1. O is the point of intersection of two equal chords ABand CD such that OB = OD, then triangles OAC and ODB are

(A) Equilateral but not similar

(B) Isosceles but not similar

(C) Equilateral and similar

(D) Isosceles and similar

Answer: (D)

Explanation:

Since O is the point of intersection of two equal chords AB and CD such that OB = OD,

As chords are equal and OB = OD, so AO will also be equal to OC

Also ∠AOC = ∠DOB = 45⁰

Now in triangles OAC and ODB

AO/OB = CO/OD

And ∠AOC = ∠DOB = 45⁰

So triangles are isosceles and similar

2. D and E are respectively the midpoints on the sides AB and AC of a triangle ABC and BC = 6 cm. If DE || BC, then the length of DE (in cm) is

(A) 2.5                                                  

(B) 3

(C) 5                                                                      

(D) 6

Answer:  B

Explanation:

By midpoint theorem,

If D and E are respectively the midpoints on the sides AB and AC of a triangle ABC, DE||BC and BC = 6 cm

So, DE will be half of BC i.e. 3cm

3. In triangle PQR, if PQ = 6 cm, PR = 8 cm, QS = 3 cm, and PS is the bisector of angle QPR, what is the length of SR?

(A) 2                                                                     

(B) 4

(C) 6                                                                      

(D) 8

Answer:  (B)

Explanation:

Since, PS is the angle bisector of angle QPR

So, by angle bisector theorem,

QS/SR = PQ/PR

⇒ 3/SR = 6/8

⇒ SR = (3 X 8)/6 cm = 4 cm

4. The lengths of the diagonals of a rhombus are 16 cm and 12cm. Then, thelength of the side of the rhombus is

(A) 9 cm

(B) 10 cm

(C) 8 cm                                                               

(D) 20 cm

Answer:(B)

Explanation:

The diagonals of rhombus bisect each other at right angle, so side of rhombus is the hypotenuse for the triangles formed.

Therefore,

By Pythagoras theorem

(16/2)² + (12/2)² = Side²

⇒ 8² + 6² = Side²

⇒ 64 + 36 = Side²

⇒ Side = 10 cm

5. A flag pole 18 m high casts a shadow 9.6 m long. Find the distance of the top of the pole from the far end of the shadow.

(A) 25.6                                                                               

(B) 20.4

(C) 23.7                                                                

(D) 32.5

Answer:(B)

Explanation:

According to given question

The far end of shadow is represented by point A,

Therefore we need to Find AC

By Pythagoras theorem,

(18)² + (9.6)² = (AC)²

⇒ AC² = 416.16

⇒ AC = 20.4 m (approx)

6. Diagonals of a trapezium PQRS intersect each other at the point O, PQ || RS and PQ = 3 RS, Then the ratio of areas of triangles POQ and ROS is:

(A) 1:9                                                                  

(B) 9:1

(C) 3:1                                                                  

(D) 1:3

Answer:(B)

Explanation:According to given Question

Since

SR || PQ,

So, ∠OSR= ∠OQP (alternate interior angles)

Also ∠SOR= ∠POQ (vertically opposite angles)

So triangles SOR and POQ are similar,

Therefore,

ar(POQ)/ar(SOR) = (PQ/SR)²

ar(POQ)/ar(SOR) = (3 SR/SR)²

ar(POQ)/ar(SOR) = 9/1

7. ABCD is a trapezium in which AB|| DC and P, Q are points on ADand BC respectively such that PQ || DC. If PD = 18 cm, BQ = 35 cm andQC = 15 cm, find AD.

(A) 55cm                                                             

(B) 57cm

(C) 60cm                                                             

(D) 62cm

Answer:(C)

Explanation:

According to question

ABCD is a trapezium in which AB || DC and P and Q are points on AD and BC, respectively such that PQ || DC. If PD = 18 cm, BQ = 35 cm and QC = 15 cm,

In triangle ABD

DP/AP = OD/OB

In triangle BDC

BQ/QC = OB/OD

This implies

DP/AP = QC/BQ

18/AP = 15/35

AP = (18 x 35)/15

AP = 42

Therefore, AD = AP + DP = 42 + 18 = 60cm

8. Areas of two similar triangles are 36 cm² and 100 cm². If the length of a side of the larger triangle is 20 cm, then the length of the corresponding side of the smaller triangle is:

(A) 12cm                                                                             

(B) 13cm

(C) 14cm                                                                             

(D) 15cm

Answer:(A)

Explanation:

Let the side of smaller triangle be x cm.

ar(Larger Triangle)/ar(Smaller Triangle) = (side of larger triangle/side of smaller triangle)²

100/36 = (20/x)²

x = √144

X = 12 cm

9. In the figure if ∠ACB = ∠CDA, AC = 8 cm and AD = 3 cm, find BD.

(A) 53/3 cm                                                        

(B) 55/3 cm

(C) 64/3 cm                                                        

(D) 35/7 cm

Answer:(B)

Explanation:

In triangle ACB and ADC

∠A=∠A

∠ACB = ∠CDA

Therefore triangle ACB and ADC are similar,

Hence

AC/AD = AB/AC

AC² = AD X AB

8² = 3 x AB

⇒ AB = 64/3

This implies,

BD = 64/3 – AD

⇒ BD = 55/3

10. If ABCD is parallelogram, P is a point on side BC and DP when produced meets AB produced at L, then select the correct option

(A) DP/BL = DC/PL

(B) DP/PL = DC/BL

(C) DP/PL = BL/DC

(D) DP/PL = AB/DC

Answer: (B)

Explanation:

In ΔALD, we have

BP || AD

∴ LB/BA = LP/PD

⇒ BL/AB = PL/DP

⇒ BL/DC = PL/DP [∵ AB = DC

⇒ DP/PL = DC/BL

11. In the figure given below DE || BC. If AD = x, DB = x – 2, AE = x + 2 and EC = x – 1, the value of x is:

(A) 4                                                                                                     

(B) 8

(C) 16                                                                                                   

(D) 32

Answer: (A)

Explanation:

In triangle ABC, we have DE || BC

∴ AD/DB = AE/EC (By Thale’s Theorem)

⇒ x/x – 2 = (x + 2)/(x – 1)

⇒ x (x – 1) = (x – 2)(x + 2)

⇒ x² – x = x² – 4

⇒ x = 4

12. The length of altitude of an equilateral triangle of side 8cm is

(A) √3 cm                                                            

(B) 2√3 cm

(C) 3√3 cm                                                          

(D) 4√3 cm

Answer:(D)

Explanation:

The altitude divides the opposite side into two equal parts,

Therefore, BD = DC = 4 cm

In triangle ABD

AB² = AD² + BD²

8² = AD² + 4²

AD² = 64 – 16

AD² = 48

AD = 4√3 cm

13. If ΔABC ~ ΔDEF, AB = 4 cm, DE = 6 cm, EF = 9 cm and FD = 12 cm, find the perimeter of ABC.

(A) 18    cm                                                         

(B) 20 cm

(C) 21    cm                                                         

(D) 22 cm

Answer:(A)

Explanation:

According to question,

ΔABC ~ ΔDEF,

 AB = 4 cm, DE = 6 cm, EF = 9 cm and FD = 12 cm,

Therefore,

AB/DE = BC/EF = AC/DF

4/6 = BC/9 = AC/12

⇒ 4/6 = BC/9

⇒ BC = 6 cm

And

4/6 = AC/12

⇒ AC = 8 cm

Perimeter = AB + BC + CA

= 4 + 6 + 8

= 18 cm

14. A 5 m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4 m high. If the foot of the ladder is moved 1.6 m towards the wall, then the distance by which the top of the ladder would slide upwards on the wall is:

(A) 2 m                                                                                

(B) 1.2 m

(C) 0.8 m                                                                             

(D) 0.5 m

Answer:(C)

Explanation:

Let AC be the ladder of length 5m and BC = 4m be the height of the wall where ladder is placed. If the foot of the ladder is moved 1.6m towards the wall i.e. AD = 1.6 m, then the ladder is slided upward to position E i.e. CE = x m.

In right triangle ABC

AC² = AB² + BC²

⇒5² = AB² + 4²

⇒ AB = 3m

⇒ DB = AB – AD = 3 – 1.6 = 1.4m

In right angled ΔEBD

ED² = EB² + BD²

⇒ 5² = EB² + (1.4)²

⇒ EB = 4.8m

EC = EB – BC = 4.8 – 4 = 0.8m

Hence the top of the ladder would slide upwards on the wall at distance 0.8 m.

15. Corresponding sides of two similar triangles are in the ratio of 2 : 3. If the area of the smaller triangle is 48 cm², then the area of the larger triangle is:

(A) 108 m²                                                                          

(B) 107 m²

(C) 106 m²                                                                          

(D) 230 m²

Answer:(A)

Explanation:

According to given Question

ar(Larger Triangle)/ar(Smaller Triangle) = (side of larger triangle/side of larger triangle)²

ar(Larger Triangle)/48 = (3/2)²

ar(Larger Triangle) = (9 x 48 )/4

ar(Larger Triangle) = 108 cm²

Important Link

Quick Revision Notes : Triangles

NCERT Solution : Triangles

MCQs: Triangles

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