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Class 10 Maths MCQ Questions of Constructions with Answers?

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Class 10 Maths MCQ Question of Constructions are given here with answers. Solving these objective questions will help students to score better marks in the board exam. All these MCQ Questions are available online as per the CBSE syllabus and exam pattern. 

Practicing MCQ Questions is the only key to success for better revision. You can also start practice with class 10 sample papers for Exams. Students can solve Class 10 sample papers and MCQ Questions with Answers to know their preparation level.
 

Practice Class 10 Maths MCQ Question of Constructions

1. A point O is at a distance of 10 cm from the centre of a circle of radius 6 cm. How many tangents can be drawn from point O to the circle?
(a) 1
(b) 3
(c) Infinite
(d) 2

2. In division of a line segment AB, any ray AX making angle with AB is

(a) right angle
(b) obtuse angle
(c) any arbitrary angle
(d) acute angle

3. To divide a line segment AB in the ratio 3:4, 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) 5
(b) 7
(c) 9
(d) 11

4. To divide a line segment AB of length 7.6 cm in the ratio 5 : 8, a ray AX is drawn first such that ∠BAX forms 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) A5
(b) A6
(c) A10
(d) A13

5. To construct a triangle similar to a given ΔPQR with its sides 5/8 of the similar sides of ΔPQR, draw a ray QX such that ∠QRX is an acute angle and X lies on the opposite side of P with respect to QR. Then locate points Q1, Q2, Q3, … on QX at equal distances, and the next step is to join:

(a) Q10 to C
(b) Q3 to C
(c) Q8 to C
(d) Q4 to C

6. 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

7. To divide a line segment AB in the 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

8. To draw a pair of tangents to a circle which are inclined to each other at an angle of 35°. It is required to draw tangents at the end points of those two radii of the circle, the angle between which is

(a) 105°
(b) 70°
(c) 140°
(d) 145°

9. To draw a pair of tangents to a circle which are inclined 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°

10. To construct a triangle similar to a given ΔPQR with its sides, 9/5 of the corresponding sides of ΔPQR draw a ray QX such that ∠QRX is an acute angle and X is on the opposite side of P with respect to QR. The minimum number of points to be located at equal distances on ray QX is:

(a) 5

(b) 9

(c) 10

(d) 14

11. To divide a line segment PQ in the ratio m : n, where m and n are two positive integers, draw a ray PX so that ∠PQX is an acute angle and then mark points on ray PX at equal distances such that the minimum number of these points is:

(a) m + n
(b) m – n
(c) m + n – 1
(d) Greater of m and n

12. To draw a pair of tangents to a circle which are inclined to each other at an angle of 45°, it is required to draw tangents at the endpoints of those two radii of the circle, the angle between which is:

(a) 135°
(b) 155°
(c) 160°
(d) 120°

13.  A pair of tangents can be constructed from a point P to a circle of radius 3.5 cm situated at a distance of ___________ from the centre.

(a) 3.5 cm
(b) 2.5 cm
(c) 5 cm
(d) 2 cm

14. To construct a triangle similar to a given ΔABC with its sides 8/5 of the corresponding sides of ΔABC draw a ray BX such that ∠CBX is an acute angle and X is on the opposite side of A with respect to BC. The minimum number of points to be located at equal distances on ray BX is:

(A) 5                                                  
(B) 8
(C) 13                             
(D) 3

15. Which theorem criterion we are using in giving the just the justification of the division of a line segment by usual method ?

(a) SSS criterion
(b) Area theorem
(c) BPT
(d) Pythagoras theorem

16. A point P is at a distance of 8 cm from the centre of a circle of radius 5 cm. How many tangents can be drawn from point P to the circle?

(a) 0
(b) 1
(c) 2
(d) Infinite

17. A line segment drawn perpendicular from the vertex of a triangle to the opposite side is called the

(a) Bisector
(b) Median
(c) Perpendicular
(d) Altitude

18. If the scale factor is 3/5, then the new triangle constructed is _____ the given triangle.

(a) smaller the
(b) greater than
(c) overlaps
(d) congruent to

19. By geometrical construction, which one of the following ratios is not possible to divide a line segment?

(a) 1 : 10
(b) √9 : √4
(c) 10 : 1
(d) 4 + √3 : 4 – √3

20. In constructions, the scale factor is used to construct ______ triangles.

(a) right
(b) equilateral
(c) similar
(d) congruent

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Answer:

1. Answer: (a) 2

2. Answer: (d) acute angle

3. Answer: (b) 7

Explanation: We know that to divide a line segment in the ratio m: n, first draw a ray AX which makes an acute angle BAX, then we are required to mark m + n points at equal distances from each other.

Here, m = 3, n = 4

So, the minimum number of these points = m + n = 3 + 4 = 7

4. Answer: (d) A13

Explanation: The minimum points located in the ray AX is 5 + 8 = 13. Hence, point B will join point A13.

5. Answer: (c) Q8 to C

Explanation: Here we locate points Q1, Q2, Q3, Q4, Q5, Q6, Q7 and Q8 on QX at equal distances and in the next step join the last point Q8 to R.

6. Answer: (D) 12

Explanation: We know that to divide a line segment in the ratio m : n, first draw a ray AX which makes an acute angle BAX , then marked m+n points at equal distances from each other.

Here m = 5, n = 7

So minimum number of these point = m + n = 5 + 7 = 12

7. Answer: (B) A11

Explanation: Here minimum 4+7=11 points are located at equal distances on the ray AX and then B is joined to last point, i.e., A11.

8. Answer: (d) 145°

9. Answer: (d) 120°

10. Answer: (b) 9

Explanation: To draw a triangle similar to a given triangle with its sides m/n of the similar sides of a given triangle, the minimum number of points to be located at an equal distance is equal to m or n, whichever is greater.

Here, m/n = 9/5

9 > 5, therefore the minimum number of points to be located is 9.

11. Answer: (a) m + n

Explanation: To divide a line segment PQ in the ratio m : n, where m and n are two positive integers, draw a ray PX so that ∠PQX is an acute angle and then mark points on ray PX at equal distances such that the minimum number of these points is m + n.

12. Answer: (a) 135°

Explanation: To draw a pair of tangents to a circle which are inclined to each other at an angle of 45°, it is required to draw tangents at the endpoints of those two radii of the circle, the angle between which is 135°.

13. Answer: (c) 5 cm

Explanation: The pair of tangents can be drawn from an external point only, so its distance from the centre must be greater than the radius. Since only 5cm is greater than the radius of 3.5 cm. So the tangents can be drawn from the point situated at a distance of 5 cm from the centre.

14. Answer: (B)

Explanation: To construct a triangle similar to a given triangle with its sides m/n of the corresponding sides of given triangle ,the minimum number of points to be located at equal distance is equal to the greater of m and n in m/n.

Here, m/n = 8/5

So the minimum number of points to be located at equal distance on ray BX is 8.

15. Answer: (c) BPT

16. Answer: (c) 2

Explanation: From the given,

Distance of a point from the centre of the circle > Radius of the circle

So, the point lies outside the circle.

Hence, we can draw 2 tangents to the circle from the point P.

17. Answer: (d) Altitude

18. Answer: (a) smaller than

Explanation: If the scale factor is 3/5, then the new triangle constructed is smaller than the given triangle since in 3/5, the numerator is less than the denominator. 

19. Answer: (d) 4 + √3 : 4 – √3

Explanation: Since the ratio 4 + √3 : 4 – √3 can not be simplified in the form of integers like other given ratios.

20. Answer: (c) similar

Explanation: In constructions, the scale factor is used to construct similar triangles.

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