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Page No 143:

Question 1:

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°

Answer:




PA and PB are two tangents drawn from P to circle with centre O.

∠APB = 60°

Now, 

∠OAP = 90°     (Radius is perpendicular to the tangent at the point of contact)

∠OBP = 90°     (Radius is perpendicular to the tangent at the point of contact)

In quadrilateral OAPB,

∠AOB + ∠OAP + ∠OBP + ∠APB = 360º         (Angle sum property of quadrilateral)

⇒ ∠AOB + 90º + 90º + 60º = 360º

⇒ ∠AOB = 360º − 240º = 120º

Thus, in order 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 120°.

Hence, the correct answer is option (d).

Page No 143:

Question 2:

To divide a line segment AB in the ratio 5:7, first 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:


We know that, in order to divide a line segment AB in the ratio m : n (m, n are positive integers), first draw a ray AX such 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 m + n.

Here, m = 5 and n = 7

m + n = 5 + 7 = 12

Thus, to divide a line segment AB in the ratio 5 : 7, first 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 12.

Hence, the correct answer is option (d).

Page No 143:

Question 3:

To divide a line segment AB in the ratio m : n (m, n are positive integers), draw a ray AX such 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 than m and n
(b) m + n
(c) m + n – 1
(d) mn

Answer:


To divide a line segment AB in the ratio m : n (m, n are positive integers), firstly draw a ray AX such that ∠BAX is an acute angle. Then, along AX mark off m + n points at equal distances.



Here, AP : PB = m : n

Thus, to divide a line segment AB in the ratio m : n (m, n are positive integers), draw a ray AX such 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 m + n.

Hence, the correct answer is option (b).



Page No 144:

Question 4:

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:




PA and PB are two tangents drawn from P to circle with centre O.

∠APB = 35°

Now, 

∠OAP = 90°     (Radius is perpendicular to the tangent at the point of contact)

∠OBP = 90°     (Radius is perpendicular to the tangent at the point of contact)

In quadrilateral OAPB,

∠AOB + ∠OAP + ∠OBP + ∠APB = 360º         (Angle sum property of quadrilateral)

⇒ ∠AOB + 90º + 90º + 35º = 360º

⇒ ∠AOB = 360º − 215º = 145º

Thus, in order 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 145º.

Hence, the correct answer is option (d).

Page No 144:

Question 5:

To construct a triangle similar to a given ∆ABC with its side 8/5 of the corresponding side of ∆ABC 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:


In order to construct a triangle similar to a given triangle with its sides mn of the corresponding sides of the triangle, the minimum number number of points to be located at equal distances on the ray is m or n, whichever is greater.

If m > n, then minimum points to be located at equal distances on the ray is m.

If n > m, then minimum points to be located at equal distances on the ray is n.

Here, m = 8 and n = 5

8 > 5

Thus, in order to construct a triangle similar to a given ∆ABC with its side 85 of the corresponding side of ∆ABC, 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 8.

Hence, the correct answer is option (b).

Page No 144:

Question 6:

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:


To divide a line segment AB in the ratio m : n (m, n are positive integers), firstly draw a ray AX such that ∠BAX is an acute angle. Now, along AX mark off m + n points at equal distances.
Suppose these m + n points marked on ray AX be A1, A2, A3, ..., Am, Am + 1, ..., Am + n. Then, join B to Am + n.

Here, m = 4 and n = 7

m + n = 4 + 7 = 11

So, B is joined to the point A11.

Thus, in order 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 A11.

Hence, the correct answer is option (b). 

Page No 144:

Question 7:

To construct a triangle similar to a given ∆ABC with its sides 3/7 of the corresponding sides of ∆ABC, 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:


In order to construct a triangle similar to a given triangle with its sides mn of the corresponding sides of the triangle, the minimum number number of points to be located at equal distances on the ray is m or n, whichever is greater.

If m > n, then minimum points to be located at equal distances on the ray is m.

If n > m, then minimum points to be located at equal distances on the ray is n.

Here, m = 3 and n = 7

7 > 3

So, seven points B1, B2, B3, B4, B5, B6, B7 are marked at equal distance on BX. Then B7 is joined to C.

Thus, in order to construct a triangle similar to a given ∆ABC with its sides 37 of the corresponding sides of ∆ABC, 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,...., B7 on BX at equal distances and next step is to join B7 to C.

Hence, the correct answer is option (c).

Page No 144:

Question 8:

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, .... and B1, B2,..... are located at equal distances on rays 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:


To divide a line segment AB in the ratio m : n, draw a ray AX such that ∠BAX is an acute angle. Then draw a ray BY parallel to AX. The points A1, A2, ..., Am and B1, B2, ..., Bn are located at equal distances on rays AX and BY respectively. Then the points Am and Bn are joined to divide the line segment AB in the ratio m : n.



Here, m = 5 and n = 6

So, the points A1, A2, ..., A and B1, B2, ..., B6 are located at equal distances on rays AX and BY respectively. Then the points A5 and B6 are joined.

Thus, in order 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, ..., A5 and B1, B2, ..., B6 are located at equal distances on rays AX and BY respectively. Then the points joined are A5 and B6.

Hence, the correct answer is option (a).

Page No 144:

Question 9:

To divide a line segment AB internally in the ratio 5 : 2, first a ray AX is drawn so that ∠BAX is an acute angle and then points A1, A2, A3, ... are located at equal distance on ray AX and point B is joined to 
(a) A6
(b) A7
(c) A3
​(d) A2

Answer:

Given, a line segment AB in the ratio 5 : 2
A : B = 5 : 2
Minimum number of points located at equal distances on the ray AX = Sum of ratio
                                                                                                              = 5 + 2
                                                                                                              = 7
Here, A1A2A3, ... are located at equal distances on the ray AX.
Point B is joined to the last point is A7.

Hence, the correct answer is option (b).

Page No 144:

Question 10:

To divide a line segment AB internally in the ratio 4 : 7 first a ray AX is drawn so that ∠BAX is an acute angle and then at equal distances points are marked on ray AX such that the minimum number of these points is 
(a) 9
(b) 10
(c) 11
​(d) 12  ​

Answer:

Given, a line segment AB in the ratio 4 : 7.

Now, draw a ray AX which makes an acute angle ∠BAX, then mark a + b points at equal distance.

Here, a = 4 and b = 7.

Therefore, the minimum number of these points = a + b
                                                                              = 4 + 7
                                                                              = 11
Hence, the correct answer is option (c).

Page No 144:

Question 11:

To divide a line segment AB in the ratio 3 : 2, draw a ray AX such that ∠BAX is an acute angle, then draw ray BY parallel to AX and then locate points A1, A2, A3 ... and B1, B2, B3 ... are equal distances on ray AX and BY respectively. Then the points to be joined are
(a) A3 and B2
(b) Aand B3
(c) Aand B3
​(d) Aand B1

Answer:

Given, a line segment AB in the ratio 3 : 2.
a : b = 3 : 2

To divide a line segment AB in the ratio 3 : 2, the third division of AX and the second division of BY are joined.
Thus, A3 and B2 are joined. 

Hence, the correct answer is option (a).

Page No 144:

Question 12:

To construct a triangle similar to a given triangle such that ∠CBX is an acute angle and X lines on the opposite side of A with respect to BC. Then locate points X1, X2, X3... at equal distance on BX. The points to be joined in the next step are

(a) X4 and C
(b) X1 and C
(c) Xand C
​(d) X4 and C 

Answer:

Ans

Page No 144:

Question 13:

To construct a triangle similar to a given ΔABC with its sides 75 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. Then locate points X1X2X3... at equal distance on BX. The points to be joined in the next step are
(a) X7 and C
(b) X5 and C
(c) Xand C
​(d) X12 and C 

Answer:

Steps:
1. Locate points X1, X2, X3, X4, X5, X6 and X7 on BX at equal distances.

2. Join the last point X5 (the 5th point, 5 being smaller of 5 and 7 in 75) to C.



Hence, the correct answer is option (b).



Page No 145:

Question 14:

If two a 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 
(a) 100°
(b) 60°
(c) 90°
​(d) 120°

Answer:

Given: The two radii of a circle are inclined to each other at 120°.
Now, make the figure as per the given information.

Here, OP and OQ are the tangents to the circle.
Since the tangent to a circle is perpendicular to the radius through the point of contact.
Thus, OP ⊥ PR and  OQ ⊥ QR

In quadrilateral PRQO, by using the sum of angles of a quadrilateral, we get
∴ ∠POQ + ∠OQR + ∠QRP + ∠RPO = 360°
⇒ 120° + 90° + θ + 90° = 360°
θ = 60°
Thus, the angle between them should be 60°.

Hence, the correct answer is option (b).

Page No 145:

Question 15:

To construct a cyclic quadrilateral ABCD in which ∠B = 90°, if a circle on which points A, B, C and D lie, has to be drawn, then the centre of this circle is
(a) the mid-point of diagonal AC
(b) the mid-point of diagonal BD
(c) the point of intersection of diagonals AC and BD
​(d) a  point which lies neither on AC nor on BD

Answer:

Given that, ABCD is a cyclic quadrilateral with ∠B = 90° and the points A, B, C and D lie on the circle.
Since ABCD is a cyclic quadrilateral, the sum of its opposite angles is 180°. As ∠B = 90°, the angle opposite to it, i.e, ∠D is also a right angle.
⇒ ∠D = 90°



The two angles ∠B and ∠D are formed by the chord AC on the circumference of the circle. Now, the only chord that subtends a right angle on the circumference of the circle is the diameter. Thus, AC is the diameter of the circle.

This means that the centre of the circle is the mid-point of the diagonal AC of the cyclic quadrilateral ABCD.

Hence, the correct answer is option (a).

Page No 145:

Question 16:

To draw a pair of tangents to a circle which are right angles to each other, it is required to draw tangent at end points of the two radii of the circle, which are inclined at an angle of 
(a) 45°
(b) 120°
​(c) 60°
​(d) 90°

Answer:

Given: The pair of tangents to a circle which are inclined to each other at 90°.
Now, make the figure as per the given information.

Here, OP and OQ are the tangents to the circle.
Since the tangent to a circle is perpendicular to the radius through the point of contact.
Thus, OP ⊥ PR and  OQ ⊥ QR

In quadrilateral PRQO, by using the sum of angles of a quadrilateral, we get
∴ ∠POQ + ∠OQR + ∠QRP + ∠RPO = 360°
⇒ 90° + 90° + θ + 90° = 360°
θ = 90°
Thus, the angle between them should be 90°.
Hence, the correct answer is option (d).

Page No 145:

Question 17:

To draw a pair of tangents to a circle which are inclined to each other at angle x°, it is require to draw tangent at the end points of those two radii of the circle, the angle between which is 
(a) 90° – x°
(b) 90° + x°
​(c) 180° – x°
​(d) 180° + x°

Answer:

First, make the figure as per the given information.

Here, OP and OQ are the tangents to the circle.
Since the tangent to a circle is perpendicular to the radius through the point of contact.
Thus, OP ⊥ PR and  OQ ⊥ QR

In quadrilateral PRQO, by using the sum of angles of a quadrilateral, we get
∴ ∠POQ + ∠OQR + ∠QRP + ∠RPO = 360°
x° + 90° + θ + 90° = 360°
θ = 180° − x°

Thus, the angle between them should be 180° − x°.
Hence, the correct answer is option (c).

Page No 145:

Question 18:

​If you draw a pair of tangents to a circle C(O, r) from point P such that OP = 2r, then the angle between the two tangents is 
(a) 90°
(b) 30°
​(c) 60°
​(d) 45°

Answer:




In ∆AOP,
sinAPO=AOPOsinAPO=r2rsinAPO=12sinAPO=sin30°            sin30°=12  APO=30°


In ∆BOP,
sinBPO=BOPOsinBPO=r2rsinBPO=12sinBPO=sin30°        sin30°=1BPO=30°

APB=APO+BPOθ=30°+30°=60°

Thus, the angle between tangents is 60°.

Hence, the correct answer is option (c).

Page No 145:

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
(a) a circle with AB as diameter 
(b) a circle with AM as diameter 
​(c) a circle with MB as diameter 
​(d) extend AB to P such that BP = MB and draw a circle with MP as diameter

Answer:

Draw circles with radii 3 cm and 2 cm, such that the distance between their centres A and B is 6 cm.
A perpendicular bisector of AB is drawn intersecting AB at M.

Next step is to construct a circle with centre M and AB as diameter.


Thus, the next step is to draw a circle with AB as diameter.
Hence, the correct answer is option (a).

Page No 145:

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
(a) mid-point of q
(b) mid-point of p
​(c) mid-point of q – r
​(d) mid-point of pq

Answer:

Step 1:
Draw a circle of radius p with centre as O.

Step 2:

Draw a circle of  radius q taking O as its centre. Locate a point P on this circle and join OP.

Step 3:

Bisect OP. Let M be the mid-point of PO, i.e. q.



Hence, the correct answer is option (a).

Page No 145:

Question 21:

​​To draw a tangents at point B to the circumcircle of an isosceles right ΔABC right angled at B, we need to draw through B
(a) a line parallel to AC
(b) a line perpendicular to AB
​(c) a line perpendicular to BC
​(d) a line inclined at 60° to AB

Answer:

In â–³ABC,

 AB = BC                    (Given)
∴  ∠ACB = ∠BAC     .....(1) (Angles opposite to equal sides are equal)
 
According to alternate segment theorem, the angle between the tangent and chord at the point of contact is equal to the angles made by the chord in the corresponding alternative segment.
DBE is the tangent and BC is the chord.
∴  ∠EBC = ∠BAC     .....(2) 
 
From (1) and (2),
⇒ ∠ACB = ∠EBC
If a transversal intersects two lines such that a pair of alternate interior angles are equal, then two lines are parallel.
∴ DE ∥ AC
∴  The tangent at B to the circumcircle of â–³ABC is parallel to AC.



Hence, the correct answer is option (a).



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