∆DEF DE 2 their corresponding sides.
∆ ABC ~ ∆ DEF
2. AB BC 2 BP BP
⇒ = = = ……II.
DE EF 2 EQ EQ Corresponding sides of similar ∆ s are proportional
AB BP
= and ∠ A = ∠ D
3. DE EQ From II and the fact the ∆ ABC ~ ∆ DEF
⇒ ∆ APB ~ ∆ DQE
BP AP By SAS-similarity axiom
⇒ = …….III
EQ DQ
AB AP
⇒ =
DE DQ From II and III.
AB 2 AP 2
⇒ = ……..IV
DE 2 DQ 2
Area ∆ABC AP 2
⇒ =
Area ∆DEF DQ 2 From I and IV.
4.
Hence, proved
Corollary-3 : The areas of two similar triangles proportional to the squares on their corresponding angle bisector
segments.
Given : ∆ ABC ~ ∆ DEF and AX, DY are their
Area of ∆ABC AX 2
To prove : =
Area of ∆DEF DY 2
Proof :
STATEMENT REASON
Area ∆ABC AB 2
1. = Area of two similar ∆ s are proportional to the squares on
Area ∆DEF DE 2 their corresponding sides.

2. ∆ ABC ~ ∆ DEF Given

⇒ ∠A = ∠D
1 1
⇒ ∠A = ∠D 1 1
2 2 ⇒ ∠ BAX = ∠ A and ∠ EDY = ∠ D
⇒ ∠ BAX = ∠ EDY 2 2

3. In ∆ ABX and ∆ EDY, we have Given

∠ BAX = ∠ EDY From 2.
∠B = ∠E ∆ ABC ~ ∆ DEF
By AA similarity axiom
∴ ∆ ABX ~ ∆ DEY
AB AX AB 2 AX 2
⇒ = ⇒ =
DE DY DE 2 DY 2
Area ∆ABC AX 2
4. =
Area ∆DEF DY 2 From 1 and 3.
 PRAYAS Education

Ex.11 It is given that ∆ ABC ~ ∆ PQR, area ( ∆ ABC) = 36 cm2 and area ( ∆ PQR) = 25 cm2. If QR = 6 cm, find length
of BC.
P
A

B C Q R
Sol. We know that the areas of similar triangles are proportional to the squares of their corresponding sides.
Area of (∆ABC ) BC 2
∴ =
Area of (∆PQR) QR 2
Let BC = x cm. Then.
36 x 2 36 x 2 36 × 36  6 × 6  36
= 2 ⇔ = ⇔ x2 = ⇔ x= = = 7.2
25 6 25 36 25  5  5
Hence BC = 7.2 cn

Ex.12 P and Q are points on the sides AB and AC respectively of ∆ ABC such that PQ║BC and divides ∆ ABC into
parts, equal in area. Find PB : AB.

Sol. Area ( ∆ APQ) = Area (trap. PBCQ) [Given]

⇒ Area ( ∆ APQ) = [Area ( ∆ ABC) – Area ( ∆ APQ)
⇒ 2 Area ( ∆ APQ) = Area ( ∆ ABC)
Area of (∆APQ) 1
⇒ = …(i)
Area of (∆ABC ) 2
Now, in ∆ APQ and ∆ ABC, we have
∠ PAQ = ∠ BAC [Common ∠ A]
∠ APQ = ∠ ABC [PQ║BC, corresponding ∠ s are equal]
∴ ∆ APQ ~ ∆ ABC.

We known that the areas of similar ∆ s are proportional to the squares of their corresponding sides.
Area of (∆APQ) AP 2 AP 2 1
∴ = ⇒ = [Using (i)]
Area of (∆ABC ) AB 2 AB 2 2
AP 1
⇒ = i.e., AB = 2 . AP
AB 2
⇒ AB = 2 (AB – PB) ⇒ 2 PB = ( 2 – 1) AB
PB ( 2 − 1)
⇒ = .
AB 2
∴ PB : AB = ( 2 – 1) : 2
 PRAYAS Education

Ex.13 Two isosceles triangles have equal vertical angles and there areas are in the ratio 16 : 25. Find the ratio of their
corresponding heights.
Sol. Let ∆ ABC and ∆ DEF be the given triangles in which AB = AC, DE = DF, ∠ A = ∠ D and
Area of (∆ABC ) 16
= Draw AL ⊥ BC and DM ⊥ EF
Area of (∆DEF ) 25
AB DE
Now, = 1 and =1 [ AB = AC and DE = DF]
AC DF
AB AC
⇒ = .
DE DF
∴ In ∆ ABC and ∆ DEF, we have
AB AC
= and ∠ A = ∠ D
DE DF
⇒ ∆ ABC ~ ∆ DEF [By SAS similarity axiom]
But, the ratio of the areas of two similar ∆ s is the same as the ratio of the square of their corresponding heights.
Area of (∆ABC ) AL2 16  AL2  AL 4
= ⇒ =  
2 
⇒ =
Area of (∆DEF ) DM 2
25  DM  DM 5
∴ AL : DM = 4 : 5, i.e., the ratio of their corresponding heights = 4 : 5,
Ex.14 If the areas of two similar triangles are equal, prove that they are congruent.
Sol. Let ∆ ABC ~ ∆ DEF and area ( ∆ ABC) = area ( ∆ DEF).
Since the ratio of the areas of two similar ∆ s is equal to the ratio of the squares on their corresponding sides, we
have
Area of (∆ABC ) AB 2 AC 2 BC 2
= = =
Area of (∆DEF ) DE 2 DF 2 EF 2
AB 2 AC 2 BC 2
⇒ = = =1 [ Area ( ∆ ABC) = Area ( ∆ DEF)]
DE2 2 DF 2
2 EF 2
2
⇒ AB = DE , AC = DF and BC = EF2
2 2

⇒ AB = DE, AC = DF and BC = EF
∴ ∆ ABC ≅ ∆ DEF [By SSS congruence]
Ex.15 In fig, the line segment XY is parallel to side AC of ∆ ABC and it divides the triangle into two parts of equal
AX
areas. Find the ratio . [NCERT]
AB
Sol, We are given that XY║AC.
⇒ ∠ 1 = ∠ 3 and ∠ 2 = ∠ 4 [Corresponding angles]
⇒ ∆ BXY~ ∆ BAC [AA similarity]
ar (∆BXY ) ( BY ) 2
⇒ = [By theorem] …(i)
ar (∆BAC ) ( BA) 2
Also, we are given that
1 ar (∆BXY ) 1
ar ( ∆ BXY) = × ar (∆BAC ) ⇒ = …(ii)
2 ar (∆BAC ) 2
2
 BX  1 BX 1
From (i) and (ii), we have   = ⇒ = …(iii)
 BA  2 BA 2
AX AB − BX BX BX 1
Now, = =1− =1− =1− [By (iii)]
AB AB AB BA 2
2 −1 2 − 2
= =
2 2
AX 2 − 2
Hence, = =
AB 2
 PRAYAS Education

Theorem-4 [Pythagoras Theorem] : In a right angled triangle, the square on the hypotenuse is equal to the sum of
the squares on the other two sides.
Given : A ∆ ABC in which ∠ B = 900.
To prove : AC2 = BA2 + BC2.
Construction : From B, Draw BD ⊥ AC
Proof :

STEMENT REASON
1. In ∆ ADB and ∆ ABC, we have :
∠ BAD = ∠ CAB = ∠ A Common
∠ ADB = ∠ ABC Each = 900
∴ ∆ ADB ~ ∆ ABC By AA axiom of similarity
AD AB
⇒ = Corr. sides of similar ∆ s are proportional
AB AC
⇒ AB 2 = AD × AC ..(i)
2. In ∆ CDB and ∆ CBA, we have : ∆ CDB = ∆ CBA Each = 900
∠ BCD = ∠ ACB
∴ ∆ CDB ~ ∆ CBA Common
By AA axiom of similarity
DC BC
⇒ =
BC AC Corr. sides of similar ∆ s are proportional
⇒ BC 2 = DC × AC ..(ii)
Adding (i) and (ii), we get
3.
AB 2 + BC 2 = AD × AC + DC × AC
 AD + DC = AC
= (AD + DC) × AC = AC 2
Hence, AB2 + BC2 = AC2
Theorem-5 [Converse of pythagoras Theorem] : In a triangle if the square of one sides is equal to the sum of the
squares of the squares of the other two sides, then the triangle is right angled.
Given : A ∆ ABC in which AB2 + BC2 = AC2
To prove : ∠ B = 900.
Construction : Draw a ∆ DEF in which
CE = AB, EF = BC and ∠ E = 900

Proof :
STATEMENT REASON
1. In ∆ DEF, we have : ∠ = 900
∴ DE 2 + EF 2 = DF 2 By Phthagoras Theorem
⇒ AB 2 + BC 2 = DF 2  DE = AB and EF = BC
 AB2 + BC2 = AC2 (Given)
⇒ AC 2 = DF 2

2. In ∆ ABC and ∆ DEF, we have : By construction

AB = DE By construction
BC = DF Proved above
∴ ∆ ABC ≅ ∆ DEF By SSS congruence
⇒ ∠B = ∠E c.p.c.t
⇒ ∠ E = 900  ∠ E = 900

Hence, ∠ B = 900
 PRAYAS Education

3
Ex.16 If ABC is an equilateral triangle of side a, prove that its altitude = a.
2
Sol. ∆ ABC is an equilateral triangle.
We are given that AB = BC = CA = a. AD is the altitude, i.e., AD ⊥ BC.
Now, in right angled triangles triangles ABD and ACD, we have
AB = AC [Given]

and AD = AD [Common side]

⇒ ∆ ABD = ∆ ACD [By RHS congruence]

1 a
⇒ BD = CD ⇒ BD = DC = BC =
2 2
From right triangle ABD,
2
a
2 2
AB = AD + BD 2
⇒ a – AD +  
2 2

2
a2 3 2
⇒ AD2 = a 2 − = a
4 4
3
⇒ AD = a.
2
Ex.17 In a ∆ ABC, obtuse angled at B, if AD is perpendicular to CB produced, prove that :
AC2 = AB2 + BC2 + 2BC × BD

Sol. In ∆ ADB, ∠ D = 900.

∴ AD2 + DB2 = AB2 …(i) [By Pythagoras Theorem]

In ∆ ADC, ∠ D = 900.

∴ AC2 = AD2 + DC2 [By Pythagoras Theorem]

= AD2 + (DB + BC)2

= AD2 + DB2 + BC2 + 2DB × BC
= AB2 + BC2 + 2BC × BD [Using (i)]
Hence, AC2 = AB2 + BC2 + 2BC × BD.

Ex.18 In the given figure, ∠ B = 900. D and E are any points on AB and BC respectively. Prove that :
AE2 + CD2 = AC2 + DE2.

Sol. In ∆ ABE, ∠ B = 900

∴ AE2 = AB2 + BE2 …(i)
In ∆ DBC, ∠ B = 900.
∴ CD2 = BD2 + BC2 …(ii)
Adding (i) and (ii), we get :
AE2 + CD2 = (AB2 + BC2) + (BE2 + BD2)
= AC2 + DE2 [By Pythagoras Theorem]
2 2 2 2
Hence, AE + CD = AC + DE .
 PRAYAS Education

Ex.19 A point O in the interior of a rectangle ABCD is joined with each of the vertices A, B, C and D.
Prove that : OA2 + OC2 = OB2 + OD2

Sol. Through O, draw EOF║AB. Then, ABFE is a rectangle.

In right triangles OEA and OFC, we have :
OA2 = OE2 + AE2
OC2 = OF2 + OF2
∴ OA2 + OC2 = OE2 + OF2 + AE2 + CF2
Again, in right triangles OFB and OED, we have :
OB2 = OF2 + BF2
OD2 = OE2 + DE2
∴ OB2 + OD2 = OF2 + OE2 + BF2 + DE2
= OE2 + OF2 + AE2 + CF2 …(i) [ BF = AE & DE = CF]
From (i) and (ii), we get
OA2 + OC2 = OB2 + OD2.
Ex.20 In the given figure, ∆ ABC is right-angled at C.
Let BC = a, CA = b, AB = c and CD = p, where CD ⊥ AB.
1 1 1
Prove that : (i) cp = ab (ii) 2
= 2 + 2
p a b
1 1
Sol. (i) Area of ∆ ABC = BC × CD cp.
2 2
1 1
Also, area of ∆ ABC = BC × AC ab.
2 2
1 1
∴ cp = ab. ⇒ cp = ab
2 2
ab
(ii) cp = ab ⇒ p =
c
2 2
a b
⇒ p2 = 2
c
1 c2 a2 + b2
⇒ = = [  c 2 = a 2 + b2 ]
p 2 a 2b 2 a 2b 2
1 1 1
⇒ 2
= 2 = 2
p a b
Ex.21 Prove that in any triangle, the sum of the square of any two sides is equal to twice the square of half of the third
side together with twice the square of the median which bisects the third side. (Appollonius Theorem)
Sol. Given : A ∆ ABC in which AD is a median.
2
1 
To prove : AB + AC = 2AD + 2  BC  or AB2 + AC2 = 2(AD2 + BD2)
2 2 2

2 
Construction : Draw AE ⊥ BC.
Proof :  AD id median
∴ BD = DC
Now, AB2 + AC2 = (AE2 + BE2) + (AE2 + CE2) = 2AE2 + BE2 + CE2
= 2[AD2 – DE2] + BE2 + CE2
= 2AD2 – 2DE2 + (BD + DE)2 + (DC – DE)2
= 2AD2 – 2DE2 + (BD + DE)2 + (DC – DE)2
2
1 
= 2(AD + BD ) = = 2AD + 2  BC 
2 2 2

2 
Hence, Proved.
 PRAYAS Education

 SYNOPSIS

 SIMILAR TRIANGLES. Two triangles are said to be similar if

(i) Their corresponding angles are equal and (ii) Their corresponding sides are proportional.

 All congruent triangles are similar but the similar triangles need not be congruent.

 Two polygons of the same numbers of sides are similar, if

(i) their corresponding angels are equal and
(ii) their corresponding sides are in the same ratio.

 BASIC PROPORTIONALITY THEOREM. In a triangle, a line drawn parallel to one side, to intersect the other
sides in distinct points, divides the two sides in the third side.

 CONVERSE OF BASIC PROPORTIONALITY THEOREM. If a line divides any two sides of a triangle in
the same ratio, the line must be parallel to the third side.

 AAA-SIMILARITY. If in two triangles, corresponding angles are equal, i.e., the two corresponding angles are
equal, then the triangles are similar.

 SSS-SIMILARITY. If the corresponding sides of two triangles are proportional, then they are similar.

 SSS-SIMILARITY. If in triangles one pair of corresponding sides proportional and the included angles are equal
then the two triangles are similar.

 The ratio of the areas of similar triangles is equal to the ratio of the squares of their to the sum of the squares

 PYTHAGORAS THEOREM. In a right triangle, if the square of one side is equal to the sum of the squares of
the other two sides.

 CONVERSE OF PYTHAGORAS THEOREM. In a triangle, if the square of one side is equal to the sum of the
squares of the other two sides then the angle opposite to the first side is a right angle.
 PRAYAS Education

EXERCISE – 1 (FOR SCHOOL/BOARD EXAMS)

OBJECTIVE TYPE QUESTIONS
CHOOSE THE CORRECT ONE

1. Triangle ABC is such that AB = 3 cm, BC = 2 cm and CA = 2.5 cm. Triangle DEF is similar to ∆ ABC. If EP 4
cm, then the perimeter of ∆ DEF is :
(A) 7.5 cm (B) 15 cm (C) 22.5 cm (D) 30 cm
2. In ∆ ABC, AB = 3 cm, AC = 4 cm and AD is the bisector of ∠ A. Then, BD : DC is :
(A) 9 : 16 (B) 16 : 9 (C) 3 : 4 (D) 4 : 3
3. In a equilateral triangle ABC, if AD ⊥ BC, then :
(A) 2AB2 = 3AD2 (B) 4AB2 = 3AD2 (C) 3AB2 = 4AD2 (D) 3AB2 = 2AD2
4. ABC is a triangles and DE is drawn parallel to BC cutting the other sides at D and E. If AB = 3.6 cm, AC = 2.4 cm
and AD = 2.1 cm, then AE is equal to :
(A) 1.4 cm (B) 1.8 cm (C) 1.2 cm (D) 1.05 cm
5. The line segments joining the mid points of the sides of a triangle from four triangles each of which is :
(A) similar to the original triangle (B) congruent to the original triangle.
(C) an equilateral triangle (D) an isosceles triangle.
6. In ∆ ABC and ∆ DEF, ∠A = 50 , ∠B = 70 , ∠C = 60 0 , ∠D = 60 0 , ∠E = 70 0 , ∠F = 50 0 , then ∆ ABC is similar
0 0

to :
(A) ∆ DEF (B) ∆ EDF (C) ∆ DFE (D) ∆ FED
7. D, E, F are the mid points of the sides BC, CA and AB respectively of ∆ ABC. Then ∆ DEF is congruent to
triangle
(A) ABC (B) AEF (C) BFD, CDE (D) AFE, BFD, CDE
8. If in the triangles ABC and DEF, angle A is equal to angle E, both are equal to 400, AB : ED = AC : EF and angle
F is 650, then angel B is :-
(A) 350 (B) 650 (C) 750 (D) 850
9. In a right angled ∆ ABC, right angled at A, if AD ⊥ BC such that AD = p, if BC = a, CA = b and AB = c, then :
1 1 1
(A) p2 = b2 + c2 (B) 2
= 2+ 2
p b c
p p
(C) = (D) p2 = b2 c2
a b
AX
10. In the adjoining figure, XY is parallel to AC. If XY divides the triangle into equal parts, then the value of =
AB
1 1
(A) (B)
2 2
2 +1 2 −1
(C) (D)
2 2
11. The ratio of the corresponding sides of two similar triangles is 1 : 3. The ratio of their corresponding heights is :
(A) 1 : 3 (B) 3 : 1 (C) 1 : 9 (D) 9 : 1
12. The areas of two similar triangles are 49 cm2 and 64 cm2 respectively. The ratio of their corresponding sides is :
(A) 49 : 64 (B) 7 : 8 (C) 64 : 49 (D) None of these
13. The areas of two similar triangles are 12 cm2 and 48 cm2. If the height of the similar one is 2.1 cm, then the
corresponding height of the bigger one is :
(A) 4.41 cm (B) 8.4 cm (C) 4.2 cm (D) 0.525 cm
 PRAYAS Education

14. In the adjoining figure, ABC and DBC are two triangles on the same base BC, AL ⊥ BC and DM ⊥ BC. Then,
area (∆ABC )
is equal to ;
area (∆DBC )
AO AO 2
(A) (B)
OD OD 2
AO OD 2
(C) (D)
AD AO 2
15. In the adjoining figure, AD : DC = 2 : 3, then ∠ ABC is equal to :

(A) 300 (B) 400 (C) 450 (D) 1100

16. In ∆ ABC, D and E are points on AB and AC respectively such that DE ║ BC. If AE = 2 cm, EC = 3 cm and BC
= 10 cm, then DE is equal to ;
20
(A) 5 cm (B) 4 cm (C) 15 cm (D) cm
3
17. In the given figure, ∠ ABC = 900 and BM is a median, AB = 8 cm and BC = 6 cm. Then, length BM is equal to :

(A) 3 cm (B) 4 cm (C) 5 cm (D) 7 cm

18. If D, E, F are respectively the mid points of the sides BC, CA and AB of ∆ ABC and the area of ∆ ABC is 24 sq.
cm, then the area of ∆ DFE is :-
(A) 24 cm2 (B) 12 cm2 (C) 8 cm2 (D) 6 cm2
19. In a right angled triangle, if the square of the hypotenuse is twice the product of the other two sides, then one of
the angles of the triangle is :-
(A) 150 (B) 300 (C) 450 (D) 600
20. Consider the following statements :
1. If three sides of a triangles are equal to three sides of another triangle, then the triangles are congruent.
2. If three angles of a triangles are respectively equal to three angles of another triangle, then the two
triangles are congruent.
Of these statements,
(A) 1 is correct and 2 is false (B) both 1 and 2 are false
(C) both 1 and 2 are correct (D) 1 is false and 2 is correct

(OBJECTIVE) ANSWER KEY EXERCISE

Que. 1 2 3 4 5 6 7 8 9 10
Ans. B C C A A D D C B B
Que. 11 12 13 14 15 16 17 18 19 20
Ans. A B C A B B C D C A
 PRAYAS Education

EXERCISE – 2 (FOR SCHOOL/BOARD EXAMS)

OBJECTIVE TYPE QUESTIONS
VERY SHORT ANSWER TYPE QUESTIONS
1. In the given figure, XY║BC.
Given that AX = 3 cm, XB = 1.5 cm and BC = 6 cm.
Calculate :
AY
(i) (ii) XY
YC
2. D and E are points on the sides AB and AC respectively of ∆ ABC. For each of the following cases, state whether
DE║BC :
(i) AD = 5.7 cm, BD = 9.5 cm, AE = 3.6 cm, and EC = 6 cm
(ii) AB = 5.6 cm, AD = 1.4 cm, AC = 9.6 cm, and EC = 2.4 cm.
(iii) AB = 11.7 cm, BD = 5.2 cm, AE = 4.4 cm, and AC = 9.9 cm.
(iv) AB = 10.8 cm, BD = 4.5 cm, AC = 4.8 cm, and AE = 2.8 cm.
3. In ∆ ABC, AD is the bisector of ∠ A. If BC = 10 cm, BD = 6 cm and AC = 6 cm, find AB.

4. AB and CD are two vertical poles height 6 m and 11 m respectively. If the distance between their feet is 12 m, find
the distance between their tops.

5. ∆ ABC and ∆ PQR are similar triangles such that are ( ∆ ABC) = 49 cm2 and area ( ∆ PQR) = 25 cm2. If AB = 5.6
cm, find the length of PQ.
6. ∆ ABC and ∆ PQR are similar triangles such that are ( ∆ ABC) = 28 cm2 and area ( ∆ PQR) = 63 cm2. If PR = 8.4
cm, find the length of AC.
7. ∆ ABC ~ ∆ DEF. If BC = 4 cm, EF = 5 cm and area ( ∆ ABC) = 32 cm2, determine the area of ∆ DEF.
8. The areas of two similar triangles are 48 cm2 and 75 cm2 respectively. If the altitude of the first triangle be 3.6 cm,
find the corresponding altitude of the other.
9. A rectangular field is 40 m long and 30 m broad. Find the length of its diagonal.
10. A man goes 15 m due west and then 8 m due north. How far is he from the starting point?
11. A ladder 17 m long reaches the window of a building 15 m above the ground. Find the distance of the foot of the
ladder from the building.
 PRAYAS Education

SHORT ANSWER TYPE QUESTIONS

1. In the given fig, DE║BC. E
(i) If AD = 3.6 cm, AB = 9 cm and AE = 2.4 cm, find EC. D
AD 3
A
(ii) If = and AC = 5.6 cm, find AE. B C
DB 5
(iii) If AD = x cm, DB = (x – 2) cm, AE = (x + 2) cm and EC = (x – 1) cm, find the value of x.
2. In the given figure, BA║DC. Show that ∆ OAB ~ ∆ ODC. If AB = 4 cm, CD = 3 cm, OC = 5.7 cm and OD = 3.6
cm, find OA and OB.

3. In the given figure, ∠ ABC = 900 and BD ⊥ AC. If AB = 5.7 cm, BD = 3.8 cm and CD = 5.4 cm, find BC.

4. In the given figure, ∆ ABC ~ ∆ PQR and AM, PN are altitude, whereas AX and PY are medians. Prove that
AM AX
=
PN PY

5. In the given figure, BC║DE, area ( ∆ ABC) = 25 cm2, area (trap. BCED) = 24 cm2 and DE = 1 4 cm. Calculate the
length of BC.

6. In ∆ ABC, ∠ C = 900. If BC = a, AC = b and AB = c, find :

(i) c when a = 8 cm and b = 6 cm.
(ii) a when c = 25 cm and b = 7 cm.
(iii) b when c = 13 cm and a = 5 cm.
7. The sides of a right triangle containing the right angle are (5x) cm and (3x – 1) cm. If the area of triangle be 60
cm2, calculate the length of the sides of the triangle.
8. Find the altitude of an equilateral triangle of side 5 3 cm.
9. In the adjoining figure (not drawn to scale), PS – 4 cm, SR = 2 cm, PT = 3 cm and QT = 5 cm.
(i) Show that ∆ PQR ~ ∆ PST. (ii) Calculate ST, if QR = 5.8 cm.
 PRAYAS Education

10. In the given figure, AB║PQ and AC║PR. Prove that BC║QR.

11. In the given figure, AB and DE are perpendicular to BC. If AB = 9 cm, DE = 3 cm and AC = 24 cm, calculate AD.

12. In the given figure, DE║BC. If DE = 4 cm, BC = 6 cm and area ( ∆ ADE) = 20 cm2, find the area of ∆ ABC.

13. A ladder 15 m long reaches a window which is 9 m above the ground on one side of the street. Keeping its foot at
the same point, the ladder is turned is turned to the other side of the street to reach a window 12 m high. Find the
width of the street.

14. In the given figure, ABCD is a quadrilateral in which BC = 3 cm, AD = 13 cm, DC = 12 cm and ∠ ABD
= ∠ BCD = 900. Calculate the length of AB.

15. In the given figure, ∠ PSR = 900, PQ = 10 cm, QS = 6 cm and RQ = 9 cm, calculate the length of PR.

16. In a rhombus PQRS, side PQ = 17 cm and diagonal PR = 16 cm. Calculate the area of the rhombus.
17. From the given figure, find the area of trapezium ABCD.

18. In a rhombus ABCD, prove that AC2 + BD2 = 4AB2.

 PRAYAS Education

19. A ladder 13 m long rests against a vertical wall. If the foot of the ladder is 5 m from the foot of the wall, find the
distance of the other end of the ladder from the ground.
LONG ANSWER TYPE QUESTIONS
1. In the given figure, it is given that ∠ ABD = ∠ CDB = ∠ PQB = 900. If AB = x units, CD = y units and PQ = z
1 1 1
units, prove that + =
x y z

2. In the adjoining figures, ABCD is a parallelogram, P is a point on side BC and DP when produced meets AB
produced at L. Prove that : (i) DP : PL = DC : BL (ii) DL : DP = AL : DC.

3. In the given figure, ABCD is a parallelogram, E is a point on BC and the diagonal BD intersects AE at F. Prove
that DF × FE = FB × FA.

4. In the adjoining figure, ABCD is a parallelogram in which AB = 16 cm BC = 10 cm and L is a point on AC such

that CL : LA = 2 : 3. If BL produced meets CD at M and AD produced at N, prove that :
(i) ∆ CLB ~ ∆ ALN (ii) ∆ CLM ~ ∆ ALB

5. In the given figure, medians AD and BE of ∆ ABC meet at G and DF║BE. Prove that
(i) EF = FC (ii) AG : GD = 2 : 1.

6. In the given figure, the medians BE and CF of ∆ ABC meet at G. Prove that :
(i) ∆ GEF ~ ∆ GBC and therefore, BG = 2 GE. (ii) AB × AF = AE × AC.
 PRAYAS Education

7. In the given figure, DE║BC and BD – DC.

(i) Prove that DE bisects ∠ ADC.
(ii) If AD = 4.5 cm, AE = 3.9 cm and DC = 7.5 cm, find CE.
(iii) Find the ratio AD : DB.
8. O is point inside a ∆ ABC. The bisectors of ∠ AOB. ∠ BOC and ∠ COA meet the sides AB, BC and in points D,
E and F respectively. Prove that AD-BE-CF = DB.EC.FA

9. In the figure, DE║BC.

(i) Prove that ∆ ADE and ∆ ABC are similar.

1
(ii) Given that AD = BD. Calculate DE, if BC = 4.5 cm.
2
10. In the adjoining figure, ABCD is a trapezium in which AB║DC and AB = 2 DC. Determine the ratio of areas of
∆ AOB and ∆ COD

11. In the adjoining figure, LM is parallel to BC. AB = 6 cm, AL = 2 cm and AC = 9 cm. Calculate :
(i) the length of CM.

Area (∆ALM )
(ii) the value of
Area (trap.LBCM )

12. In the given figure, DE║BC. and DE : BC = 3 : 5. Calculate the ratio of the areas of ∆ ADE and the trapezium
BCED.

13. In ∆ ABC, D and E are mid-points of AB and AC respectively. Find the ratio of the areas of ∆ ADE and ∆ ABC.

14. In a ∆ PQR, L and M are two points on the base QR, such that ∠ LPQ = ∠ QRP and ∠ RPM = ∠ RQP. Prove that
(i) ∆ PQL – ∆ RPM (ii) QL.RM = PL.PM (iii) PQ2 = QL.QR
 PRAYAS Education

15. In the adjoining figures, the medians BD and CE of a ∆ ABC meet at G.

Prove that:
(i) ∆ EGD ~ ∆ CGB
(ii) BG = 2 GD from (i) above.

16. In the adjoining figure, PQRS is a rarallelog’am with PQ = 15 cm and RQ = 10 cm. L is a point on RP such that
RL : LP = 2 : 3. QL produced meets RS at M and PS produced at N. Find the length of PN and RM.

17. In ∆ PQR, LM║QR and PM : MR = 3 : 4. Calculate:

PL LM
(i) and then :
PQ QR
Area (∆ALM )
(ii)
Area (∆MNR)
Area (∆LQM )
(iii)
Area (∆LQN )

18. In ∆ ABC, ∠ B = 900 and D is the mid point of BC.

Prove that :
(i) AC2 = AD2 + 3CD2
(ii) BC2 = 4(AD2 – AB2)

19. In ∆ ABC, if AB = AC and D is a point on BC. Prove that BC2 – AD2 = BD × CD.

SIMILAR TRIANGLE ANSWER KEY EXERCISE (X)-CBSE

VELY SHORT ANSWER TYPE QUESTIONS
1
1. (i) (ii) 4 cm 2. (i) Yes, (ii) No, (iii) No, (iv) Yes 3. 9 cm 4. 13 m 5. PQ = 4 cm 6. AC = 5.6 cm
22
7. 50 cm 8. 4.5 cm 9. 50 m 10. 17 m 11. 8m
SHORT ANSWER TYPE QUESTIONS
1. (i) 3, 6 cm, (ii) 2.1 cm, (iii) x = 4 2. OA = 4.8 cm, OB = 7.6 cm 3. 8.1 cm 5. 10 cm 6. (i) 10 cm, (ii) 24 cm,
(iii) 12 cm 7. 15 cm,8 cm, 17cm 8. 7.5 cm 9. 2.9 cm 11. 16 cm 12. 45 cm2 13. 21m 14. 4 cm 15. 17 cm
16. 240 cm2 17. 14 cm2 19. 12 m
LONG ANSWER TYPE QUESTIONS
1
7. (ii) 6.5 cm, (ii) 3 : 8 9. DE = 1.5 cm 10. 4 : 1 11. (i) 6 cm, (ii) 12. 9 : 16 13. 1 : 4
8
PL LM 3
16. PN = 15 cm, RM = 10 cm 17. (i) = = (ii) 3 : 7 (iii) 10 : 7
PQ QR 7
 PRAYAS Education

EXERCISE – 3 (FOR SCHOOL/BOARD EXAMS)

PERVIOUS YEARS BOARD QUESTIONS
VERY SHORT ANSWER TYPE QUESTIONS
1. ∆ ABC and ∆ DEF are similar, BC = 3 cm, EF = 4 cm and area of ∆ ABC = 54 cm2. Determine the area of
∆ DEF. Delhi-1996
2. In ∆ ABC, CE ⊥ AB, BD ⊥ AC and & BD intersect at P, considering triangles BEP and CPD. Prove that
BP × PD = EP × PC. Delhi-1996C
3. A right triangle has hypotenuse of length q cm and one side of length p cm. if (q – p) = 2, express the length of
third side of the right triangle in terms of q. Al-1996C
4. In the given figure, ABC is a triangle in which AB = AC. D and E are points on the sides AB and AC respectively,
such that AD = AE. Show that the points B, C, E and D are concyclic. Al-1996C
5. In a ∆ ABC, AB = AC and D is a point on side AC, such that BC2 = AC × CD. Prove that BD = BC. Al-1997
6. ∆ ABC is right angled at B. On side AC, a point D is taken such that AD = DC and AB = BD. Find the measure of
∠ CAB Delhi-1998
7. In a ∆ ABC, P and Q are points on the sides AB and AC respectively such that PQ is parallel to BC. Prove that
median AD, drawn from A to BC, bisects PQ. Al-1998
8. Two poles of height 7 m and 12 m stand on a plane ground. If the distance between their feet is 12 m, find the
distance between their tips. Al-1998C
9. In a ∆ ABC, D and E are points on AB & AC respectively such that DE is parallel to BC and AD : DB = 2 : 3.
Determine Area ( ∆ ADE) : Area ( ∆ ABC). Foreign-1999
10. In the given figure, ∠ A = ∠ B and D & E are points on AC and BC respectively such that AD = BE, show that
DE║AB. Delhi-1999
11. In figure, ∠ 1 = ∠ 2 and ∠ 3 = ∠ 4. Show that PT. QR = PR. ST. Foreign-2000

1
12. In figure, LM║NQ and LN║PQ. If MP = MN, find the ratio of the areas of ∆ LMN and ∆ QNP.
3
Foreign-2000

13. ABC is an isosceles triangle right angled at B. Two equilateral triangles BDC and AEC are constructed with side
1
BC and AC. Prove that area of ∆ BCD = area of ∆ ACE. Delhi-2001
22 2
14. The areas of two similar triangles are 81 cm and 49 cm respectively. If the altitude of the first triangle 6.3 cm,
find the corresponding altitude of the other. Al-2001
15. L and M are the mid-points of AB and BC respectively of ∆ ABC, right-angled at B. prove that 4LC2 = AB2 +
4BC2. Al-2001;Foreign-2001
 PRAYAS Education

16. The areas of two similar triangles are 121 cm2 and 64 cm2 respectively. If the median of the first triangle is 12.1
cm. find the corresponding median of the other. Al-2001
17. In an equilateral triangle ABC, AD is the altitude drawn from A on side BC. Prove that 3AB2 = 4AD2.
Delhi-2002
18. (i) Prove that the equilateral triangle described on the two sides of a right angled triangle are together equal to
the equilateral triangle on the hypotenuse in terms of their areas. Al-2002
(ii) P is a point in the interior of ∆ ABC, X, Y and Z are point on lines PA.PB and PC respectively such that
XY║AB and XZ║BC. Prove that YZ║BC. Al-2002 : Delhi-2003 [NCERT]
(iii) D and E are points on the sides AB and AC respectively of ∆ ABC such that DE is parallel to BC and
AD : DB = 4 : 5. CD and BE intersect each other at F. Find the ratio of the areas of ∆ DEF and ∆ BCE
Al-2000 : Al-2003
(iv) P. Q are respectively points on sides AB and AC of triangle ABC. If AP = 2 cm. PB = 4 cm. AQ = 3 cm
and QC = 6 cm. prove that BC = 3PQ. Foreign-2003
CA CB
19. D is a point on the side BC of ∆ ABC such that ∠ ADC = ∠ BAC. Prove that = . Delhi-2002:[NCERT]
CD CA
AO BO
20. ABCD is a trapezium in which AB║DC. The diagonals AC and BD intersect at O. Prove that =
OC DO
Al-2004:[NCERT]
BD AD
21. In a ∆ ABC, AD ⊥ BC and = . Prove that ABC is a right triangle, right angled at A. Foreign-2004
AD DC
22. In a right angled triangle ABC, ∠ A = 900 and AD ⊥ BC. Prove that AD2 = BD × CD. Delhi-2004C, 2006
23. In fig., AB║DE and BD║EF. Prove that DC2 = CF × AC. Al-2004C : Delhi-
2007

24. If one diagonal of a trapezium divides the other diagonal in the ratio of 1 : 2. prove that one of the parallel sides is
double the other. Foreign-2005
25. In ∆ ABC, AD ⊥ BC, prove that AB + CD = AC + DB .
2 2 2 2
Delhi-2005C, Al-2006 [NCERT]
26. Prove that the sum of the squares of the sides of a rhombus is equal to sum of the squares of its diagonals.
Al-2005C [NCERT]
27. In figure, S and T trisect the side QR of a right triangle PQR. Prove that 8PT2 = 3PR2 + 5PS2.

OR
If BL and CM are medians of a triangle ABC right-angled at A, then prove that 4(BL2 + CM2) = 5BC2.
Al-2006 C; Foreign-2009
 PRAYAS Education

28. In the fig, P and Q are points on the sides AB and AC respectively of ∆ ABC such that AP = 3.5 cm,
PB = 7 cm, AQ = 3 cm and QC = 6 cm. If PQ = 4.5 cm find BC. Delhi-2008

29. In fig. ∠ M = ∠ N = 460 Express x in terms of a, b and c where a, b and c are lengths of LM, MN and NK
respectively. Delhi-2009

30. In figure, ∆ ABC is a right triangle, right-angled at A and AC ⊥ BD. Prove that AB2 = BC. BD. Al-2009

2
31. In a ∆ ABC, DE║BC. If DE = BC and area of ∆ ABC = 81 cm2, find the area of ∆ ADE. Foregin-2009
3
SHORT ANSWER TYPE QUESTIONS
1. P and Q are points on the sides CA and CB respectively of a ∆ ABC right-angled at C. prove that
AQ2 + BP2 = AB2 + PQ2. Delhi-1996, 2007
2. ABC is a right triangle, right angled at B. AD and CE are the two medians drawn from A and C respectively. If
3 5
AC = 5 cm and AD = cm, find the length of CE. Al -1997
2
3. In ∆ ABC, if AD is the median, show that AB2 + AC2 = 2 [AD2 + BD2]. Delhi-1997, 98
4. In the given figure, M is the mid-point of the side CD of parallelogram ABCD. BM. When joined meets AC is L
and AD produced in E. Prove that EL = 2BL. Al-1998; Delhi-1999, Al-2009

5. ABC is a right triangle, right-angled at C. if p is the length of the perpendicular from C to AB and a, b, c have the
1 1 1
usual meaning, then prove that (i) pc = ab (ii) 2
= 2 = 2 Delhi-1998, 98 C
p a b
6. In an equilateral triangle PQR, the side QR is trisected at S. Prove that 9PS2 = 7PQ2. Al-1998, 98C [NCERT]
7. If the diagonals of a quadrilateral divide each other proportionally, prove that it is trapezium. Foreign-1999
8. In an isosceles triangle ABC with AB = AC, BD is a perpendicular from B to the side AC. Prove that BD2 – CD2 =
2CD. AD. Foreign-1999
 PRAYAS Education

ar.∆ABC AO
9. ABC and DBC are two triangles on the same base BC. If AD intersect BC at O. Prove that =
ar.∆DBC DO
Al-1999C; Delhi-2005
10. In ∆ ABC, ∠ A is acute. BD and CE are perpendiculars on AC and AB respectively. Prove that AB × AE =
AC × AD. Al-2003
11. Points P and Q are on sides AB and AC of a triangle ABC in such a way that PQ is parallel to side BC. Prove that
the median AD drawn from vertex A to side BC bisects the segment PQ. Foreign -2003
12. If the diagonals of a quadrilateral divide each other proportionally, prove that it is a trapezium.
OR
Two ∆ s ABC and DBC are on the same base BC and on the same side of BC in which ∠ A = ∠ D = 900. If CA
and BD meet each other at E, show that AE.ED. Delhi-2008
13. D and E are points on the sides CA and CB respectively of ∆ ABC right-angled at C. prove that AE2 + BD2
= AB2 + DE2.
OR
BE AC
In fig. DB ⊥ BC, DE ⊥ AB and AC ⊥ BC. Prove that = . Al-2008
DE BC

14. E is a point on the side AD produced of a ║gm ABCD and BE intersects CD at F. Show that ∆ ABC ~ ∆ CFB.
Foreign-2008
15. In fig, ∆ ABC is right angled at C and DE ⊥ AB. Prove that ∆ ABC ~ ∆ ADE and hence find the lengths of AE
and DE.

In fig, DEFG is a square and ∠ BAC = 900. Show that DE2 = BD × EC Delhi-2009

1
16. In fig, AD ⊥ BC and BD = CD. Prove that 2CA2 = 2AB2 + BC2. Al-2009
3
 PRAYAS Education

17. In fig, two triangles ABC and DBC lie on the same side of base BC. P is a point on BC such that PQ║BA and
PR║BD. Prove that QR║AD. Foreign-2009

LONG ANSWER TYPE QUESTIONS

1. In a right triangle ABC, right-angled at C, P and Q are points on the sides CA and CB respectively which divide
these sides in the ratio 1 : 2. Prove that Al-1996C
(i) 9AQ2 = 9AC2 + 4BC2 (ii) 0BP2 = 9BC2 + 4AC2 (iii) 9(AQ2 + BP2) = 13AB2.
2. The ratio of the areas of similar triangles is equal to the ratio of the square on the corresponding sides, prove.
Using the above theorem, prove that the area of the equilateral triangle described on the side of a square is half the
area of the equilateral triangle described on its diagonal. Delhi-1997C; 2005C; Foreign-2003
3. Perpendiculars OD, OE and OF are drawn to sides BC, CA and AB respectively from a point O in the interior of a
∆ ABC. Prove that :
(i) AF2 + BD2 + CE2 = OA2 + OB2 + OC2 – OD2 – OE2 – OF2.
(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2. Delhi-1997C, [NCERT]
4. In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares on the other two sides.
Prove. Using the above theorem, determine the length of AD in terms of b and C. Al-1997 C

5. If a line is drawn parallel to one side of a triangle, other two sides are divided in the same ratio, Prove. Using this
result to prove the following : In the given figure, if ABCD is a trapezium in which AB║DC║EF, then
AE BF
= . Foreign-1998
ED FC

6. State and prove Pythagoras. Use the theorem and calculate are ( ∆ PMR) from the given figure.
Delhi-1998C, 2006
 PRAYAS Education

7. In a right-angled triangle, the square of hypotenuse is equal to the sum of the squares of the two sides. Given that
∠ B of ∆ ABC is an acute angle and AD ⊥ BC. Prove that AC2 = AB2 + BC2 – 2BC. BD. Delhi-1999
8. In a right triangle, prove that the square on the hypotenuse is equal to the sum of the squares on the other two
sides. Using above, solve the following : In quadrilateral ABCD, find the length of CA, if CD ⊥ DB, CD = 6 m,
DB = 12 m and AB = 11 m. Delhi-2000
9. Prove that the ratio of the areas of two similar triangles is equal to the squares of their corresponding sides. Using
the above, do the following

area (∆ABC ) 9 AD
In fig, ∆ ABC and ∆ PQR are isosceles triangles in which ∠ A = ∠ P. If = ,find . Al-2000
area (∆PQR) 16 PS
10. In a right-angled triangle, prove that the square on the hypotenuse is equal to the sum of the squares on the other
two sides. Using the above result, find the length of the second diagonal of a rhombus whose side is 5 cm and one
of the diagonals is 6 cm. Al-2001
11. In a triangle, if the square on one side is equal to the sum of the squares on the other two sides prove that the angle
opposite the first side is a right angel.
Using the above theorem and prove that following : In triangle ABC, AD ⊥ BC and BD = 3CD. Prove that 2AB2 =
2AC2 + BC2. Al-2003
12. In a right triangle, prove that the square on hypotenuse is equal to sum of the squares on the other two sides.Using
the above result, prove that following : PQR is a right triangle right angled at Q. If S bisects QR, show that PR2 = 4
PS2 – 3 PQ2. Delhi-2004C
13. If a line is drawn parallel to one side of a trial prove that the other two sides are divided in the same ratio. Using
the above result, prove from fig. that AD = BE if ∠ A = ∠ B and DE║AB. Al-2004C

14. Prove that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.
Apply the above theorem on the following : ABC is a triangle and PQ is a straight line meeting AB in P and AC in
Q. If AP = 1 cm, PB = 3 cm, AQ = 1.5 cm, QC = 4.5 cm, prove that area of ∆ APQ is one-sixteenth of the area of
∆ ABC. Delhi-2005
15. If a line is drawn parallel to one side of a triangle, prove that the other two sides are divided in the same ratio. Use
BE BC
the above to prove the following : In the given figure DE║AC and DC║AP. Prove that = . Al-2005
EC CP
 PRAYAS Education

16. In a triangle if the square on one side is equal to the sum of squares on the other two sides, prove that the angle
opposite to the first side is a right angle. Using the above theorem to prove the following :
In a quadrilateral ABCD, ∠ B = 900. If AD2 = AB2 + BC2 + CD2, prove that ∠ ACD = 900. Al-2205
17. If a line is drawn parallel to one side of a triangle, to intersect the other two sides in distinct points, prove that the
other two sides are divided in the same ratio. Using the above, prove the following : In figure, DE║AC and BD =
CE. Prove that ABC is an isosceles triangle. Delhi-2007, 2009

18. Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding
sides. Using the above for the following : If the areas of two similar triangles are equal, prove that they are
congruent. Al-2007
19. Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding
sides. Using the above result, prove the following :
BX 2 −1
In a ∆ ABC, XY is parallel to BC and it divides ∆ ABC into two parts of equal area. Prove that =
AB 2
Delhi-2008
20. Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding
sides. Using the above, do the following :
The diagonals of a trapezium ABCD, with AB║DC, intersect each other at the point O. If AB = 2 CD, find the
ratio of the area of ∆ AOB to the area of ∆ COD. Al -2008
21. If a line is drawn parallel to one side of a triangle, to intersect the other two sides in distinct points, prove that the
other two sides are divided in the same ratio. Using the above, prove the following : In the fig, AB║DE and BC
║EF. Prove that AC║DF. Foreign-2008

2. Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding
sides. Using the above, do the following : In a trapezium ABCD, AC and BD are intersecting at O, AB║DC and
AB = 2 CD. If area of ∆ AOB = 84 cm2, find the area of ∆ COD. Delhi-2009

SIMILAR TRIANGLE ANSWER KEY EXERCISE (X)-CBSE

VERY SHORT ANSWER TYPE QUESTIONS

2. 96 cm2 3. 2 (q − 1) 6. 600 8. 13 m 9. 4.25 12. 9 : 4 14. 4.9 cm 16. 8.8 cm 18. (iii) 16 : 81
 ac 
28. 13.5 cm 19.   31. 36 cm
2

 b + c 
SHORT ANSWER TYPE QUESTIONS
15 36
2. 2 5 cm 15. AE = , DE =
13 13
LONG ANSWER TYPE QUESTIONS
bc
4. 6. 24 cm2 8. 13 cm 9. 3 : 4 10. 8 cm 21. 4 : 1 23. 21 cm2
b +c
2 2
 PRAYAS Education

EXERCISE – 1 (FOR OLYMPIADS)

CHOOSE THE CORRECT ONE
1. In a triangle ABC, if AB, BC and AC are the three sides of the triangle, then which of the statements is necessarily
true?
(A) AB + BC < AC (B) AB + BC > AC (C) AB + BC = AC (D) AB2 + BC2 = AC2
2. The sides of a triangle are 12 cm, 8 cm and 6 cm respectively, the triangle is :
(A) acute (B) obtuse (C) right (D) can’t be determined
3. In an equilateral triangle, the incentre, circumcentre, orthocenter and centroid are:
(A) concylic (B) coincident (C) collinear (D) none of these
4. In the adjoining figure D is the midpoint of a ∆ ABC. DM and DN are the perpendiculars on AB and AC
respectively and DM = DN, then the ∆ ABC is :

(A) right angled

(B) isosceles
(C) equilateral
(D) scalene

5. Triangle ABC is such that AB = 9 cm, BC = 6 cm, AC = 7.5 cm, Triangle DEF is similar to ∆ ABC, If EF = 12 cm
then DE is :
(A) 6 cm (B) 16 cm (C) 18 cm (D) 15 cm
6. In ∆ ABC, AB = 5 cm, AC = 7 cm. If AD is the angle bisector of < A. Then BD : CD is :
(A) 25 : 49 (B) 49 : 25 (C) 6 : 1 (D) 5 : 7
7. In a ∆ ABC, D is the mid-point of BC and E is mid-point of AD, BF passes through E. What is the ratio of AF : FC

(A) 1 : 1
(B) 1 : 2
(C) 1 : 3
(D) 2 : 3

8. In a ∆ ABC, AB = AC and AD ⊥ BC, then :

(A) AB < AD (B) AB > AD (C) AB = AD (D) AB ≤ AD
9. The difference between altitude and base of a right angled triangle is 17 cm and its hypotenuse is 25 cm. What is
the sum of the base and altitude of the triangle is ?
(A) 24 cm (B) 31 cm (C) 36 cm (D) can’t be determined
10. If AB, BC and AC be the three sides of a triangle ABC, which one of the following is true ?
(A) AB – BC = AC (B) (AB – BC) > AC (C) (AB – BA) < AC (D) AB2 – CB2 = AC2
11. In the adjoining figure D, E and F are the mid-points of the sides BC, AC and AB respectively. ∆ DEF is
congruent to triangle :
(A) ABC
(B) AEF
(C) CDE , BFD
(D) AFE , BFD and CDE

12. In the adjoining figure ∠ BAC = 600 and BC = a, AC = b and AB = c, then :

(A) a2 = b2 + c2
(B) a2 = b2 + c2 – bc
(C) a2 = b2 + c2 + bc
(D) a2 = b2 + 2bc
 PRAYAS Education

13. If the medians of a triangle are equal, then the triangle is:
(A) right angled (B) isosceles (C) equilateral (D) scalene
14. The incentre of a triangle is determined by the:
(A) Medians (B) angle bisectors
(C) perpendicular bisectors (D) altitudes
15. The point of intersection of the angle bisectors of a triangle is :
(A) orthocenter (B) centroid (C) incentre (D) circumcentre
16. A triangle PQR is formed by joining the mid-points of the sides of a triangle ABC, ‘O’ is the circumcentre of
∆ ABC, then for ∆ PQR, the point ‘O’ is :
(A) incentre (B) circumcentre (C) orthocenter (D) centroid
17. If AD, BE, CF are the altitudes of ∆ ABC whose orthocenter is H, then C is the orthocenter of :
(A) ∆ ABH (B) ∆ BDH (C) ∆ ABD (D) ∆ BEA
18. In an equilateral ∆ ABC, if a, b and c denote the lengths of perpendiculars from A, B and C respectively on the
opposite sides, then:
(A) a > b > c (B) a > b < c (C) a = b = c (D) a = c ≠ b
19. Any two of the four triangles formed by joining the midpoints of the sides of a given triangle are:
(A) congruent (B) equal in area but not congruentAB > AD
(C) unequal in area and not congruent (D) none of these
20. The internal bisectors of ∠ B and ∠ C of ∆ ABC meet at O. If ∠ A = 800 then ∠ BOC is :
(A) 500 (B) 1600 (C) 1000 (D) 1300
21. The point in the plane of a triangle which is at equal perpendicular distance from the sides of the triangle is :
(A) centroid (B) incentre (C) circumcentre (D) orthocenter
22. Incentre of a triangle lies in the interior of :
(A) an isosceles triangle only (B) a right angled triangle only
(C) any equilateral triangle only (D) any triangle
23. In a triangle PQR, PQ = 20 cm and PR = 6 cm, the side QR is :
(A) equal to 14 cm (B) less than 14 cm (C) greater than 14 cm (D) none of these
24. If ABC is a right angled triangle at B and M, N are the mid-points of AB and BC, than 4 (AN2 + CM2) is equal to-
5
(A) 4AC2 (B) 6 AC2 (C) 5 AC2 (D) AC2
4
BD
25. ABC is a right angle triangle at A and AD is perpendicular to the hypotence. Then is equal to :
CD
2 2
 AB   AB  AB AB
(A)   (B)   (C) (D)
 AC   AD  AC AD

26. Let ABC be an equilateral triangle. Let BE ⊥ CA meeting CA at E, then (AB2 + BC2 + CA2) is equal to :
(A) 2BE2 (B) 3 BE2 (C) 4 BE2 (D) 6 BE2
27. If D, E and F are respectively the mid-points of sides of BC, CA and AB of a ∆ ABC. If EF = 3 cm, FD = 4 cm,
and AB = 10 cm, then DE, BC and CA respectively will be equal to :
10
(A) 6, 8 and 20 cm (B) 4, 6 and 8 cm (C) 5, 6 and 8 cm (D) , 9 and 12 cm
3
 PRAYAS Education

28. In the right angle triangle ∠ C = 900. AE and BD are two medians of a triangle ABC meeting at F. The ratio of the
area of ∆ ABF and the quadrilateral FDCE is :
(A) 1 : 1 (B) 1 : 2 (C) 2 : 1 (D) 2 : 3
29. The bisector of the exterior ∠ A of ∆ ABC intersects the side BC produced to D. Here CF is parallel to AD.
AB BD
(A) =
AC CD
AB CD
(B) =
AC BD
AB BC
(C) =
AC CD
(D) None of these
30. The diagonal BD of a quadrilateral ABCD bisects ∠ B and ∠ D, then :
AB AD
(A) =
CD BC
AB AD
(B) =
BC CD
(C) AB = AD × BC
(D) None of these
31. Two right triangles ABC and DBC are drawn on the same hypotenuse BC on the same side of BC. If AC and DB
intersects at P, then
AP BP
(A) =
PC DP
(B) AP × DP = PC × BP
(C) AP × PC× = BP × DP
(D) AP × BP× = PC × PD
32. In figure, ABC is a right triangle, right angled at B. AD and CE are the two medians drawn from A and C
3 5
respectively. If AC = 5 cm and AD = cm, find the length of CE:
2
(A) 2 5 cm
(B) 2.5 cm
(C) 5 cm
(D) 4 2 cm
33. In a ∆ ABC, AB = 10 cm, BC = 12 cm and AC = 14 cm. Find the length of median AD. If G is the centroid, find
length of GA :
5 5 10 8 8
(A) 7, 7 (B) 5 7 ,4 7 (C) , 7 (D) 4 7 , 7
3 9 3 3 3
34. The three sides of a triangles are given. Which one of the following is not a right triangle ?
(A) 20, 21, 29 (B) 16, 63, 65
(C) 56, 90, 106 (D) 36, 35, 74
 PRAYAS Education

35. In the figure AD is the external bisector of ∠ EAC, intersects BC produced to D. If AB = 12 cm, AC = 8 cm and
BC = 4 cm, find CD.
(A) 10 cm
(B) 6 cm
(C) 8 cm
(D)9 cm
36. In ∆ ABC, AB2 + AC2 = 2500 cm2 and median AD = 25 cm, find BC.
(A) 25 cm (B) 40 cm (C) 50 cm (D) 48 cm
37. In the given figure, AB = BC and ∠ BAC = 150. AB = 10 cm. Find the area of ∆ ABC.
(A) 50 cm2
(B) 40 cm2
(C) 25 cm2
(D) 32 cm2
DE 2
38. In the given figure, if = and if AE = 10 cm. Find AB
BC 3
(A) 16 cm
(B) 12 cm
(C) 15 cm
(D) 18 cm

39. In the figure AD = 12 cm. AB = 20 cm and AE = 10 cm. Find EC.

(A) 14 cm
(B) 10 cm
(C) 8 cm
(D) 15 cm

40. In the given fig, BC = AC = AD, ∠ EAD = 810. Find the value of x.
(A) 450
(B) 540
(C) 630
(D) 360

41. What is the ratio of inradius to the circumradius of a right angled triangle?
(A) 1 : 2 (B) 1 : 2 (C) 2 : 5 (D) Can’t be determined

Que. 1 2 3 4 5 6 ANSWER
7 8 KEY 9 10 11 12 13 14 15
Ans. B B B B C D B B B C D B C B C
Que. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Ans. C A C A D B D C C A C C A A B
Que. 31 32 33 34 35 36 37 38 39 40 41
Ans. C A D D C C C C A B D
PRAYAS Education