# Class 10 Maths Chapter 11 Constructions MCQs

**Class 10 Maths Chapter 11 MCQs (Constructions)** are available online here with answers. All these multiple-choice questions are prepared as per the latest CBSE syllabus (2021 – 2022) and NCERT guidelines. Solving these objective type questions will help students to score good marks in the board exam, which they can verify with the help of detailed explanations provided here. To get all class 10 Maths chapter-wise MCQs, Click here.

## Class 10 Maths MCQs for Constructions

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**1. 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

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

**2. 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 A _{1}, A_{2}, A_{3}, ….are located at equal distances on the ray AX and the point B is joined to:**

(a) A_{5}

(b) A_{6}

(c) A_{10}

(d) A_{13}

Answer: **(d) A _{13}**

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

**3. 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 Q _{1}, Q_{2}, Q_{3}, … on QX at equal distances, and the next step is to join:**

(a) Q_{10} to C

(b) Q_{3} to C

(c) Q_{8} to C

(d) Q_{4} to C

Answer: **(c) Q _{8} to C**

Explanation: Here we locate points Q_{1}, Q_{2}, Q_{3}, Q_{4}, Q_{5}, Q_{6}, Q_{7} and Q_{8} on QX at equal distances and in the next step join the last point Q_{8} to R.

**4. 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

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.

**5. To construct a pair of tangents to a circle at an angle of 60° to each other, it is needed to draw tangents at endpoints of those two radii of the circle, the angle between them should be:**

(a) 100°

(b) 90°

(c) 180°

(d) 120°

Answer:** (d) 120°**

Explanation: The angle between the two radii should be 120° because the figure produced by the intersection point of pair of tangents and the two endpoints of those two radii and the centre of the circle, is a quadrilateral. Hence, the sum of the opposite angles should be 180°.

**6. 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

Answer: **(a) m + n**

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.

**7. 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°

Answer: **(a) 135°**

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°.

**8. 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

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.

**9. To construct a triangle ABC and then a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle. A ray AX is drawn where multiple points at equal distances are located. The last point to which point B will meet the ray AX will be:**

(a) A_{1}

(b) A_{2}

(c) A_{3}

(d) A_{4}

Answer: **(c) A _{3}**

Explanation: The greater of 2 or 3 will be the maximum number of points. Hence, the last point will be A_{3}.

**10. To construct a triangle similar to a given ΔPQR with its sides 3/7 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 Q _{1}, Q_{2}, Q_{3}, … on QX at equal distances, and the next step is to join:**

(a) Q_{10} to C

(b) Q_{3} to C

(c) Q_{7} to C

(d) Q_{4} to C

Answer: **(c) Q _{7} to C**

Explanation: Here we locate points Q_{1}, Q_{2}, Q_{3}, Q_{4}, Q_{5}, Q_{6} and Q_{7} and QX at equal distances and in next step join the last point Q_{7} to R.

**11. 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

Answer: **(a) smaller than**

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.

**12. 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:

Given ratio is 5 : 6.

That means, we have to locate 5 points such as A_{1}, A_{2}, A_{3}, A_{4}, A_{5} on the ray AX and 6 points such as B_{1}, B_{2}, B_{3}, B_{4}, B_{5}, B_{6}, on BY.

Also, we need to join A5 and B6.

**13. 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

Answer: **(d) 4 + √3 : 4 – √3**

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

**14. By geometrical construction, is it possible to divide a line segment in the ratio 1/√3 : √3?**

(a) Yes

(b) No

(c) Cannot be determined

(d) None of these

Answer: **(a) Yes**

Given ratio is 1/√3 : √3

Multiplying by √3, we get

(1/√3) × √3 : √3 × √3

1 : 3

Thus, the simplified ratio contains only integers.

Hence, the geometrical construction makes it possible to divide the given line segment.

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

(a) right

(b) equilateral

(c) similar

(d) congruent

Answer: **(c) similar**

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

**16. In the division of a line segment AB, any ray AX making angle with AB is _______.**

(a) an acute angle

(b) a right angle

(c) an obtuse angle

(d) reflex angle

Answer: **(a) an acute angle.**

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

**17. 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

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.

**18. To divide a line segment AB in the ratio p : q, 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 9. Here, the possible ratio of p : q is**

(a) 3 : 5

(b) 4 : 7

(c) 2 : 9

(d) 5 : 4

Answer: **(d) 5 : 4**

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 need to mark m + n points at equal distances from each other.

By verifying all the given options, the ratio which satisfies p + q = 9 is 5 : 4.

**19. A line segment drawn perpendicular from the vertex of a triangle to the opposite side is known as**

(a) altitude

(b) median

(c) bisector of side

(d) radius of incircle of the triangle

Answer: **(a) altitude**

A line segment drawn perpendicular from the vertex of a triangle to the opposite side is known as altitude.

**20. ****If the line segment is divided in the ratio 3 : 7, then how many parts does it contain while constructing the point of division?**

(a) 3

(b) 7

(c) 4

(d) 10

Answer: **(d) 10**

Explanation:

The line segment is divided in the ratio 3: 7 means, it contains 3 parts on one side and 7 parts on the other side of the point of division. Hence, there will be a total of (3 + 7) parts, i.e. 10 parts.

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