Math Ncert Exemplar 2019 Solutions for Class 9 Maths Chapter 10 - Circles

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Page No 99:

Question 1:

AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the centre of the circle is :
(A) 17 cm
(B) 15 cm
(C) 4 cm
(D) 8 cm

Answer:

Given that, AD = 34 cm and AB = 30 cm. First, draw OL $⊥$ AB.
As, the perpendicular form the center of a circle to a chord bisects the chord.

$∴ AL = LB = \frac{1}{2} AB = 15 cm$
Now, $△ OLA$ is right angled, then
OA² = OL² + AL² [by Pythagoras theorem]
$\begin{array}{l} \Rightarrow {(17)}^{2} = {OL}^{2} + {(15)}^{2} \\ \Rightarrow 289 = {OL}^{2} + 255 \\ \Rightarrow {OL}^{2} = 289 - 225 = 64 \\ ∴ OL = 8 cm [taking positive square root, because length is always positive] \end{array}$

Thus, the distance of the chord form the center is 8 cm.
Hence, the correct answer is option D.

Page No 99:

Question 2:

In the figure, if OA = 5 cm, AB = 8 cm and OD is perpendicular to AB, then CD is equal to:

(A) 2 cm
(B) 3 cm
(C) 4 cm
(D) 5 cm

Answer:

The perpendicular from the center of a circle to a chord bisects the chord.
∴ AC = CB = $\frac{1}{2}$ AB
= $\frac{1}{2}$ $\times$ 8
= 4 cm
It is given that OA = 5 cm.
$\Rightarrow {AO}^{2} = {AC}^{2} + {OC}^{2} \Rightarrow {(5)}^{2} = {(4)}^{2} + {OC}^{2} \Rightarrow 25 = 16 + {OC}^{2} \Rightarrow {OC}^{2} = 25 - 16 = 9$
$∴ OC = 3 cm$ [âˆµLength is always positive, taking a positive square root]
Thus,
OA = OD = radius of a circle
OD = 5 cm
CD = OD − OC = 5 − 3 = 2 cm
Hence, the correct answer is option A.

Page No 99:

Question 3:

If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is :
(A) 6 cm
(B) 8 cm
(C) 10 cm
(D) 12 cm

Answer:

Given that, AB = 12 cm and BC = 16 cm.
In the circle, BC $⊥$ AB. Thus, AC is the diameter of circle [âˆµ diameter of a circle subtends a right angle to the circle]

In right angled $△ ABC$ , using Pythagoras theorem,
$\Rightarrow {AC}^{2} = {AB}^{2} + {BC}^{2} \Rightarrow {AC}^{2} = {(12)}^{2} + {(16)}^{2} \Rightarrow {AC}^{2} = 144 + 256 \Rightarrow {AC}^{2} = 400 \Rightarrow AC = \sqrt{400} = 20 cm$
[taking positive square root, because diameter is always positive]
$\begin{array}{rcl} ∴ Radius of circle & = & \frac{1}{2} (AC) \\ = & \frac{1}{2} \times 20 \\ = & 10 cm \end{array}$
Thus, the radius of circle is 10 cm.
Hence, the correct answer is option C.

Page No 99:

Question 4:

In the figure, if ∠ABC = 20°, then ∠AOC is equal to:

(A) 20°
(B) 40°
(C) 60°
(D) 10°

Answer:

Given that, ∠ABC = 20°.

The angle subtended by an arc at the centre is twice the angle subtended at the rest of the circle.
⇒ ∠AOC = 2∠ABC
⇒ ∠AOC = 2 × 20°
= 40°

Hence, the correct answer is option B.

Page No 99:

Question 5:

In the figure, if AOB is a diameter of the circle and AC = BC, then ∠CAB is equal to:

(A) 30º
(B) 60º
(C) 90º
(D) 45º

Answer:

The diameter subtends a right angle to the circle.
$∴ ∠ BCA = 90 ° . . . . . (1)$

As, AC = BC
$\Rightarrow ∠ ABC = ∠ CAB . . . . . (2)$ [âˆµ angles opposite to equal sides are equal]

In $△ ABC,$ by angle sum property of a triangle
$∠ CAB + ∠ ABC + ∠ BCA = 180 °$

Using (1) and (2),
$\Rightarrow ∠ CAB + ∠ CAB + ∠ 90 ° = 180 °$
$\Rightarrow 2 ∠ CAB = 180 ° - 90 ° \Rightarrow ∠ CAB = \frac{90 °}{2} ∴ ∠ CAB = 45 °$
Hence, the correct answer is option D.

Page No 100:

Question 6:

In the figure, if ∠OAB = 40°, then ∠ACB is equal to :

(A) 50°
(B) 40°
(C) 60°
(D) 70°

Answer:

In $△$ OAB, OA = OB = radius of the circle
$∴ ∠ OAB = ∠ OBA (∵ angles opposite to equal sides are equal) \Rightarrow ∠ OBA = 40 °$

Also, by angle sum property of triangles in $△$ AOB,
$∠ AOB + ∠ OBA + ∠ BAO = 180 °$
$\Rightarrow ∠ AOB + 40 ° + 40 ° = 180 °$
$\Rightarrow ∠ AOB = 180 ° - 80 ° = 100 °$

The angle subtended by an arc at the centre of a circle is double the angle subtended by it at the rest of the circle.
$∠ AOB = 2 ∠ ACB \Rightarrow 100 ° = 2 ∠ ACB \Rightarrow ∠ ACB = \frac{100 °}{2} = 50 °$
Hence, the correct answer is option A.

Page No 100:

Question 7:

In the figure, if ∠DAB = 60°, ∠ABD = 50°, then ∠ACB is equal to:

(A) 60°
(B) 50°
(C) 70°
(D) 80°

Answer:

Given that, $∠ DAB = 60 °, ∠ ABD = 50 °$ .
As, $∠ ADB = ∠ ACB$ [âˆµ angles in same segment of a circle are equal]

In $△ ABD,$
$∠ ABD + ∠ ADB + ∠ DAB = 180 °$ [by angle sum property of a triangle]
$\Rightarrow 50 ° + ∠ ADB + 60 ° = 180 ° \Rightarrow ∠ ADB = 180 ° - 110 ° = 70 °$
$\Rightarrow ∠ ACB = 70 °$

Hence, the correct answer is option C.

Page No 100:

Question 8:

ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC = 140°, then ∠BAC is equal to:
(A) 80°
(B) 50°
(C) 40°
(D) 30°

Answer:

Given that, ABCD is a cyclic quadrilateral and $∠ ADC = 140 °$ .

The sum of the opposite angles in a cyclic quadrilateral is 180°, then
$∠ ADC + ∠ ABC = 180 ° \Rightarrow 140 ° + ∠ ABC = 180 ° \Rightarrow ∠ ABC = 180 ° - 140 ° ∴ ∠ ABC = 40 °$

As, AB is a diameter of the circle and $∠ ACB$ is an angle in a semi-circle.
$∴ ∠ ACB = 90 °$

In $△ ABC,$ by angle sum property of a triangle
$∠ BAC + ∠ ACB + ∠ ABC = 180 °$
$\Rightarrow ∠ BAC + 90 ° + 40 ° = 180 ° \Rightarrow ∠ BAC = 180 ° - 130 ° = 50 °$
Hence, the correct answer is option B.

Page No 100:

Question 9:

In the figure BC is a diameter of the circle and ∠BAO = 60º. Then ∠ADC is equal to :

(A) 30º
(B) 45º
(C) 60º
(D) 120º

Answer:

In $△ AOB,$
$∠ OBA = ∠ BAO$ [âˆµ angles opposite to equal sides are equal]
$\Rightarrow ∠ OBA = 60 ° [As ∠ BAO = 60 °]$

Now, angles in the same segment AC are equal.
$∴ ∠ ABC = ∠ ADC$
$\Rightarrow ∠ ADC = 60 °$
Hence, the correct answer is option C.

Page No 101:

Question 10:

In the figure, ∠AOB = 90° and ∠ABC = 30°, then ∠CAO is equal to:

(A) 30°
(B) 45°
(C) 90°
(D) 60°

Answer:

In $△$ OAB, by angle sum property of a triangle,
$∠ OAB + ∠ ABO + ∠ BOA = 180 °$
$∠ OAB + ∠ OAB + 90 ° = 180 °$ [âˆµ angles opposite to equal sides are equal]
$\Rightarrow 2 ∠ OAB = 180 ° - 90 ° \Rightarrow ∠ OAB = \frac{90 °}{2} = 45 ° . . . . . (1)$

Now, the angle subtended by an arc on the centre of the circle is twice the angle subtended by it at any other point on the circle.
$∴ ∠ ACB = \frac{1}{2} ∠ AOB \Rightarrow ∠ ACB = \frac{1}{2} \times 90 ° = 45 °$

In $△ ACB$ ,
$∠ ACB + ∠ CBA + ∠ CAB = 180 °$
$\Rightarrow 45 ° + 30 ° + ∠ CAB = 180 °$
$\Rightarrow ∠ CAB = 180 ° - 75 ° \Rightarrow ∠ CAB = 105 ° \Rightarrow ∠ CAO + ∠ OAB = 105 ° \Rightarrow ∠ CAO + 45 ° = 105 °$

$∴ ∠ CAO = 105 ° - 45 ° = 60 °$
Hence, the correct answer is option D.

Page No 101:

Question 1:

State whether the following statement is True or False.
Two chords AB and CD of a circle are each at distances 4 cm from the centre. Then AB = CD.

Answer:

The statement is true.
This is because the chords equidistant from the centre of circle are equal in length.

Page No 101:

Question 2:

State whether the following statement is True or False.
Two chords AB and AC of a circle with centre O are on the opposite sides of OA. Then ∠OAB = ∠OAC .

Answer:

The statement is false.
In the figure below, AB and AC are two chords of a circle.
Join OB and OC.

In $△ OAB and △ OAC,$
OA = OA [common side]
OB = OC [radius of circle]

It cannot be seen that any angle or the third side of the two triangles is equal and thus, $△ OAB$ is not congruent to $△ OAC .$
$∴ ∠ OAB \neq ∠ OAC .$

Page No 101:

Question 3:

State whether the following statement is True or False.
Two congruent circles with centres O and O′ intersect at two points A and B. Then ∠AOB = ∠AO′B.

Answer:

The statement is true.
Consider the figure given below:

Join AB, OA and OB, O’A and BO’.

In $△ AOB and △ AO ’ B,$
OA = AO’ [both circles have same radius]
OB = BO’ [both circles have same radius]
AB = AB [common chord]

$\Rightarrow △ AOB = △ AO ’ B$ [by SSS congruence rule]
Hence, $∠ AOB = ∠ AO ’ B$ [by CPCT]

Page No 101:

Question 4:

State whether the following statement is True or False.
Through three collinear points, a circle can be drawn.

Answer:

The statement is false.
This is because a circle can pass through only two collinear points, not three.

Page No 101:

Question 5:

State whether the following statement is True or False.
A circle of radius 3 cm can be drawn through two points A, B such that AB = 6 cm.

Answer:

The statement is true.

Consider the diameter of the circle as 6 cm.
⇒ AB = 6 cm
Then, radius of a circle = $\frac{AB}{2} = \frac{6}{2} = 3 cm .$

Page No 102:

Question 6:

State whether the following statement is True or False.
If AOB is a diameter of a circle and C is a point on the circle, then AC²+ BC²= AB².

Answer:

The statement is true.

The diameter of a circle subtends a right angle at any other point on the circle.

If AOB is a diameter of a circle and C is a point on the circle, then $△$ ACB is right angled at C.

In right angled $△$ ACB, by Pythagoras theorem
AC² + BC² = AB²

Page No 102:

Question 7:

State whether the following statement is True or False.
ABCD is a cyclic quadrilateral such that ∠A = 90°, ∠B = 70°, ∠C = 95° and ∠D = 105°.

Answer:

The statement is false.

The sum of opposite angles in a cyclic quadrilateral is 180°.
Now,
$∠ A + ∠ C = 90 ° + 95 ° = 185 ° \neq 180 °$
and
$∠ B + ∠ D = 70 ° + 105 ° = 175 ° \neq 180 °$

Here, the sum of opposite angles is not equal to 180°. Therefore, ABCD is not a cyclic quadrilateral.

Page No 102:

Question 8:

State whether the following statement is True or False.
If A, B, C, D are four points such that ∠BAC = 30° and ∠BDC = 60°, then D is the centre of the circle through A, B and C.

Answer:

The statement is false.

Consider the following case:

Since, the angle subtended by an arc at the centre is double the angle subtended by it at any other point on the circle.
∠BAC = 30°

But now,

Here, ∠BDC = 60°.
⇒ reflex(∠BDC) = 300°
Again, the angle subtended by an arc at the centre is double the angle subtended by it at any other point on the circle.
∠BAC = 150° ≠ 30°

Page No 102:

Question 9:

State whether the following statement is True or False.
If A, B, C and D are four points such that ∠BAC = 45° and ∠BDC = 45°, then A, B, C, D are concyclic.

Answer:

The statement is true.
As, $∠ BAC = 45 ° and ∠ BDC = 45 °$

Angles in the same segment of a circle are equal. As a result, A, B, C, and D are all concyclic.

Page No 102:

Question 10:

State whether the following statement is True or False.
In the figure, if AOB is a diameter and ∠ADC = 120°, then ∠CAB = 30°.

Answer:

The statement is true.
Join CA and CB.

As, ADCB is a cyclic quadrilateral. Thus, the sum of opposite angles of cyclic quadrilateral is 180°.
$∠ ADC + ∠ CBA = 180 °$
$\Rightarrow ∠ CBA = 180 ° - 120 ° = 60 ° [As, ∠ ADC = 120 °]$

In $△ ACB$ , by angle sum property of a triangle],
$∠ CAB + ∠ CBA + ∠ ACB = 180 °$
$\Rightarrow ∠ CAB + 60 ° + 90 ° = 180 °$ [âˆµ angle formed by the diameter to the circle is 90° i.e., $∠ ACB = 90 °$ ]
$\Rightarrow ∠ CAB = 180 ° - 150 ° = 30 °$

Page No 103:

Question 1:

If arcs AXB and CYD of a circle are congruent, find the ratio of AB and CD.

Answer:

Let AXB and CYD be the two arcs of a circle arcs with centre O and radius r.

Now, when the two chords are obtained, i.e., AB and CD for the congruent arcs AXB and CYD, their repective chords will be equal.

Hence, AB : CD is equal to 1 : 1.

Page No 103:

Question 2:

If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, prove that arc PXA ≅ Arc PYB.

Answer:

Let PQ be the perpendicular bisector of AB, which intersects it at M and always passes through the centre of the circle O.
To prove: arc PXA ≅ arc PYB
Construction: Join AP and BP.

In $△$ APM and $△$ BPM,
AM = MB (âˆµ the perpendicular drawn from the centre of a circle to a chord, bisects it)
$∠ PMA = ∠ PMB = 90 °$
PM = PM (common)
Thus, $△ APM ≅ △ BPM$ [by Side Angle Side Congruency]

∴ PA = PB
$\Rightarrow arc PXA ≅ arc PYB .$

Page No 103:

Question 3:

A, B and C are three points on a circle. Prove that the perpendicular bisectors of AB, BC and CA are concurrent.

Answer:

Consider three non-collinear points on a circle, i.e., A, B and C.
Join AB, BC and CA.
Construction: Draw TU, PQ and RS, which are the perpendicular bisectors of AB, BC and CA respectively.

To prove: The perpendicular bisectors are concurrent.
Proof: The points A, B and C are non-collinear.
Thus, the perpendicular bisectors TU, PQ and RS are not parallel and they will intersect at a point. Let the point of intersection be O.

Now, join OC, OA and OB.

Consider $△$ AOU and $△$ BOU.
AU = BU (âˆµ TU is the perpendicular bisector of AB)
∠AUO = ∠BUO = 90°
OU = OU (common)
∴ $△ AUO ≅ △ BUO$ by SAS congruence criterion.
⇒ OA = OB (by CPCT)

Similarly, $△ BUO ≅ △ BPO$ and $△ BPO ≅ △ CPO$ .
⇒ OA = OB = OC (by CPCT)

Let OA = OB = OC = r.
This means that O is the only point equidistant from points on the circle A, B and C. Thus, O is the centre of the circle.
Therefore, the perpendicular bisectors of AB, BC and CA are concurrent.
Hence, proved.

Page No 103:

Question 4:

AB and AC are two equal chords of a circle. Prove that the bisector of the angle BAC passes through the centre of the circle.

Answer:

Given: AB and AC are two equal chords whose centre is O
$∴ AB = AC$
To prove: Center O lies on the bisector of $∠ BAC .$
Construction: Join BC. Draw bisector AD of $∠ BAC \Rightarrow ∠ BAD = ∠ CAD$

Proof: In $△ BAO and △ CAO$ ,
AB = AC (given)
$∠ BAO = ∠ CAO$ (by construction)
AO = AO (common)
$∴ △ BAO ≅ △ CAO$ [by SAS congruence rule]
$\Rightarrow$ BO = CO [by CPCT]
and
$∠ BOA = ∠ COA$ [by CPCT]

Also, BO = CO and $∠ BOA = ∠ COA = 90 °$
Thus, AO is the perpendicular bisector of the chord BC.
Therefore, the bisector of $∠ BAC$ i.e., AD passes through the centre O.
Hence proved.

Page No 103:

Question 5:

If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel.

Answer:

Given that, AB and CD are two chords of a circle with centre O. PQ is the diameter of the circle bisecting the chord AB and CD at points L and M respectively.

To prove: AB $∥$ CD

Proof:
Now, L is the mid-point of AB. The line joining the centre of a circle to the mid-point of a chord is perpendicular to the chord.
$∴ OL ⊥ AB$ .
$\Rightarrow ∠ ALO = 90 ° . . . . . (1)$

Also, OM $⊥$ CD.
$∴ ∠ OMD = 90 ° . . . . . (2)$

From (1) and (2),
$∠ ALO = ∠ OMD = 90 °$

Since, these are alternating angles.
Therefore, AB $∥$ CD.
Hence, proved.

Page No 103:

Question 6:

ABCD is such a quadrilateral that A is the centre of the circle passing through B, C and D. Prove that $∠ CBD + ∠ CDB = \frac{1}{2} ∠ BAD$ .

Answer:

In a circle, consider a quadrilateral ABCD with centre A.
To prove: $∠ CBD + ∠ CDB = \frac{1}{2} ∠ BAD$
Construction: Join AC and BD.

Proof: In a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
As, arc CD subtends $∠ DAC$ at the centre and $∠ CBD$ at a point B.
$∴ ∠ DAC = 2 ∠ CBD . . . . . (1)$

Again, arc BC subtends $∠ CAB$ at the centre and $∠ CDB$ at a point D in the remaining part of the circle.
$∴ ∠ CAB = 2 ∠ CDB . . . . . (2)$

On adding (1) and (2),
$∠ DAC + ∠ CAB = 2 ∠ CBD + 2 ∠ CDB$
$\Rightarrow ∠ BAD = 2 (∠ CBD + ∠ CDB)$
$\Rightarrow ∠ CDB + ∠ CBD = \frac{1}{2} ∠ BAD$
Hence, proved.

Page No 103:

Question 7:

O is the circumcentre of the triangle ABC and D is the mid-point of the base BC. Prove that ∠BOD = ∠A.

Answer:

Consider a triangle $△$ ABC inside a circle with centre O. Also, D is the mid-point of BC.
To prove: $∠ BOD = ∠ A or ∠ BOD = ∠ BAC$
Construction: Join OB, OD and OC.

Proof: In $△$ BOD and $△$ COD,
OB = OC [radius of circle]
BD = DC [D is the mid-point of BC]
Also, OD = OD [Common]
$∴ △ BOD ≅ △ COD$ [by SSS congruence rule]
Then, $∠ BOD = ∠ COD$ [CPCT] .....(1)

In a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
$∴ 2 ∠ BAC = ∠ BOC$
$\Rightarrow ∠ BAC = \frac{2}{2} ∠ BOD$ [As $∠ BOC = 2 ∠ BOD]$ [Using (1)]
$\Rightarrow ∠ BAC = ∠ BOD$

Hence, proved.

Page No 103:

Question 8:

On a common hypotenuse AB, two right triangles ACB and ADB are situated on opposite sides. Prove that ∠BAC = ∠BDC.

Answer:

Given that, two right angled triangles $△ ACB and △ ADB$ have a common hypotenuse AB.
To prove: $∠ BAC = ∠ BDC$
Construction: Join CD.

Proof: Since, $△ ACB and △ ADB$ are right angled triangles.
Therefore,
∠C + ∠D = 90° + 90°
⇒ ∠C + ∠D = 180°
Thus, ADBC is a cyclic quadrilateral as sum of opposite angles of a cyclic quadrilateral is 180°.

Also, ∠BAC and ∠BDC lie in the same segment BC and angles in the same segment of a circle are equal.
∴ ∠BAC = ∠BDC.

Hence, proved.

Page No 103:

Question 9:

Two chords AB and AC of a circle subtends angles equal to 90º and 150º, respectively at the centre. Find ∠BAC, if AB and AC lie on the opposite sides of the centre.

Answer:

In $△ BOA,$ OB = OA [radius of circle]
$∴ ∠ OAB = ∠ OBA . . . . . (1)$ [âˆµ angles opposite to equal sides are equal]

In $△$ OAB, by angle sum property of a triangle,
$∠ OBA + ∠ OAB + ∠ AOB = 180 °$
$\Rightarrow ∠ OAB + ∠ OAB + 90 ° = 180 ° [from (1)] \Rightarrow 2 ∠ OAB = 180 ° - 90 ° \Rightarrow ∠ OAB = \frac{90 °}{2} = 45 °$

In $△ AOC, AO = OC$ [radius of a circle]
$∴ ∠ OCA = ∠ OAC . . . . . (2)$ [âˆµ angles opposite to equal sides are equal]

Now, by angle sum property of a triangle,
$∠ AOC + ∠ OAC + ∠ OCA = 180 °$
$\Rightarrow 150 ° + 2 ∠ OAC = 180 ° [from (2)] \Rightarrow 2 ∠ OAC = 180 ° - 150 ° \Rightarrow 2 ∠ OAC = 30 ° \Rightarrow ∠ OAC = 15 ° ∴ ∠ BAC = ∠ OAB + ∠ OAC = 45 ° + 15 ° = 60 °$
Hence, $∠ BAC$ is 60°.

Page No 103:

Question 10:

If BM and CN are the perpendiculars drawn on the sides AC and AB of the triangle ABC, prove that the points B, C, M and N are concyclic.

Answer:

In ΔABC,
$BM ⊥ AC and CN ⊥ AB$ .
To prove: Points B, C, M and N are concyclic.
Construction: Construct a circle passing through the points B, C, M and N.

Proof:
Consider BC as the diameter of the circle.
Then, BC subtends a 90° at any other point on the circle.
Here, ∠BNC = 90°.
Thus, point N lies on the circle.

Similarly, the point M also lies on the circle.
Therefore, BCMN form a concyclic quadrilateral.
Hence, proved.

Page No 103:

Question 11:

If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral so formed is cyclic.

Answer:

$△$ ABC is an isosceles triangle such that AB = AC and DE âˆ¥ BC.
To prove: Quadrilateral BCDE is a cyclic quadrilateral.
Construction: Draw a circle that passes through the points B, C, D and E.

Proof: In $△ ABC$ ,
AB = AC [âˆµ equal sides of an isosceles triangle]
$\Rightarrow ∠ ACB = ∠ ABC . . . . . (1)$ [angle opposite to the equal sides are equal]

As, DE âˆ¥ BD.
$\Rightarrow ∠ ADE = ∠ ACB . . . . . (2)$ [corresponding angles]

On adding $∠ EDC$ on both sides in (2),
$∠ ADE + ∠ EDC = ∠ ACB + ∠ EDC$
$\Rightarrow 180 ° = ∠ ACB + ∠ EDC$ [âˆµ $∠ ADE and ∠ EDC$ from linear pair axiom]
$\Rightarrow ∠ EDC + ∠ ABC = 180 °$
Hence, as the sum of opposite angle is 180°. BCDE is a cyclic quadrilateral.

Page No 104:

Question 12:

If a pair of opposite sides of a cyclic quadrilateral are equal, prove that its diagonals are also equal.

Answer:

Let ABCD be a cyclic quadrilateral where AD = BC.
Construction: Join AC and BD.

To prove: AC = BD
Proof: In $△$ AOD and $△$ BOC,
$∠ OAD = ∠ OBC and ∠ ODA = ∠ OCB$ [âˆµ angles subtended in the same segment are equal]
Also, AD = BC.
$∴ △ AOD ≅ △ BOC$ [by ASA congruence rule]

By CPCT,
AO = OB .....(1)
OC = OD .....(2)

Adding (1) and (2),
AO + OC = OB + OD
⇒ AC = BD
Hence, proved.

Page No 104:

Question 13:

The circumcentre of the triangle ABC is O. Prove that ∠OBC + ∠BAC = 90º.

Answer:

A circle is circumscribed on a ΔABC having centre O.
To prove: $∠ OBC + ∠ BAC = 90 °$
Construction: Join BO and CO.

Proof : Let $∠ OBC = ∠ OCB = θ$ .
In $△ OBC$ , by angle sum property of triangles,
$△ OBC, ∠ BOC + ∠ OCB + ∠ CBO = 180 ° . . . . . (1)$
$\Rightarrow ∠ BOC + θ + θ = 180 °$
$\Rightarrow ∠ BOC = 180 ° - 2 θ$

In a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
$∴ ∠ BOC = 2 ∠ BAC \Rightarrow ∠ BAC = \frac{∠ BOC}{2} = \frac{180 ° - 2 θ}{2} = 90 ° - θ \Rightarrow ∠ BAC + θ = 90 ° ∴ ∠ BAC + ∠ OBC = 90 ° [from (1)]$
Hence, proved.

Page No 104:

Question 14:

A chord of a circle is equal to its radius. Find the angle subtended by this chord at a point in major segment.

Answer:

As, AB is a chord of a circle, which is equal to the radius of the circle.
⇒ AB = BO
Construction: Join AC, OA and BC.

Since, OA = OB = Radius of circle
$\Rightarrow$ OA = AB = BO
Thus, $△ OAB$ is an equilateral triangle.
$\Rightarrow ∠ AOB = 60 °$ [All angles of an equilateral triangle is 60°]

In a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at remaining part of the circle.
$\Rightarrow ∠ AOB = 2 ∠ ACB \Rightarrow ∠ ACB = \frac{60 °}{2} = 30 °$
Hence, the angle subtended by this chord at a point in major segment is 30°.

Page No 104:

Question 15:

In the figure, ∠ADC = 130° and chord BC = chord BE. Find ∠CBE.

Answer:

Given: $∠ ADC = 130 °$ and chord BC = chord BE.
Let the points A, B, C and D form a cyclic quadrilateral.

As, the sum of opposite angles of a cyclic quadrilateral $△$ DCB is 180°.
Then, $∠ ADC + ∠ OBC = 180 °$
$\Rightarrow 130 ° + ∠ OBC = 180 ° \Rightarrow ∠ OBC = 180 ° - 130 ° = 50 °$

In $△$ BOC and $△$ BOE,
BC = BE [given that the chords are equal]
∠BCO = ∠BEO [angles opposite to equal sides are equal)
and OB = OB [common side]
⇒ $△$ BOC $≅$ $△$ BOE
$\Rightarrow ∠ OBC = ∠ OBE = 50 ° [CPCT]$

$\Rightarrow ∠ CBE = ∠ CBO + ∠ EBO = 50 ° + 50 ° = 100 °$

Hence, $∠ CBE = 100 °$ .

Page No 104:

Question 16:

In the figure, ∠ACB = 40º. Find ∠OAB.

Answer:

Given that, $∠ ACB = 40 °$ .
The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
$∴ ∠ AOB = 2 ∠ ACB \Rightarrow ∠ AOB = 80 ° . . . . . (1)$

In $△ AOB$ ,
AO = OB [radius of a circle]
$\Rightarrow ∠ OBA = ∠ OAB . . . . . (2)$ [âˆµ angles opposite to the equal sides of a triangle are equal]

Also, the sum of all three angles in a triangle AOB is 180°.
$∴ ∠ AOB + ∠ OBA + ∠ OAB = 180 ° \Rightarrow 80 ° + ∠ OAB + ∠ OAB = 180 ° [From (1) and (2)] \Rightarrow 2 ∠ OAB = 180 ° - 80 ° \Rightarrow 2 ∠ OAB = 100 ° ∴ ∠ OAB = \frac{100 °}{2} = 50 °$
Hence, $∠ OAB$ is 50°.

Page No 104:

Question 17:

A quadrilateral ABCD is inscribed in a circle such that AB is a diameter and ∠ADC = 130°. Find ∠BAC.

Answer:

Given that, quadrilateral ABCD is inscribed in a circle having centre O. Also, $∠ ADC = 130 °$ .

As, ABCD is a quadrilateral inscribed in a circle, therefore ABCD become a cyclic quadrilateral.

Therefore, the sum of opposite angles of a cyclic quadrilateral is 180°.
Then,
$∠ ADC + ∠ ABC = 180 °$
$\Rightarrow 130 ° + ∠ ABC = 180 ° \Rightarrow ∠ ABC = 50 °$

Now, AB is a diameter of a circle. Thus, the angle subtended by AB at any point of the circle is right angle.
$∴ ∠ ACB = 90 °$

In $△ ABC$ , by angle sum property of a triangle,
$∠ BAC + ∠ ACB + ∠ ABC = 180 °$
$\Rightarrow ∠ BAC + 90 ° + 50 ° = 180 °$
$\begin{array}{rcl} \Rightarrow & ∠ BAC = 180 ° - (90 ° + 50 °) \\ = & 180 ° - 140 ° \\ = & 40 ° \end{array}$

Hence, $∠ BAC$ is 40°.

Page No 104:

Question 18:

Two circles with centres O and O′ intersect at two points A and B. A line PQ is drawn parallel to OO′ through A(or B) intersecting the circles at P and Q. Prove that PQ = 2OO′.

Answer:

Given, two circles having centres O and O′ intersect at points A and 8.
Construction: Join OO, OP, O′Q, OM and O′N. Draw line PQ parallel to OO′.

To prove: PQ = 2OO′
Proof:
In $△ OPB$ ,
BM = MP [OM is the perpendicular bisector of PB]
and
in $△$ O′BQ,
BN = NQ [O′ N is the perpendicular bisector of BQ]
$∴ BM + BN = PM + NQ$

Adding (BM + BN) on both the sides,
$\Rightarrow 2 (BM + BN) = BM + BN + PM + NQ$
⇒ 2OO′ = (BM + MP) + (BN + NQ) [âˆµ OO′ = MN = MB + BN]
= BP + BQ
= PQ
â€‹⇒ 2OO′ = PQ
Hence, proved.

Page No 104:

Question 19:

In the figure, AOB is a diameter of the circle and C, D, E are any three points on the semi-circle. Find the value of ∠ACD + ∠BED.

Answer:

Join AE.
Write D
Since, A, C, D and E are four point on a circle, therefore ACDE is a cyclic quadrilateral.
$\Rightarrow$ $∠ ACD + ∠ AED = 180 °$ [sum of opposite angles in a cyclic quadrilateral is 180°] .....(1)
Now, the diameter of a circle subtends a right angle at any point on the circle.
$∴ ∠ AEB = 90 ° . . . . . (2)$

On adding (1) and (2),
$(∠ ACD + ∠ AED) + ∠ AEB = 180 ° + 90 ° = 270 °$
$\Rightarrow ∠ ACD + ∠ BED = 270 °$

Hence, $∠ ACD + ∠ BED$ is 270°.

Page No 105:

Question 20:

In the figure, ∠OAB = 30º and ∠OCB = 57º. Find ∠BOC and ∠AOC.

Answer:

Given that, $∠ OAB = 30 ° and ∠ OCB = 57 °$ .
In $△ AOB$ .
OA = OB [radius of a circle]
$\Rightarrow ∠ OBA = ∠ BAO = 30 °$ [angles opposite to equal sides are equal]

In $△ AOB$ , by angle sum property of a triangle,
$\Rightarrow ∠ AOB + ∠ OBA + ∠ BAO = 180 °$
$\Rightarrow ∠ AOB + 30 ° + 30 ° = 180 °$
$\begin{array}{rcl} ∴ ∠ AOB & = & 180 ° - 2 (30 °) \\ = & 180 ° - 60 ° \\ = & 120 ° . . . . . (1) \end{array}$

Now, in $△ OCB,$
OC = OB [radius of a circle]
$\Rightarrow ∠ OBC = ∠ OCB = 57 °$ [angles opposite to equal sides are equal]

In $△ OCB,$ by angle sum property of a triangle,
$∠ COB + ∠ OCB + ∠ CBO = 180 °$
$∴ ∠ COB = 180 ° - (∠ OCB + ∠ OBC)$
$\begin{array}{rcl} \Rightarrow & ∠ COB = 180 ° - (57 ° + 57 °) \\ = & 180 ° - 114 ° \\ = & 66 ° . . . . . (2) \end{array}$
$\Rightarrow ∠ COB = ∠ BOC = 66 °$

From (1),
$∠ AOB = 120 °$ .

$\Rightarrow ∠ AOC + ∠ COB = 120 ° \Rightarrow ∠ AOC + 66 ° = 120 ° [from (2)] ∴ ∠ AOC = 120 ° - 66 ° = 54 °$
Hence, $∠ AOC$ is 54° and $∠ BOC$ is 66°.

Page No 106:

Question 1:

If two equal chords of a circle intersect, prove that the parts of one chord are separately equal to the parts of the other chord.

Answer:

Given: Let AB and CD are two equal chords of a circle meeting at point E.
To prove: AE = CE and BE = DE
Construction: Draw OM $⊥$ AB, ON $⊥$ CD and join OE, where O is the centre of circle.

Proof: In $△ OME and △ ONE,$
OM = ON [equal chords are equidistant from the centre]
OE = OE [common side]
Also, $∠ OME = ONE$ [each 90°]
$∴ △ OME ≅ △ ONE [∵ RHS congurence rule]$
$\Rightarrow EM = EN$ [CPCT] .....(1)

Also, AB = CD.
On dividing both sides by 2,
$\frac{AB}{2} = \frac{CD}{2} \Rightarrow AM = CN . . . . . (2)$
[âˆµPerpendicular drawn from the centre of acircle to a chord, bisects it]

On adding (1) and (2),
EM + AM = EN + CN
$\Rightarrow AE = CE . . . . . (3)$

Now, AB = CD.

On subtracting AE from both the sides,
AB $-$ AE = CD $-$ AE
⇒ BE = CD $-$ CE [from (3)]
⇒ BE = DE
Hence, proved.

Page No 106:

Question 2:

If non-parallel sides of a trapezium are equal, prove that it is cyclic.

Answer:

Given: ABCD is a trapezium with equal non-parallel sides AD and BC.
To prove: Trapezium ABCD is a cyclic.
Construction: Join BE such that BE $∥$ AD.

Proof: As, AB $∥$ DE and AD $∥$ BE
Then, the quadrilateral ABED is a parallelogram.
$∴ ∠ BAD = ∠ BED . . . . . (1)$ [opposite angles of a parallelogram are equal]
and AD = BE .....(2) [opposite sides of a parallelogram are equal]
But AD = BC .....(3)

From (2) and (3),
BC = BE
$\Rightarrow ∠ BEC = ∠ BCE$ .....(4) [angles opposite to equal sides are equal]

Also,
$∠ BEC + ∠ BED = 180 °$ [linear pair axiom]
$∴ ∠ BCE + ∠ BAD = 180 °$ [from (1) and (4)]

In a cyclic quadrilateral, the sum of opposite angles is 180°.
Thus, trapezium ABCD is a cyclic.
Hence, proved.

Page No 106:

Question 3:

If P, Q and R are the mid-points of the sides BC, CA and AB of a triangle and AD is the perpendicular from A on BC, prove that P, Q, R and D are concyclic.

Answer:

In $△$ ABC, R and Q are mid points of AB and CA respectively.
By mid-point theorem, RQ âˆ¥ BC.
Similarly, PQ âˆ¥ AB and PR âˆ¥ CA.

In quadrilateral BPQR, BP âˆ¥ RQ and PQ âˆ¥ BR. Thus, the quadrilateral BPQR is a parallelogram.

Similarly, quadrilateral ARPQ is a parallelogram.
∴ ∠A = ∠RPQ (âˆµ Opposite sides of parallelogram are equal)

Also, PR âˆ¥ AC and PC is the transversal.
∴ ∠BPR = ∠C (Corresponding angles)
∠DPQ = ∠DPR + ∠RPQ
= ∠A + ∠C .....(1)

And RQ âˆ¥ BC and BR is the transversal.
∴ ∠ARO = ∠B (Corresponding angles) .....(2)

In $△$ ABD, R is the mid point of AB and OR âˆ¥ BD.
∴ O is the mid point of AD (Converse of mid point theorem)
⇒ OA = OD

In $△$ AOR and $△$ DOR,
OA = OD
∠AOR = ∠DOR = 90° {∠ROD = ∠ODP (Alternate angles) & ∠AOR = ∠ROD = 90° (linear pair)}
OR = OR (Common)
∴ $△$ AOR $≅$ $△$ DOR (by SAS congruence criterion)

⇒ ∠ARO = ∠DRO (CPCT)
⇒ ∠DRO = ∠B (Using (2))

In quadrilateral PRQD,
∠DRO + ∠DPQ = ∠B + ( ∠A + ∠C)
= ∠A + ∠B + ∠C (Using (1))
⇒ ∠DRO + ∠DPQ = 180° (âˆµ ∠A + ∠B + ∠C = 180°)

Therefore, quadrilateral PRQD is a cyclic quadrilateral. Thus, the points P, Q, R and D are concyclic.
Hence, proved.

Page No 106:

Question 4:

ABCD is a parallelogram. A circle through A, B is so drawn that it intersects AD at P and BC at Q. Prove that P, Q, C and D are concyclic.

Answer:

Given: ABCD is a parallelogram.
Let a circle whose centre is O passes through A and B such that it intersects AD at P and BC at Q.
To prove: Points P, Q, C and D are concyclic.
Construction: Join point P to Q. Thus, PQ line segment is constructed.

Proof:
As, $∠ 1 = ∠ A$ [âˆµ exterior angle property of cyclic quadrilateral]
But $∠ A = ∠ C$ [âˆµ opposite angles of a parallelogram are equal]
∴ $∠ 1 = ∠ C$ .....(1)

But
$∠ C + ∠ D = 180 °$ [On the same side of the transversal, the sum of co-interior angles is 180°]
$\Rightarrow ∠ 1 + ∠ D = 180 °$ [from (1)]
Therefore, the quadrilateral QCDP is cyclic.
So, the points P, Q, C and D are concyclic.
Hence, proved.

Page No 106:

Question 5:

Prove that angle bisector of any angle of a triangle and perpendicular bisector of the opposite side if intersect, they will intersect on the circumcircle of the triangle.

Answer:

Given: $△ ABC$ is inscribed in a circle. The bisector of $∠ A$ and perpendicular bisector of BC intersect at point O.
To prove: The angle bisector of ∠A and perpendicular bisector of BC intersect on the circumcircle of $△$ ABC.
Construction: Join BO and OC.

Proof: Let the angle bisector of ∠A intersect the circumcircle of $△$ ABC at point D. Now, join DC and DB.

Since ,angles in the same segment are equal.
∴ ∠BCD = ∠BAD
⇒ ∠BCD = ∠BAD = $\frac{1}{2}$ ∠A .....(1)

Similarly,
∠DBC = ∠DAC = $\frac{1}{2}$ ∠A .....(2)

From (1) and (2),
∠DBC = ∠BCD
⇒ BD = DC [âˆµ sides opposite to equal angles are equal]
Thus, D lies on the perpendicular bisector of BC.
Therefore, the angle bisector of ∠A and perpendicular bisector of BC intersect on the circumcircle of $△$ ABC.
Hence, proved.

Page No 106:

Question 6:

If two chords AB and CD of a circle AYDZBWCX intersect at right angles, prove that arc CXA + arc DZB = arc AYD + arc BWC = semicircle.

Answer:

In a circle AYDZBWCX, two chords AB and CD intersect at right angles.
To prove: arc CXA + arc DZB = arc AYD + arc BWC = Semi-circle.
Construction: Draw diameter EF parallel to CD with M as the centre.

Proof: As, CD $∥$ EF.
⇒ arc EC = arc FD .....(1)
arc ECXA = arc EWB [symmetrical in relation to a circle's diameter]
⇒ arc AF = arc BF .....(2)

Now, arc ECXAYDF = Semi-circle
⇒ arc EA + arc AF = Semi-circle
⇒ arc EC + arc CXA + arc BF = Semi-circle [from (2)]
⇒ arc DF + arc CXA + arc BF = Semi-circle [from (1)]
⇒ arc DF + arc FB + arc CXA = Semi-circle
⇒ arc DZB + arc CXA = Semi-circle
As, the circle divides itself in two semi-circles. Therefore, the remaining portion of the circle is also equal to the semi-circle.
Thus, arc AYD + arc BWC = Semi-circle
Hence, proved.

Page No 106:

Question 7:

If ABC is an equilateral triangle inscribed in a circle and P be any point on the minor arc BC which does not coincide with B or C, prove that PA is angle bisector of ∠BPC.

Answer:

Given: P is any point on the minor arc BC which does not coincide with B or C and $△$ ABC is an equilateral triangle inscribed on a circle.
To prove: PA is an angle bisector of $∠ BPC$ .
Construction: Join PB and PC.

Proof: As, $△ ABC$ is an equilateral triangle.
$∠ 3 = ∠ 4 = 60 °$

Also,
$∠ 1 = ∠ 4 = 60 ° . . . . . (1)$ [âˆµ angles in the same segment AB]
$∠ 2 = ∠ 3 = 60 ° . . . . . (2)$ [âˆµ angles in the same segment AC]
$∴ ∠ 1 = ∠ 2 = 60 °$

Thus, PA is the bisector of $∠ BPC .$
Hence, proved.

Page No 106:

Question 8:

In the figure AB and CD are two chords of a circle intersecting each other at point E. Prove that $∠ AEC = \frac{1}{2}$ (Angle subtended by arc CXA at centre + angle subtended by arc DYB at the centre).

Answer:

Given, two chords AB and CD intersecting each other at point E.
To prove: $∠ AEC = \frac{1}{2}$ [angle subtended by arc CXA at centre + angle subtended by arc DYB at the centre]
Construction: Extend the line DO and BO at the points I and H on the circle. Then, join AC.

Proof: In a circle, the angle subtended by an arc at the centre is twice the angle subtended by it on the remaining part of the circle.
$∴ ∠ 1 = 2 ∠ 6 . . . . . (1) and ∠ 3 = 2 ∠ 7 . . . . . (2)$

In $△$ AOC,
OC = OA [radius of circle]
$∠ OCA = ∠ 4$ [âˆµ angles opposite to equal sides are equal]

Also, by angle sum property of triangles,
$∠ AOC + ∠ OCA + ∠ 4 = 180 °$
$\Rightarrow ∠ AOC + ∠ 4 + ∠ 4 = 180 ° \Rightarrow ∠ AOC = 180 ° - 2 ∠ 4 . . . . . (3)$

In $△ OCD$ , angles opposite to equal sides are equal.
$∴ ∠ 6 = ∠ ECO$

Then, in $△ AEC$ , by angle property sum of a triangle,
$∠ AEC + ∠ ECA + ∠ CAE = 180 °$
$\begin{array}{rcl} \Rightarrow & ∠ AEC = 180 ° - (∠ ECA + ∠ CAE) \\ \Rightarrow & ∠ AEC = 180 ° - [(∠ ECO + ∠ OCA) + ∠ CAO + ∠ OAE] \\ = & 180 ° - (∠ 6 + ∠ 4 + ∠ 4 + ∠ 5) \end{array}$
$\begin{array}{rcl} \Rightarrow & ∠ AEC = 180 ° - (2 ∠ 4 + ∠ 5 + ∠ 6) \\ = & 180 ° - (180 ° - ∠ AOC + ∠ 7 + ∠ 6) \end{array}$
[from (3) and in $△ AOB, ∠ 5 = ∠ 7,$ as angles opposite to equal sides are equal.

$\begin{array}{rcl} ∠ AEC & = & ∠ AOC - \frac{∠ 3}{2} - \frac{∠ 1}{2} [from (1) and (2)] \\ = & ∠ AOC - \frac{∠ 1}{2} - \frac{∠ 2}{2} - \frac{∠ 3}{2} + \frac{∠ 2}{2} \\ = & ∠ AOC - \frac{1}{2} (∠ 1 + ∠ 2 + ∠ 3) + \frac{∠ 8}{2} [∠ 2 = ∠ 8 ∵ vertically opposite angles] \\ = & ∠ AOC - \frac{∠ AOC}{2} + \frac{∠ DOB}{2} \end{array}$
$\Rightarrow ∠ AEC = \frac{1}{2} (∠ AOC + ∠ DOB)$
$= \frac{1}{2}$ [angle subtended by arc CXA at the center + angle subtended by arc DYB at the center]
Hence, proved.

Page No 107:

Question 9:

If bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q, prove that PQ is a diameter of the circle.

Answer:

Given that, ABCD is a cyclic quadrilateral. DP and QB are the bisectors of ∠D and ∠B.
To prove: PQ is the diameter of the circle.
Construction: Join QD and QC.

Proof: ABCD is a cyclic quadrilateral. The bisectors of opposite angles of the cyclic quadrilateral, ∠A and ∠C, intersect the circle circumscribing at the points P and Q respectively.
Now, the opposite angles of a cyclic quadrilateral are supplementary.
⇒ ∠A + ∠C = 180°
$\Rightarrow \frac{1}{2} ∠ A + \frac{1}{2} ∠ C = \frac{1}{2} \times 180 ° = 90 °$
⇒ ∠PAB + ∠BCQ = 90°

But ∠BCQ = ∠BAQ [âˆµ Angles in the segment of a circle are equal]
∴ ∠PAB + ∠BAQ = 90°
⇒ ∠PAQ = 90°
Thus, ∠PAQ is in semicircle.
∴ PQ is diameter of the circle.
Hence, proved.

Page No 107:

Question 10:

A circle has radius $\sqrt{2}$ cm. It is divided into two segments by a chord of length 2 cm. Prove that the angle subtended by the chord at a point in major segment is 45º.

Answer:

Given, a circle having centre O with Ab = 2 cm be a chord of a circle. The radius OM is divided by chord AB in two equal segments.
To prove: $∠ APB = 45 °$

Proof: As, the perpendiculard rawn from the centre of a circle to a chord, bisects it.
∴ AN = NB = 1 cm
and OB = $\sqrt{2} cm$

In $△ ONB$ , by Pythagoras theorem,
${OB}^{2} = {ON}^{2} + {NB}^{2}$
$\Rightarrow {(\sqrt{2})}^{2} = {ON}^{2} + {(1)}^{2} \Rightarrow {ON}^{2} = 2 - 1 = 1 \Rightarrow ON = 1 cm$
[using positive square root, as distance is always positive]

Also,
$∠ ONB = 90 °$ [âˆµ ON is the perpendicular bisector of the chord AB]
$∴ ∠ NOB = ∠ NBO = 45 °$

Similarly, $∠ AON = 45 °$ .

Then, $∠ AOB = ∠ AON + ∠ NOB$
= 45° + 45°
= 90°

The angle subtended by an arc at the centre of the circle if twice the angle subtended by it at any other point on the circle.
$\begin{array}{rcl} ∴ ∠ APB & = & \frac{1}{2} ∠ AOB \\ = & \frac{90 °}{2} \\ = & 45 ° \end{array}$
Hence, proved.

Page No 107:

Question 11:

Two equal chords AB and CD of a circle when produced intersect at a point P. Prove that PB = PD.

Answer:

Given: Two equal chords AB and CD of a circle intersecting at a point P.
To prove: PB = PD
Construction: Join OP, draw OL $⊥$ AB and OM $⊥$ CD.

Proof: As, AB = CD.
⇒ OL = OM [âˆµ equal chords are equidistant from the centre]

In $△ OLP and △ OMP$ ,
OL = OM
$∠ OLP = ∠ OMP$ [right angle i.e. 90°]
and OP = OP [common side]
$∴ △ OLP ≅ △ OMP$ [by RHS congruence rule]
⇒ LP = MP [by CPCT] .....(1)

Now, AB = CD.
$\Rightarrow \frac{1}{2} AB = \frac{1}{2} CD$ [dividing by 2 on both sides]
$\Rightarrow BL = DM . . . . . (2)$
[perpendicular draw from centre to the circle bisects the chord ⇒ AL = LB and CM = MD]

On subtracting $2 from 1$ , we get
LP $-$ BL = MP $-$ DM
$\Rightarrow$ PB = PD
Hence, proved.

Page No 107:

Question 12:

AB and AC are two chords of a circle of radius r such that AB = 2AC. If p and q are the distances of AB and AC from the centre, prove that 4q²= p²+ 3r².

Answer:

Given: In a circle of radius r, there are two chords AB and AC such that AB = 2AC. Then, the distance of AB and AC from the centre are p and q respectively.
To prove: 4q²= p²+ 3r²,
Proof: Let AC = a, then AB = 2a.

From the centre O, perpendiculars are drawn to the chords AC and AB which intersect them at points M and N respectively.
$∴ AM = MC = \frac{a}{2}$ and AN = NB = a

In $△ OAM$ , by Pythagoras theorem,
${AO}^{2} = {AM}^{2} + {MO}^{2}$
$\Rightarrow {AO}^{2} = {(\frac{a}{2})}^{2} + q^{2} . . . . . (1)$

In $△ OAN$ , by Pythagoras theorem,
${AO}^{2} = {(AN)}^{2} + {(NO)}^{2}$
$\Rightarrow {AO}^{2} = {(a)}^{2} + p^{2} . . . . . (2)$

From (1) and (2),
${(\frac{a}{2})}^{2} + q^{2} = a^{2} + p^{2}$
$\Rightarrow \frac{a^{2}}{4} + q^{2} = a^{2} + p^{2}$
$\Rightarrow a^{2} + 4 q^{2} = 4 a^{2} + 4 p^{2}$
$\Rightarrow 4 q^{2} = 3 a^{2} + 4 p^{2} \Rightarrow 4 q^{2} = p^{2} + 3 (a^{2} + p^{2})$
$\Rightarrow 4 q^{2} = p^{2} + 3 r^{3}$ [In right angled triangle $△ OAN, r^{2} = a^{2} + p^{2}$ ]
Hence, proved.

Page No 107:

Question 13:

In the figure, O is the centre of the circle, ∠BCO = 30°. Find x and y.

Answer:

Given, O is the centre of the circle and $∠ BCO = 30 °$ . Join OB and AC and let the perpendicular from the centre of the circle intersect the chord BC at E.

In $△ BOC,$
CO = BO [radius of circle]
$∴ ∠ OBC = ∠ OCB = 30 °$ [âˆµ Angles opposite to equal sides are equal]

By applying angle sum property of a triangle,
$\begin{array}{rcl} \Rightarrow & ∠ BOC = 180 ° - (∠ OBC + ∠ OCE) \\ = & 180 ° - (30 ° + 30 °) \\ = & 120 ° \end{array}$

In a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
$\Rightarrow ∠ BOC = 2 ∠ BAC$
$∴ ∠ BAC = \frac{120 °}{2} = 60 °$

Now, consider $△$ ABE and $△$ ACE.
AE = AE (common)
$∠ AEB = ∠ AEC = 90 °$
BE = EC (the perpendicular from the centre of the circle to a chord, bisects it)
$∴ △$ ABE $≅ △$ ACE (by SAS congruence)
$\Rightarrow ∠ BAE = ∠ CAE = 30 ° (by CPCT)$ [AE is an angle bisector of angle A]
$\Rightarrow x = 30 °$

In $△$ ABE, by angle sum property of a triangle,
$∠ BAE + ∠ EBA + ∠ AEB = 180 °$
$\Rightarrow 30 ° + ∠ EBA + 90 ° = 180 °$
$\begin{array}{rcl} ∴ ∠ EBA & = & 180 ° - (90 ° + 30 °) \\ = & 180 ° - 120 ° \\ = & 60 ° \end{array}$

Thus, $∠ EBA = 60 °$ .
$\Rightarrow ∠ ABD + y = 60 ° \Rightarrow \frac{1}{2} ∠ AOD + y = 60 ° (∵ angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle)$
$\Rightarrow \frac{90 °}{2} + y = 60 ° [∵ ∠ AOD = 90 °]$
$\Rightarrow 45 ° + y = 60 ° \Rightarrow y = 60 ° - 45 ° ∴ y = 15 °$
Hence, x is 30° and y is 15°.

Page No 107:

Question 14:

In the figure, O is the centre of the circle, BD = OD and CD ⊥ AB. Find ∠CAB.

Answer:

In the figure below, BD = OD, CD $⊥$ AB
In $△$ OBD,
BD = OD
and OD = OB [radius of circle]
$\Rightarrow OB = OD = BD$
Then, $△$ ODB is an equilateral triangle.
$\Rightarrow ∠ BOD = ∠ OBD = ∠ ODB = 60 °$

In $△ MBC and △ MBD$ ,
MB = MB [common side]
$∠ CMB = ∠ BMD = 90 °$
CM = MD (âˆµ In a circle, the perpendicular drawn to a chord from the centre of the circle, bisects it)
$∴ △ MBC ≅ △ MBD$ [by SAS congruence rule]
$\Rightarrow ∠ MBC = ∠ MBD$ [by CPCT]
$\Rightarrow ∠ MBC = ∠ OBD = 60 ° [∵ ∠ OBD = 60 °]$

As, AB is a diameter of the circle. Therefore, angle subtended by it at any point on the circle is a right angle.
$∴ ∠ ACB = 90 °$

In $△$ ACB, by angle sum property of a triangle,
$∠ CAB + ∠ CBA + ∠ ACB = 180 °$
$\Rightarrow ∠ CAB + 60 ° + 90 ° = 180 ° \Rightarrow ∠ CAB = 180 ° - (60 ° + 90 °) = 30 °$
Hence, the value of $∠ CAB$ is 30°.

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