## Constructions Class 10 MCQ Questions with Answers

Question 1.

## To divide a line segment AB in the ratio 5 : 7, first a ray AX is drawn so that ∠BAX is an acute angle and then at equal distances points are marked on the ray AX such that the minimum number of these points is

(A) 8

(B) 10

(C) 11

(D) 12

Answer:

(D) 12

Explanation:

Minimum number of the points marked = sum of ratios = 5 + 7 = 12.

Question 2.

## To divide a line segment AB in ratio 4 : 7, a ray AX is drawn first such that ∠ BAX is an acute angle and then points A1, A2, A3, … are located at equal distances on the ray AX and the point B is joined to :

(A) A_{12}

(B) A_{11}

(C) A_{10}

(D) A_{9}

Answer:

(B) A_{11}

Explanation:

We have to divide the line segment into 7 + 4 = 11 equal parts and 11th part will bem joined to B, here A_{12} will never appear.

Question 3.

## To divide a line segment AB in the ratio 5 : 6 draw a ray AX such that∠BAX is an acute angle, then draw a ray BY parallel to AX, and the points, A_{1}, A_{2}, A_{3},… and B_{1}, B_{2}, B_{3}, … are located at. Equal distances on ray AX and BY, respectively. Then the points joined are

(A) A_{5} and B_{6}

(B) A_{6} and B_{5}

(C) A_{4} and B_{5}

(D) A_{5} and B_{4}

Answer:

(A) A_{5} and B_{6}

Explanation:

In the figure; segment AB of given length is divided into 2 parts of ratio 5 : 6 in following steps:

(i) Draw a line-segment AB of given length.

(ii) Draw an acute angle BAX as shown in figure either upside or down side.

(iii) Draw angle ∠ABY = ∠BAX on other side of AX, that is, down side.

(iv) Divide AX into 5 equal parts by using compass.

(v) Divide BX into same distance in 6 equal parts as AX was divided.

(vi) Now, join A5 and B6 which meet AB at P. P divides AB in ratio AP : PB = 5 : 6.

Question 4.

## To draw a pair of tangents to a circle which are in¬clined to each other at an angle of 60°, it is required to draw tangents at end points of those two radii of the circle, the angle between them should be:

(A) 135°

(B) 90°

(C) 60°

(D) 120°

Answer:

(D) 120°

Explanation:

We know that tangent and radius at contact point are perpendicular to each other. So, ∠P and ∠Q in quadrilateral TPOQ formed by tangents and radii will be of 90° each.

So, the sum of

∠T+ ∠O = 180°

as T = 60° [Given]

∴∠O = 180° – 60°

= 120°

Case – Based MCQs

Attempt any four sub-parts from each question. Each sub-part carries 1 mark.

I. Read the following text and answer the below questions:

A school conducted Annual Sports Day on a triangular playground. On the ground, parallel lines have been drawn with chalk powder at a distance of 1 m. 7 flower pots have been placed at a distance of from each other along DM as shown in the figure.

Now answer the following questions:

Question 1.

## PD_{3} is parallel to:

(A) PD

(B) PE

(C) ED_{7}

(D) None of these.

Answer:

(C) ED_{7}

Explanation:

We have,

PD(C) ED_{3} || ED_{7}

_{}

Question 2.

## If ∠PD3D = 82°, then the measure of ∠ED7D is:

(A) 98°

(B) 82°

(C) 90°

(D) 45°

Answer:

(B) 82°

Explanation:

We have,

PD(C) ED_{3} || ED_{7}

Then, ∠ED_{7}D = ∠PD_{3}D

(Corresponding angles)

∴∠ED_{7}D = 82°.

Question 3.

## The ratio in which P divides DE, is:

(A) 3 : 4

(B) 7 : 3

(C) 3 : 7

(D) 2 : 5

Answer:

(A) 3 : 4

Explanation:

P divides DE in the ratio 3 : 4.

Question 4.

## The ratio of DE to DP will be:

(A) 2:5

(B) 3 : 4

(C) 3 : 7

(D) 7 : 3

Answer:

(D) 7 : 3

Explanation:

DE = 7 m

[∵ DD_{1} = D_{1}D_{2} = D_{2}D_{3} = D_{3}D_{4}

= D_{4}D_{5}= D_{5}D_{6} = D_{6}D_{7}]

and DP = 3 m

[∴ DD_{1} = D_{1}D_{2} = D_{2} D_{3}]

∴ \(\frac{D E}{D P}\) = \(\frac{7 \mathrm{~m}}{3 \mathrm{~m}}\) = \(\frac{7}{3}\)

Hence, the ratio of DE to DP is 7 : 3.

Question 5.

## The total distance used for putting 7 flower pots is:

(A) 6 m

(B) 7 m

(C) 5 m

(D) 8 m.

Answer:

(B) 7 m

Explanation:

Since, 7 flower pots have been placed at a distance of 1 m from each other, then total distance = 7 m.