Class 10 Maths Extra Questions Chapter 6 Triangles

VERY SHORT ANSWER TYPE QUESTIONS

1. Fill in the blanks :
(i) All equilateral triangles are __________ .

AB
(ii) If ABC ~ FED, then = .
ED

(iii) Circles with equal radii are _________ .
(iv) If a line is drawn parallel to one side of a triangle to intersect the other
two sides in distinct points, the other two sides are divided in the
_________ ratio.
(v) In __________ triangle, the square of the hypotenuse is equal to the sum
of the squares of the other two sides.
(vi) If two triangles are similar, their corresponding sides are __________.
(CBSE 2020)

(vii) In ABC, AB = 6 3 , AC = 12 cm and BC = 6 cm, then B = _______ .

(viii) Let ABC ~ DEF and their areas be respectively 81 cm2 and 144 cm2. If
EF = 24 cm, then length of side BC is _________ cm.
2. State True or False :
(i) All the similar figures are always congruent.
(ii) The Basic Proportionality Theorem was given by Pythagoras.
(iii) The mid-point theorem can be proved by Basic Proprotionality Theorem.
(iv) Pythagoras Theorem is valid for right angled triangle.
(v) If the sides of two similar triangles are in the ratio 4 : 9, then the areas of
these triangles are in the ratio 16 : 81.

76 Mathematics-X

3. Match the following :
Column I Column II
(a) If corresponding angles are equal in two (i) SAS similarity criterion
triangles, then the two triangles are similar.
(b) If sides of one triangle are proportional to (ii) ASA similarity criterion
the sides of the other triangle, then the two
triangles are similar.
(c) If one angle of a triangle is equal to one (iii) AAA similarity criterion
angle of the other triangle and the sides
including these angles are proportional,
then the two triangles are similar. (iv) SSS similarity criterion
PX PY 1
4. In the following figure, XY || QR and = = , then
XQ YR 2
P

X Y

Q R
1
(a) XY = QR (b) XY =
QR
3
1
(c) XY2 = QR2 (d) XY = QR
2
5. In the following figure, QA  AB and PB  AB, then AQ is
P

10 units O 9 units
A
B
6 units

Q
(a) 15 units (b) 8 units
(c) 5 units (d) 9 units

Mathematics-X 77

6. The ratio of areas of two similar triangles is equal to the
(a) ratio of their corresponding sides.
(b) ratio of their corresponding altitudes.
(c) ratio of the square of their corresponding sides.
(d) ratio of their perimeter.
7. The areas of two similar triangles are 144 cm2 and 81 cm2. If one median of the
first triangle is 16 cm, length of corresponding median of the second triangle is
(a) 9 cm (b) 27 cm
(c) 12 cm (d) 16 cm
8. In a right triangle ABC, in which C = 90° amd CD  AB. If BC = a, CA = b,
AB = c and CD = p, then A
1 1 1 1 1 1
(a) 2
 2 2 (b) 2
 2 2
p a b p a b c
D
1 1 1 1 1 1 b
 2 2  2 2 p
(c) 2 (d) 2
p a b p a b

C a B

AB 1
9. If ABC ~ DEF, ar(DEF) = 100 cm2 and = , then ar(ABC) is
DE 2
(a) 50 cm2 (b) 25 cm 2

(c) 4 cm2 (d) 200 cm2
10. If the three sides of a triangle are a, 3a and 2a , then the measure of hte
angle opposite to longest side is
(a) 45° (b) 30°
(c) 60° (d) 90°
11. A vertical pole of length 3 m casts a shadow of 7 m and a tower casts a shadow
of 28 m at a time. The height of tower is
(a) 10 m (b) 12 m
(c) 14 m (d) 16 m
12. The length of the diagonals of a rhombus are 16 cm and 12 cm. Then, the
length of the side of the rhombus is (NCERT Exempler)
(a) 9 cm (b) 10 cm
(c) 8 cm (d) 20 cm

78 Mathematics-X

13. If ABC ~ EDF and ABC is not similar to DEF, then which of the following
is not true? (NCERT Exempler)
(a) BC.EF = AC.FD (b) AB.EF = AC.DE
(c) BC.DE = AB.EF (d) BC.DE = AB.FD
14. Write the statement of pythagoras theorem.
15. Write the statement of Basic Proportionality Theorem.
16. Is the triangle with sides 12 cm, 16 cm and 18 cm a right triangle?

Area ( ABC) 9
17. If ABC ~ QRP, = , AB = 18 cm, BC = 15 cm, then find
Area (  PQR) 4
the length of PR. (CBSE 2018)
18. In the given Fig., M = N = 46°, Express x in terms of a, b and c.
L

P
a
x
46° 46°
M b c K
19. In the given Fig. AHK ~ ABC. If AK = 10 cm, BC = 3.5 cm and HK = 7 cm,
find AC. (CBSE 2010)
H

A
C K

B
20. It is given that DEF ~ RPQ. Is it true to say that D = R and F = P?
21. If the corresponding Medians of two similar triangles are in the ratio 5 : 7.
Then find the ratio of their sides.
22. An aeroplane leaves an airport and flies due west at a speed of 2100 km/hr. At
the same time, another aeroplane leaves the same place at airport and flies due
Mathematics-X 79

south at a speed of 2000 km/hr. How far apart will be the two planes after 1
hour?
23. The areas of two similar triangles ABC and DEF are 225 cm2 and 81 cm2
respectively. If the longest side of the larger triangle ABC be 30 cm, find the
longest side of the smaller triangle DEF.
24. In the given figure, if ABC ~ PQR, find the value of x?

A
R x Q

6 cm 5 cm 3.75 cm 4.5 cm

B C
4 cm
P

PX PY 1
25. In the given figure, XY || QR and = = , find XY : QR.
XQ YR 2

P

X Y

Q R

26. In the given figure, find the value of x which will make DE || AB ?
(NCERT Exempler)
A B
3x + 19 3x + 4
E D

x+3 x
C
27. If ABC ~ DEF, BC = 3EF and ar (ABC) = 117cm2 find area (DEF).

80 Mathematics-X

28. If ABC and DEF are similar triangles such that A = 45° and F = 56°,
then find the value of C.
29. If the ratio of the corresponding sides of two similar triangles is 2 : 3, then find
the ratio of their corresponding attitudes.

BC 1 ar(  PRQ)
30. It is given that ABC ~ PQR with = , then find the value of .
QR 3 ar(  ACB)

SHORT ANSWER TYPE QUESTIONS-I

31. In the given Fig. PQ = 24 cm, QR = 26 cm, PAR = 90°, PA = 6 cm and AR =
8 cm, find QPR.
Q P

A

R

BE BC
32. In the given Fig., DE || AC and DC || AP Prove that = (CBSE 2020)
EC CP

33. In ABC, AD  BC such that AD2 = BD × CD. Prove that ABC is right
angled triangle.
34. In the given Fig., D and E are points on sides AB and CA of ABC such that
B = AED. Show that ABC ~ AED.
A

D
E

B C

Mathematics-X 81

35. In the given fig., AB || DC and diagonals AC and BD intersects at O. If OA =
3x – 1 and OB = 2x + 1, OC = 5x – 3 and OD = 6x – 5, find the value of x.
D C
6x – 3
5 5x –
1 O 2x +
3x – 1
A B

36. In the given Fig. PQR is a triangle, right angled at Q. If XY || QR, PQ = 6 cm,
PY = 4 cm and PX : XQ = 1 : 2. Calculate the lengths of PR and QR.

P

X Y

Q R

37. In the given figure, AB || DE. Find the length of CD.

B

6 cm
5 cm

A
C
E

3 cm

D

38. In the given figure, ABCD is a parallelogram. AE divides the line segment BD
in the ratio 1 : 2. If BE = 1.5 cm find BC.

D C

O E

A B

82 Mathematics-X

39. In the given figure, ODC ~ OBA, BOC = 115° and CDO = 70°. Find,
(i) DOC, (ii) DCO, (iii) OAB, (iv) OBA.
D C
70°

O 115°

A B
40. Perimeter of two equilateral triangles ABC and PQR are 144 m and 96 m,
Find ar (ABC) : ar (PQR).

SHORT ANSWER TYPE QUESTIONS-II

QR QT
41. In the given figure, = and 1 = 2 then prove that PQS ~ TQR.
QS PR
(NCERT)
T

P

1 2
Q
S R
42. In equilateral ABC, AD  BC. Prove that 3BC2 = 4AD2.

BC 2 BD
43. In ABC, ACB = 90° and CD  AB. Prove that = . (HOTS)
AC 2 AD
44. In the adjoining figure ABC and DBC are on the same base BC. AD and BC
area ( ABC) AO
intersect at O. Prove that = . (CBSE 2020)
area ( DBC) DO

A C

O

B D
Mathematics-X 83

45. If AD and PS are medians of ABC and PQR respectively where ABC ~
AB AD
PQR, Prove that = .
PQ PS
46. In the given figure, DE || AC. Which of the following is correct?
ab B
ay
x or x
ay ab a

E D
x
b
C y A

47. Prove that the sum of the squares of the sides of a rhombus is equal to the sum
of the squares of its diagonals.
(NCERT, CBSE 2019, 2020)
48. A street light bulb is fixed on a pole 6 m above the level of the street. If a
woman of height 1.5 m casts a shadow of 3 m, find how far she is away from
the base of the pole.
(NCERT Exempler)
49. Two poles of height a metrs and b metres are p metres apart. Prove that the
height of the point of intersection of the lines joining the top of each pole to the
ab
foot of the opposite pole is given by metres.
ab
50. In the given figure AB || PQ || CD, AB = x, CD = y and PQ = z. Prove that
1 1 1
  .
x y z
A

C
x P
y
z
B D
Q

84 Mathematics-X

AD AE
51. In the given figure D = E and = . Prove that BAC is an isoscles
DB EC
triangle. (CBSE 2020)

52. In the figure, a point O inside ABC is joined to its vertices. From a point D on
AO, DE is drawn parallel to AB and from a point E on BO, EF is drawn parallel
to BC. Prove that DF || AC.

A

D
O
E F
B C

53. Two triangles BAC and BDC, right angled at A and D respectively are drawn
on the same base BC and on the same side of BC. If AC and DB intersect at P.
PRove that AP × PC = DP × PB. (CBSE 2019)

54. Hypotenuse of a right triangle is 25 cm and out of the remaining two sides, one
is larger than the other by 5 cm, find the length of the other two sides.
(NCERT Exempler)

Mathematics-X 85

BE BC
55. In the given figure DE || AC and = . Prove that DC || AP..
EC CP

A

D

B E C P

56. In a quadrilateral ABCD, B = 90°, AD2 = AB2 + BC2 + CD2. Prove that
ACD = 90°.

57. In the given figure, DE || BC, DE = 3 cm, BC = 9 cm and ar (ADE) = 30 cm2.
Find ar (BCED).

A

D E
3cm

B 9 cm C

86 Mathematics-X

1
58. In an equilateral ABC, D is a point on side BC such that BD = BC. Prove
3
that 9A D 2 = 7AB2. (NCERT, CBSE 2018, 2020)
59. In PQR, PD  QR such that D lies on QR. If PQ = a, PR = b, QD = c and DR =
d and a, b, c, d are positive units. Prove that (a + b) (a – b) = (c + d) (c – d).
(NCERT Exempler)
60. Prove that the ratio of the areas of two similar triangles is equal to the ratio of
the squars of their corresponding sides. (CBSE 2010, 2018, 2019)
61. In the given figure, the line segment XY is Parallel to AC of ABC and it

AX 2 1
divides the triangle into two parts of equal areas. Prove that = .
AB 2

A

X

B Y C

62. Through the vertex D of a parallelogram ABCD, a line is drawn to intersect the
DA FB FC
sides BA and BC produced at E and F respectively. Prove that = = .
AE BE CD
63. Prove that if in a triangle, the square on one side is equal to the sum of the
squares on the other two sides, then the angle opposite to the first side is a right
angle.
(CBSE 2019, 2020)
64. Prove that in a right angle triangle, the square of the hypotenuse is equal the
sum of the squares of other two sides. (CBSE 2018, 2019, 2020)

Mathematics-X 87

65. If a line is drawn parallel to one side of a triangle to intersect the other two
sides in distinct points, then prove that the other two sides are divided in the
same ratio.
(CBSE 2019, 2020)
66. In an Obtuse ABC (B is obtuse), AD is perpendicular to CB produced.
Then prove that A C2 = AB2 + BC2 + 2BC × BD.
67. In figure BN and CM are medians of a ABC right angled at A. Prove that
4(BN2 + CM2) = 5 BC2 (CBSE (C) 2020)

C

N

A B
M

68. Sides AB and AC and median AD of ABC are respectively proportional to
sides PQ and PR and median PM of PQR. Show that ABC ~ PQR.
(CBSE 2020)
69. In figure if ABC ~ DEF and their sides of lengths (in cm) are marked along
them, then find the lengths of sides of each triangle. (CBSE 2020)

A D

2x – 1 3x 18 6x

B C E F
2x + 2 3x + 9

70. The Perimeters of two similar triangles are 30 cm and 20 cm repsectively. If
one side of the first triangle is 9 cm long. Find the length of the corresponding
side of the second triangle. (CBSE 2020)
71. In figure PQR is right-angled at P.M is a point on QR such that PM is
perpendicular to QR. Show that PQ2 = QM × QR. (CBSE 2020)

88 Mathematics-X

R

M

P Q

ANSWERS AND HINTS

VERY SHORT ANSWER TYPE QUESTIONS-I

AB BC
1. (i) Similar (ii) = (iii) Congruent
FE E D
(iv) Same (v) Right (vi) Proportional
(vii)90° (viii) 18 cm
2. (i) False (ii) False (iii) True
(iv) True (v) True
3. (a) (iii) AAA similarity criterion.
(b) (iv) SSS similarity criterion.
(c) (i) SAS similarity criterion.

1
4. (B) XY = QR
3
5. (A) 15 units
6. (C) Ratio of the square of their corresponding sides.
7. (C) 12 cm

1 1 1
8. (A) 2
 2 2
p a b
9. (B) 25 cm2
10. (D) 90°
11. (B) 12 m
12. (B) 10 cm

Mathematics-X 89

13. (C) BC.DE = AB.EF
14. See point 3(iii) of Key Points
15. See point 3(i) of Key Points.
16. No, because (12)2 + (16)2  (18)2
17. 10 cm
18. KPN ~ KLM
x c

a bc
ac
x=
bc
AK HK 10 7
19. =  =  AC = 5 cm
AC BC AC 3.5
20. D = R (True)
F = P (False)
21. 5 : 7

N

2100 Km
22. W
0
E AB = (2100) 2  (2000) 2 = 2900 km
A

2000 Km
B
5

B

23. Let longest side of the DEF be x cm.
2
225  30 
=  
81 x
x = 18 cm

90 Mathematics-X

AB BC 6 4
24. =  =  x = 3cm
PQ QR 4.5 x
25. PXY ~ PQR

PX XY 1
= =
PQ QR 3
 XY : QR = 1 : 3

x3 x
26. = (By B.P.T.)
3 x  19 3x  4
x=2

ar (ABC)  BC  2  3EF  2  3  2
27. =   =   =  
ar (DEF) EF  EF  1

117
=9  ar(DEF) = 13 cm2
ar(DEF)
28. F = C = 56°
29. 2 : 3
30. 1/9

31. PR = (6) 2  (8) 2 = 10 cm.
As QR2 = PQ2 + PR2 , therefore QPR = 90°.
(By Converse of Pythagoras theorem)
AD EC
32. DE || AC, = ...(1) [ BPT]
DB BE
AD CP
DC || AP, = ...(2) [ BPT]
DB BC
From (1) and (2), we get

BE BC
=
EC CP

Mathematics-X 91

33. In ADC, AD2 = AC2 – DC2 ...(1)
A

B D C
In ADB, AD2 = AB2 – BD2 ...(2)
Adding (1) and (2), we have
2AD2 = AC2 + AB2 – BD2 – DC2
2AD2 + BD2 + DC2 = AC2 + AB2
2BD × CD + BD2 + DC2 = AC2 + AB2
(BD + DC)2 = AC2 + AB2 (Using (a + b)2 = a2 + b2 + 2ab)
BC2 = AC2 + AB2
By converse of Pythagoras Theorem, ABC is a right angled triangle.
34. B = AED (Given)
A = A (Common)
 ABC ~ AED [AA similarity criterion]
DE DO DE OC
35. Draw EO || AB, = (In ADB) and = (In ACD)
EA OB EA OA

D C
6x
– 5 3
E 5x –
1 O
2x +
3x – 1
A B

3x  1 2x  1 1
=  x= or 2
5x  3 6x  5 2
1
But x = is neglected because (5x – 3) get negative value.
2
So, x = 2 is the required value.

92 Mathematics-X

PX PY 1 4
36. =  =  YR = 8 cm
XQ YR 2 YR
 PR = 8 + 4 = 12 cm

QR = (12) 2  (6) 2 = 6 3 cm
37. ABC ~ EDC (AA Similarity criterion)

6 5
=
3 CD
CD = 2.5 cm
38. BOE ~ DOA (AA Similarity criterion)

BO BE
=
DO DA

1 1.5
=
2 DA
DA = 3 cm
BC = DA = 3 cm (Opposite sides of a parallelogram)
39. (i) 65°
(ii) 45°
(iii) 45°
(iv) 70°

ar ( ABC)  144  2 9
ar ( PQR)  96 
40. = =
4

 ar (ABC) : ar (PQR) = 9 : 4
41. In PQR, 1 = 
PR = PQ [Opposite sides of equal angles]

QR QT
 = and 1 = 1 (Common)
QS PQ

PQS ~ TQR (SAS Similarity criterion)
Mathematics-X 93

42. ADB  ADC A

BD = DC
1
 BD = BC ...(1)
2

In right angled ADB, B D C
AB2 = AD2 + BD2
2
 BC 
BC2 = AD2 +   [ AB = BC = CA and from (1)]
2 
3BC2 = 4AD2
43. ABC ~ CBD
 BC2 = AB.BD ...(1)
ABC ~ ACD
 AC2 = AB.AD ...(2)
Divide (1) by (2), we get

BC 2 BD
2 =
AC AD
44. Draw AX  BC and DY  BC
1
ar (  ABC) × BC× AX
2 AX
= 1 = ...(1)
ar ( DBC) DY
× BC× DY
2
A C
Y

O
X
B D
AXO ~ DYO [AA similarity criterion]
AX AO
= ...(2) (C.P.S.T.)
DY DO
From (1) and (2), we get

94 Mathematics-X

ar ( ABC) AO
=
ar ( DBC) DO
45.

1
BC
AB BC 2 BD
As ABC ~ PQR, Hence B = Q and = = 1 =
PQ QR QR QS
2
In ABD and PQS
AB BD
= and B = Q.
PQ QS
 ABD ~ PQS (SAS Similarity criterion).
AB AD
Hence, = (C.P.S.T.)
PQ PS
46. BED ~ BCA
x a
=
y ab
ay
x=
ab
47.
D
C

o

A B
Mathematics-X 95

In right angled AOB, AB2 = OA2 + OB2 ...(1)
In right angled BOC, BC2 = OB2 + OC2 ...(2)
In right angled COD, CD2 = OC2 + OD2 ...(3)
In right angled DOA, DA2 = OD2 + OA2 ...(4)
Adding (1), (2), (3) and (4), we get
AB2 + BC2 + CD2 + DA2 = 2OA2 + 2OB2 + 2OC2 + 2OD2
2 2 2 2
1  1  1  1 
= 2  AC  2  BD  2  AC  2  BD
2 2 2 2
[ Diagonals of rhombus  bisect each other]
= AC2 + BD2
48. ABE ~ CDE
AB BE
=
CD DE

6 3 + BD
=
1.5 3
BD = 9m
ab
49. To prove : EF =
ab
Proof : AB || EF || DC
EFC ~ ABC

EF FC A
= ...(1)
AB BC
BFE ~ BCD D

am
EF BF E
= ...(2) bm
CD BC
Adding (1) and (2), we get
B F C
EF EF FC+ BF pm
+ =
AB CD BC
96 Mathematics-X

 1 1  BC
EF    =
 AB CD  BC

1 1 
EF    = 1
a b

ab
EF =
ab

50. Same as Q. 48.

AD AE
51. 
DB EC
By converse of BPT, DE || BC
 D = B and E = C (Corresponding Angles)
But D = E
So, B = C
 AB = AC
So, ABC is an isosceles triangle.

OD OE
52. In OAB, = .... (1) ( BPT)
DA EB

OE OF
In OBC, = .... (2) ( BPT)
EB FC
From (1) and (2), we get

OD OF
=
DA FC
By converse of BPT, DF || AC.
53. APB ~ DPC (AA Similarity criterion)

AP PB
= ( C.P.S.T.)
DP PC
AP.PC = DP.PB
Mathematics-X 97

54. Let sides of right angled triangle other than hypotenuse be x cm and (x + 5) cm.
By Pythagoras theorem,
(x)2 + (x + 5)2 = (25)2
x = 15 or – 20
But side is always positive, So, x = 15.
 Length of two sides is 15 cm and 20 cm.
55. Same as Q.31.
56. In right angled ABC, AC2 = AB2 + BC2 ...(1)
Given, AD2 = (AB2 + BC2) + CD2
 AD2 = AC2 + CD2 [From (1)]
By converse of Pythagoras theorem, ACD = 90°.
57. ADE ~ ABC

ar (ADE)  DE  2
ar (ABC)  BC 
=

2
30  3
ar(ABC)  9 
=

 ar(ABC) = 270 cm2
ar (BCFD) = ar (ABC) – ar (ADE)
= 270 – 30 = 240 cm2
58. Draw AE  BC
ABE  ACE
A
1
 BE = CE  BE = BC
2
In right angled AED, AE2 = AD2 – DE2 ...(1)
In right angled AEB, AE2 = AB2 – BE2 ...(2)
From (1) and (2), we have
AD2 – DE2 = AB2 – BE2
B D E C
2
1 
AD2 – (BE – BD)2 = BC2 – BC
  ( AB = AC)
2
98 Mathematics-X

2 2
1
2 1  2 BC
AD   BC  BC  = BC 
2 3  4
9AD2 = 7AB2
59. In right angled PDQ,
PD2 = a2 – c2 ...(1)
In right angled PDR
PD2 = b2 – d2 ...(2)
From (1) and (2), we have
a2 – c2 = b2 – d2 P

a2 – b2 = c2 – d2
(a – b) (a + b) = (c + d) (c – d)
a b
60. Theorem 6.6 of NCERT.
61. Given, ar BXY = ar AXYC
ar ( ABC) = ar BXY + ar AXYC
Q c D d R
= 2 ar BXY

ar(  ABC) 2
 ar(  BXY) = 1

ABC ~ XBY
2 ar (ABC)
 AB 
  = ar (BXY)
XB

AB
= 2
XB

XB 1
=
AB 2

XB 1
1 = 1
AB 2

Mathematics-X 99

AB  XB 2 1
=
AB 2

AX 2 1
= F
AB 2
62. EAD ~ EBF

EA AD
= D
EB BF C

BF AD BF  AD BF  BC CF
 = = = =
BE AE BE  AE BA DC
63. Theorem 6.9 of NCERT.
E A B
64. Theorem 6.8 of NCERT.
65. Theorem 6.9 of NCERT.
66. In ADB A
AB2 = AD2 + DB2 ...(i)
In ADC
AC2 = AD2 + DC2
= AD2 + (DB + BC)2
= AD2 + DB2 + BC2 + 2BC.BD D B C

= AB2 + BC2 + 2BC.BD (Using (i))

1
67. BN is median  AN = CN = AC
2

1
CM is median  AN = MB = AB
2
In BAC, BC2 = AB2 + AC2 ...(1)
2
 AC 
In BNC, BC2 = AB2 +  
 2 
4BN2 = 4AB2 + AC2 ...(2)

100 Mathematics-X

In MAC, (CM)2 = (AM)2 + (AC)2
4CM2 = AB2 + 4AC2 ...(3)
Adding corresponding sides of (2) and (3)
4 (BN2 + CM2) = 5AB2 + 5AC2
4 (BN2 + CM2) = 5BC2
68. In ABC and PQR

AB AC AD
= = ...(1)
PQ PR PM
Extend AD to a point E s.t. AD = DE and PM to point L s.t. PM = ML

A P

B C Q R
D M

E L

 quadrilateral of ABEC and PQLR are parallelogram
( diagonals bisect each other)
 AC = BE, AB = EC
PR = QL, PQ = LR } ....(2)
From (1) and (2)

AB BE 2 AD AE
= = =
PQ QL 2 PM PL
 ABE ~ PQL
 ABE = PQL ...(3)
Similarly, AEC ~ PLR
 CAE = RPL ...(4)
 CAB = RPQ (from 3 and 4)
 In ABC and PQR

Mathematics-X 101

AB AC
= and CAB = RPQ
PQ PR
 ABC ~ PQR

AB BC CA
69. = = ( ABC ~ DEF)
DE EF FD

2x 1 2x  2 3x
= =
18 3x  9 6x
Solving, we get x = 5
 AB = 9 cm BC = 12 cm AC = 15 cm
DE = 18 cm EF = 24 cm FD = 30 cm
70. ABC ~ DEF

AB BC AC
 = = =k
DE EF DF
 AB = kDE, BC = kEF, AC = kDF
 AB + BC + AC = k (DE + EF + DF)

30 9
 =  x = 6 cm
20 x
71. In PMR, PR2 = PM2 + RM2 ...(1)
In PMQ, PQ2 = PM2 + MQ2 ...(2)
In PQR, RQ2 = RP2 + PQ2 ...(3)
 RM2 + MQ2 + 2RM.MQ = RP2 + PQ2 ...(4) ( RQ = RM + MQ)
Adding (1) and (2), PR2 + PQ2 = 2PM2 + RM2 + MQ2 ...(5)
From (4) and (5)
PM2 = RM.MQ

102 Mathematics-X

PRACTICE-TEST
Triangles
Time : 1 Hrs. M.M. : 20
SECTION - A
1. If sides of two similar triangles are in the ratio of 8:10, then areas of these
triangles are in the ratio __________ . 1
AB BC CA
2. If in two triangles ABC and PQR, = = , then 1
QR RP PQ
(A) PQR ~ CAB (B) PQR ~ ABC
(C) CBA ~ PQR (D) BCA ~ PQR
3. ABC is an isosceles right triangle, right angled at C, then AB2 = ..........
(A) AC2 (B) 2 AC2 (CBSE 2020)
(C) 4 AC 2 (D) 3 AC 2 1
4. A line DE is drawn parallel to base BC of ABC, meeting AB in D and AC at
E .
AB
If = 4 and CE = 2 cm, find the length of AE.
BD

SECTION B
5. The lengths of diagonals of a rhombus field are 32 m and 24 m. Find the length
of the side of the field. 2
6. A man goes 24 m towards West and then 10 m towards North. How far is he
from the starting point? 2
7. Using converse of Basic Proportionality Theorem, prove that the line joining
the mid-points of any two sides of a triangle is parallel to the third side. 2
SECTION C
8. E is a point on the side AD produced of a parallelogram ABCD and BE intersect
CD at F. Show that ABE ~ DCB. 3
9. In an equilateral triangle, prove that three times the square of one side is equal
to four times the square of one of its altitude. 3
SECTION D
10. State and prove Basic Proportionality Theorem. 4
Mathematics-X 103