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) A12
(B) A11
(C) A10
(D) A9
Answer:
(B) A11
Explanation:
We have to divide the line segment into 7 + 4 = 11 equal parts and 11th part will bem joined to B, here A12 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, A1, A2, A3,… and B1, B2, B3, … are located at. Equal distances on ray AX and BY, respectively. Then the points joined are
(A) A5 and B6
(B) A6 and B5
(C) A4 and B5
(D) A5 and B4
Answer:
(A) A5 and B6
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.
PD3 is parallel to:
(A) PD
(B) PE
(C) ED7
(D) None of these.
Answer:
(C) ED7
Explanation:
We have,
PD(C) ED3 || ED7
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) ED3 || ED7
Then, ∠ED7D = ∠PD3D
(Corresponding angles)
∴∠ED7D = 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
[∵ DD1 = D1D2 = D2D3 = D3D4
= D4D5= D5D6 = D6D7]
and DP = 3 m
[∴ DD1 = D1D2 = D2 D3]
∴ \(\frac{D E}{D P}\) = \(\frac{7 \mathrm{~m}}{3 \mathrm{~m}}\) = \(\frac{7}{3}\)
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.