Gujarat Board GSEB Solutions Class 9 Maths Chapter 10 Circles Ex 10.5 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 9 Maths Chapter 10 Circles Ex 10.5

Question 1.

Infigure,A,B and Carethreepointsona circle with centre O such that ∠BOC = 300 and ∠AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ADC.

Solution:

∠ADC = \(\frac {1}{2}\)∠AOC

[The angle subtended by an arc at the centre is double the angle subtended by it any point on the remaining part of the circle.]

= \(\frac {1}{2}\)(∠AOB + ∠BOC)

= \(\frac {1}{2}\) (60° + 300)

= \(\frac {1}{2}\) (90°) = 45°

Question 2.

A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

Solution:

∴ OA = OB = AB [Given]

∴ OAB is equilateral.

∴ ∠AOB = 60°

∠ACB = \(\frac {1}{2}\)∠A0B

[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.]

∴ ∠ACB = \(\frac {1}{2}\) x 60° = 30°

Now, ADBC is a cyclic quadrilateral.

∴ ∠ADB + ∠ACB = 180°

[The sum of either pair of opposite angles of a cyclic quadrilateral is 180°]

⇒ ∠ADB + 30°= 180°

⇒ ∠ADB = 180°- 30°

∴ ∠ADB = 150°

Question 3.

In figure, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.

Solution:

Take a point S in the major arc. Join PS and RS.

∴ PQRS is a cyclic quadrilateral.

∠PQR + ∠PSR = 180°

The sum of either pair of opposite angles of a cyclic quadrilateral is 180°

100° + ∠PSR = 180°

∠PSR = 180° – 100°

⇒ ∠PSR = 80° …….(1)

Now, ∠POR = 2∠PSR

(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]

∴ ∠POR= 2 x 800 = 160° …….(2) Using (1)

In ΔOPR,

∴ OP = OR Radiiofacircle

∴ ∠OPR = ∠ORP …………(3)

Angles opposite to equal sides of a triangle are equal

Also,

∠OPR + ∠ORP + ∠POR = 180°

Sum of all the angles of a triangle is 1800

⇒ ∠OPR + ∠OPR + 160° = 180°

Using (2) and (1)

⇒ 2∠OPR + 160° = 180°

⇒ 2∠OPR = 180° – 160° = 20°

∠OPR = 10°

Question 4.

In figure, ∠ABC = 69°, ∠ACB = 31°, find ∠BDC.

Solution:

In ΔABC,

∠BAC + ∠ABC + ∠ACB = 180°

[Sum of all the angles of a triangle is 180°]

∠BAC + 69° + 31° = 180°

∠BAC + 100° = 1800

∠BAC = 180° = 100° = 80° ………..(1)

Now, ∠BDC = ∠BAC

[Angles in the same segment of a circle are equal]

∠BDC = 80° [Using (1)]

Question 5.

In figure, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠BEC = 130° and ∠ECD = 20°. Find ∠BAC.

Solution:

In ΔCDE ∠CDE + ∠DCE = ∠BEC

[Exterior angle property of a triangle]

⇒ ∠CDE + 20° = 130°

⇒ ∠CDE = 130° – 20° = 110° ………..(2)

Now, ∠BAC = ∠CDE

[Angles in the same segment of a circle are equal]

∴ ∠BAC = ∠110° [Usmg(1)]

Question 6.

ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, If AB = BC, find ∠ECD.

Solution:

∠CDB = ∠BAC

[Angles in the same segment of a circle are equal]

∠CDB = 30° ………..(1)

∠DBC = 70° ………..(2) (Given)

In ΔBCD,

∠BCD + ∠DBC + ∠CDB = 180°

(Sum of all the angles of a triangle is 180°]

⇒ ∠BCD + 70° + 300 = 180° [Using (1) and (2)1

⇒ ∠BCD + 100° = 180°

∠BCD = 180° – 100°

⇒ ∠BCD = 80° ………(3)

In ΔABC, AB = BC

∠BCA = ∠BAC

[Angles opposite to equal sides of a triangle are equal]

⇒ ∠BCA = 30° ………(4)

∠BAC = 30° (given]

Now, ∠BCD = 80° (From (3))

= ∠BCA + ∠ECD = 80°

⇒ 30°+∠ECD = 80°

⇒ ∠ECD = 80 – 30°

⇒ ∠ECD = 50°

Question 7.

If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

Solution:

In ΔOAB and ΔOCD,

OA = OC [Radii of a circle]

OB = OD [Radii of a circle]

∠AOB = ∠COD [Vertically opposite angles]

∴ ΔOAB = ΔOCD [SAS rule]

∴ AB = CD …….(1) [CPCTI

Similarly, we can show that

AD = CB ……..(2)

Adding (1) and (2), we get

ABCD is a || gm.

(A quadrilateral having opposite sides equal is a parallelogram)

Now, diagonal BD is also a diameter

∴ ∠BAD = \(\frac {1}{2}\)BOD

= \(\frac {1}{2}\) x 180° = 90°

∴ ABCD is the rectangle. (A parallelogram with an angle 90° is a rectangel)

Question 8.

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

Solution:

Given: ABCD is a trapezium whose two non-parallel sides AD and BC are equal.

To Prove: Trapezium ABCD is cyclic.

Construction: Draw BE || AD

Proof: AB || DE [Given]

AD || BE [By construction]

∴ Quadrilateral ABED is a parallelogram.

∴ ∠BAD = ∠BED ……….(1) [Opp. ∠s of a || gm]

and AD = BE [Opp. sides of a gin]

But AD = BC ………(3) [Given]

From (2) and (3),

BE = BC

∴ ∠BEC = ∠BCE ………(4)

[Angles opposite to equal sides]

∠BEC + ∠BED = 1800 [ Linear Pair Axiom]

⇒ ∠BCE + ∠BAD = 1800 [From (4) and (1)]

⇒ Trapezium ABCD is cyclic.

[ If a pair of opposite angles of a quadrilateral is 180°, then the quadrilateral is cyclic.]

Question 9.

Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see figure). Prove that

∠ACP = ∠QCD.

Solution:

∠ACP = ∠ABP …………(1)

[Angles in the same segment of a circle are equal]

And ., ∠QCD = ∠QBD …………(2)

[Angles in the same segment of a circle are equal]

But ∠ABP = ∠QBD [Vertically opposite angles]

∠ACP = ∠QCD. [From (1) and (2)]

Question 10.

If circles are drawn taking two sides of a triangle as diameter, prove that the point of intersection of these circles lies on the third side.

Solution:

Given: Circles are described with sides AB and AC of a triangle ABC as diameters. They intersect at a point D.

To Prove: D lies on the third side BC of ∠ABC.

Construction: Join AD.

Proof: Circle described on AB as diameter intersects BC in D.

∴ ∠ADB = 90°

[Angle in a semicircle Similarly, the circle described on AC as diameter passes through D.]

∠ADC = 90° ………..(2)

Now, adding (1) and (2) we get

∠ADB + ∠ADC = 180°

∴ Points B, D, C are collinear.

∴ D lies on BC.

Question 11.

ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.

Solution:

∴ AC is the common hypotenuse of two right triangles ABC and ADC.

∴ ∠ABC = 90° = ∠ADC

Here, there are two cases arising i.e., B and D are either on the same side of AC or on opposite sides.

Case I – Both the triangles are in the same semicircle.

∴ Points A, B, D and C are concyclic.

Case II – Triangles are on opposite sides, then

∠ABC + ∠ADC = 1800

∴ A, B, C and D are concyclic.

Now, in both cases, DC is a chord

∴∠CAD = ∠CBD

[∴ Angles in the same segment are equal]

Question 12.

Prove that a cyclic parallelogram is a rectangle.

Solution:

Given: ABCD is a cyclic parallelogram.

To prove: ABCD is a rectangle.

Proof: ABCD is a cyclic quadrilateral.

∴ ∠1 + ∠2 = 180° ……….(1)

[∴ Opposite angles of a cyclic quadrilateral are supplementary]

∴ ABCD is a parailelograin.

∴ ∠1 = ∠2 ………(2) [Opp. angles of a || gm]

From (1) and (2),

∠1 = ∠2 = 90°

∴ || gm ABCD is a rectangle.