Practice Test: Triangles - Class 10 MCQ

# Practice Test: Triangles - Class 10 MCQ

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## 10 Questions MCQ Test Mathematics (Maths) Class 10 - Practice Test: Triangles

Practice Test: Triangles for Class 10 2024 is part of Mathematics (Maths) Class 10 preparation. The Practice Test: Triangles questions and answers have been prepared according to the Class 10 exam syllabus.The Practice Test: Triangles MCQs are made for Class 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Practice Test: Triangles below.
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Practice Test: Triangles - Question 1

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

Detailed Solution for Practice Test: Triangles - Question 1

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.

Practice Test: Triangles - Question 2

### 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.

Detailed Solution for Practice Test: Triangles - Question 2

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)

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Practice Test: Triangles - Question 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?

Detailed Solution for Practice Test: Triangles - Question 3

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

Practice Test: Triangles - Question 4

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:

Detailed Solution for Practice Test: Triangles - Question 4

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

Practice Test: Triangles - Question 5

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:

Detailed Solution for Practice Test: Triangles - Question 5

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

Practice Test: Triangles - Question 6

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:

Detailed Solution for Practice Test: Triangles - Question 6

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

Practice Test: Triangles - Question 7

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

Detailed Solution for Practice Test: Triangles - Question 7

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

Practice Test: Triangles - Question 8

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

Detailed Solution for Practice Test: Triangles - Question 8

The altitude divides the opposite side into two equal parts,
Therefore, BD = DC = 4 cm

In triangle ABD

Practice Test: Triangles - Question 9

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

Detailed Solution for Practice Test: Triangles - Question 9

∠A=∠A
∠ACB = ∠CDA
Therefore triangle ACB and ADC are similar,
Hence
82 = 3 x AB
⇒ AB = 64/3
This implies,
⇒ BD = 55/3

Practice Test: Triangles - Question 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

Detailed Solution for Practice Test: Triangles - Question 10

In ΔALD, we have
∴ LB/BA = LP/PD
⇒ BL/AB = PL/DP
⇒ BL/DC = PL/DP [∵ AB = DC
⇒ DP/PL = DC/BL

## Mathematics (Maths) Class 10

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## Mathematics (Maths) Class 10

116 videos|420 docs|77 tests