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)

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

2. 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

Answer: (B)

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., A_11.

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)

Explanation:   Observe the following figure:

4. To construct a triangle similar to a given ΔABC with its sides 3/7 of the corresponding sides of ΔABC, first draw a ray BX such that ∠CBX is an acute angle and X lies on the opposite side of A with respect to BC. Then locate points B₁, B₂, B₃, … on BX at equal distances and next step is to join:

(A) B₁₀ to C                            

(B) B₃ to C

(C) B₇ to C              

(D) B₄ to C

Answer: (C)

Explanation: Here we locate points B₁,B₂,B₃,B₄,B₅,B₆ and B₇ on BX at equal distances and in next step join the last point B₇ to C

5. 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

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.

6. 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) 120⁰

Answer:  (D)

Explanation: The angle between them should be 120⁰ because the figure formed by the intersection point of pair of tangents, the two end points of those two radii (at which tangents are drawn) and the centre of circle, is a quadrilateral. Thus the sum of the opposite angles in this quadrilateral must be 180^o.

7. To divide a line segment AB in the ratio p : q (p, q 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

(A) greater of p and q            

(B) p + q

(C) p + q – 1                            

(D) pq

Answer:  (B)

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

Here m = p, n = q

So minimum number of these points = m + n = p + q

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°

Answer:  (D)

Explanation: The angle between them should be 145⁰ because the figure formed by the intersection point of pair of tangents, the two end points of those two radii (at which tangents are drawn) and the centre of circle, is a quadrilateral. Thus the sum of the opposite angles in this quadrilateral must be 180^o.

9. By geometrical construction, it is possible to divide a line segment in the ratio:

Answer:  (A)

Explanation:

10. 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) 5cm                                              

(B) 2cm

(C) 3cm                                            

(D) 3.5cm

Answer:  (A)

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

11. To divide a line segment AB in the ratio 5:6, draw a ray AX such that ∠BAX is an acute angle, then drawa ray BY parallel to AX and the points A₁, A₂, A₃,…. and B₁, B₂, B₃,…. are located to 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)

Explanation:

To divide line segment AB in the ratio 5:6.

Steps of construction

1. Draw a ray AX making an acute ∠BAX.

2. Draw a ray BY parallel to AX by taking ∠ABY equal to ∠BAX.

3. Divide AX into five (m = 5) equal parts AA₁, A₁A₂, A₂A₃, A₃A₄ and A₄A₅

4. Divide BY into six (n = 6) equal parts and BB₁, B₁B₂, B₂B₃, B₃B₄, B₄B₅ and B₅B₆.

5. Join B₆ A₅. Let it intersect AB at a point C.

Then, AC : BC = 5 : 6

12. A rhombus ABCD in which AB = 4cm and ABC = 60^o, divides it into two triangles say, ABC and ADC. Construct the triangle AB’C’ similar to triangle ABC with scale factor 2/3. Select the correct figure.

Answer:  (A)

13. A triangle ABC is such that BC = 6 cm, AB = 4 cm and AC = 5 cm. For the triangle similar to this triangle with its sides equal to (3/4)th of the corresponding sides of ΔABC, correct figure is:

Answer:  (D)

14. For ∆ABC in which BC = 7.5 cm, ∠B =45⁰ and AB – AC = 4, select the correct figure.

(D) None of these

Answer:  (B)

15. Draw the line segment AB = 5 cm. From the point A draw a line segment AD = 6cm making an angle of 60⁰ with AB. Draw a perpendicular bisector of AD. Select the correct figure.

(D) None of these

Answer:  (A)

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