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Test: Constructions - 2 - Class 10 MCQ


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25 Questions MCQ Test Mathematics (Maths) Class 10 - Test: Constructions - 2

Test: Constructions - 2 for Class 10 2024 is part of Mathematics (Maths) Class 10 preparation. The Test: Constructions - 2 questions and answers have been prepared according to the Class 10 exam syllabus.The Test: Constructions - 2 MCQs are made for Class 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Constructions - 2 below.
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Test: Constructions - 2 - Question 1

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:

Detailed Solution for Test: Constructions - 2 - Question 1

According to the question, point B is joined to A11

Test: Constructions - 2 - Question 2

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

Detailed Solution for Test: Constructions - 2 - Question 2

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

Test: Constructions - 2 - Question 3

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

Detailed Solution for Test: Constructions - 2 - Question 3

Basic Proportionality Theorem (BPT) criterion we are using in giving the just the justification of the division of a line segment by usual method.

Test: Constructions - 2 - Question 4

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

Detailed Solution for Test: Constructions - 2 - Question 4

According to the question, the minimum number of those points which are to be marked should be (Numerator + Denominator) i.e., 5 + 7 = 12

Test: Constructions - 2 - Question 5

To divide a line segment AB in the ration 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

Detailed Solution for Test: Constructions - 2 - Question 5

According to the question, the point B is joined to A11.

Test: Constructions - 2 - Question 6

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

Detailed Solution for Test: Constructions - 2 - Question 6

In the figure,AA1  =A1A= A2A= A3C = 1 : 4

Test: Constructions - 2 - Question 7

To divide line segment AB in the ratio A : b ( a, b are positive integers), draw a ray AX so that ∠BAX is an acute angle and then mark points on ray AX at equal distances such that the minimum number of these points is

Detailed Solution for Test: Constructions - 2 - Question 7

According to the question, the minimum number of those points which are to be marked should be (Numerator + Denominator) i.e., (a + b)

Test: Constructions - 2 - Question 8

To divide a line segment AB in the ration 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

Detailed Solution for Test: Constructions - 2 - Question 8

According to the question, the points joined are A6 to B5. Because if we have to divide a line segment AB in the ratio m : n, then we draw rays AX and BY and mark the points A1, A2, ……, Am and B1, B2, ……, Bn on rays AX and BY respectively. Then we join the point Am to Bn

Test: Constructions - 2 - Question 9

To divide a line segment AB in the ration 2 : 3, 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 tha the minimum number of these points is

Detailed Solution for Test: Constructions - 2 - Question 9

According to the question, the minimum number of those points which are to be marked should be (Numerator + Denominator) i.e., 2 + 3 = 5

Test: Constructions - 2 - Question 10

To divide a line segment AP in the ration 2 : 9, a ray AX is drawn first such that ∠BAX is an acute angle and then points A1,A2,A3... are located of equal distances on the ray AX and the points P is joined to

Detailed Solution for Test: Constructions - 2 - Question 10

According to the question, the points P is joined to A11

 

Test: Constructions - 2 - Question 11

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 :

Detailed Solution for Test: Constructions - 2 - Question 11

According to the question, the angle between the radii should be

Test: Constructions - 2 - Question 12

To draw a pair of tangents to circle which are inclined to each other at 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 :

Detailed Solution for Test: Constructions - 2 - Question 12

According to the question, the angle between the radii should be 180°−60°=120°

Test: Constructions - 2 - Question 13

A draw a pair of tangents to a circle which are inclined to each other at an angle of 65°, it is required to draw tangents at the end points of those two radii of the circle, the angle between which is :

Detailed Solution for Test: Constructions - 2 - Question 13

According to the question, the angle between the radii should be 180°−65°=115°65°

Test: Constructions - 2 - Question 14

To draw a pair tangents to a circle which are inclined to each other at an angle of 70°, it is required to draw tangents at end points of those two radii of the circle, the angle between them should be 

Detailed Solution for Test: Constructions - 2 - Question 14

According to the question, the angle between the radii should be 180°−70° = 110°70°

Test: Constructions - 2 - Question 15

If two tangents are drawn at the end points of two radii of a circle which are inclined at 120° to each other, then the pair of tangents will be inclined to each other at an angle of

Detailed Solution for Test: Constructions - 2 - Question 15

According to the question, the pair of tangents will be inclined to each other at an angle of 180°−120° = 60°

Test: Constructions - 2 - Question 16

To draw a pair of tangents to a circle which are at right angles to each other, it is required to draw tangents at end points of the two radii of the circle, which are inclined at an angle of

Detailed Solution for Test: Constructions - 2 - Question 16

According to the question, the tangents are inclined at an angle of 180°−90° = 90°

Test: Constructions - 2 - Question 17

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 end points of the two radii of the circle, which are inclined at an angle of

Detailed Solution for Test: Constructions - 2 - Question 17

According to the question, the angle between radii is 180°−45° = 135°

Test: Constructions - 2 - Question 18

To draw a pair of tangents to a circle which are inclined to each other at angle x°, it is required to draw tangents at the end points of those two radii of the circle, the angle between which is

Detailed Solution for Test: Constructions - 2 - Question 18

To draw a pair of tangents to a circle which are inclined to each other at angle x°, it is required to draw tangents at the end points of those two radii of the circle, the angle between them is180°−x°

Test: Constructions - 2 - Question 19

To draw tangents to each of the circle with radii 3 cm and 2 cm from the centre of the other circle, such that the distance between their centres A and B is 6 cm, a perpendicular bisector of AB is drawn intersecting AB at M. The next step is to draw

Detailed Solution for Test: Constructions - 2 - Question 19

According to the question, the next step is to draw is a circle with AB as diameter.

Test: Constructions - 2 - Question 20

To draw tangents to a circle of radius ‘p’ from a point on the concentric circle of radius ‘q’, the first step is to find

Detailed Solution for Test: Constructions - 2 - Question 20

To draw tangents to a circle of radius ‘p’ from a point on the concentric circle of radius ‘q’, the first step is to find midpoint of q.

Test: Constructions - 2 - Question 21

To draw a tangent at point B to the circumcircle of an isosceles right ΔABCright angled at B, we need to draw through B

Detailed Solution for Test: Constructions - 2 - Question 21

To draw a tangent at point B to the circumcircle of an isosceles right ΔABC right angled at B, we need to draw through B in parallel to AC.

Test: Constructions - 2 - Question 22

A pair of tangents can be constructed to a circle inclined at an angle of :

Detailed Solution for Test: Constructions - 2 - Question 22

A pair of tangents can be constructed to a circle inclined at an angle of less than 180°. Here it is 175°.

Test: Constructions - 2 - Question 23

A pair of tangents can be constructed to a circle from an external point, which are inclined each other at an angle θ such that :

Detailed Solution for Test: Constructions - 2 - Question 23

A pair of tangents can be constructed to a circle from an external point, which are inclined each other at an angle θ such that 0<θ<180°.

Test: Constructions - 2 - Question 24

Two distinct tangents can be constructed from a point P to a circle of radius 2r situated at a distance:

Detailed Solution for Test: Constructions - 2 - Question 24

If we have to draw the tangents from any external point of the circle, then the distance of the external point from the centre should be more than the radius of the circle.
Therefore, two distinct tangents can be constructed from a point P to a circle of radius 2r situated at a distance more than 2r from the centre.

Test: Constructions - 2 - Question 25

Two circles touch each other externally at C and AB is a common tangent to the circles. Then, ∠ACB =

Detailed Solution for Test: Constructions - 2 - Question 25

Here OA ⊥ AB and O'B ⊥ AB, then ∠OAB = ∠O'BA = 90° And ∠AOO' = ∠BO'O = 90°

 ∴ ABO'O is a rectangle.
Also, OC = OA, then ∠OCA = ∠OAC= x [Let] ∴ 90° (x + x) = 180°
⇒ x = 45° Again, O'C = O'B, then ∠OC'B = ∠O'BC = y [Let] ∴ 90° (y + y) = 180° 
⇒ y =  45°
Now, ∠OCA + ∠ACB + ∠BCO' = 180° [Straight angle]
⇒ 45° +  ∠ACB + 45° = 180°
⇒ ∠ACB = 90° 

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