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 âˆ BOC = 300 and âˆ AOB = 60Â°. If D is a point on the circle other than the arc ABC, find âˆ ADC.

Solution:

âˆ ADC = \(\frac {1}{2}\)âˆ 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.]

= \(\frac {1}{2}\)(âˆ AOB + âˆ BOC)

= \(\frac {1}{2}\) (60Â° + 300)

= \(\frac {1}{2}\) (90Â°) = 45Â°

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:

âˆ´ OA = OB = AB [Given]

âˆ´ OAB is equilateral.

âˆ´ âˆ AOB = 60Â°

âˆ ACB = \(\frac {1}{2}\)âˆ 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.]

âˆ´ âˆ ACB = \(\frac {1}{2}\) x 60Â° = 30Â°

Now, ADBC is a cyclic quadrilateral.

âˆ´ âˆ ADB + âˆ ACB = 180Â°

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

â‡’ âˆ ADB + 30Â°= 180Â°

â‡’ âˆ ADB = 180Â°- 30Â°

âˆ´ âˆ ADB = 150Â°

Question 3.

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

Solution:

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

âˆ´ PQRS is a cyclic quadrilateral.

âˆ PQR + âˆ PSR = 180Â°

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

100Â° + âˆ PSR = 180Â°

âˆ PSR = 180Â° – 100Â°

â‡’Â âˆ PSR = 80Â° …….(1)

Now, âˆ POR = 2âˆ 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]

âˆ´ âˆ POR= 2 x 800 = 160Â° …….(2) Using (1)

In Î”OPR,

âˆ´ OP = OR Radiiofacircle

âˆ´ âˆ OPR = âˆ ORP …………(3)

Angles opposite to equal sides of a triangle are equal

Also,

âˆ OPR + âˆ ORP + âˆ POR = 180Â°

Sum of all the angles of a triangle is 1800

â‡’ âˆ OPR + âˆ OPR + 160Â° = 180Â°

Using (2) and (1)

â‡’ 2âˆ OPR + 160Â° = 180Â°

â‡’ 2âˆ OPR = 180Â° – 160Â° = 20Â°

âˆ OPR = 10Â°

Question 4.

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

Solution:

In Î”ABC,

âˆ BAC + âˆ ABC + âˆ ACB = 180Â°

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

âˆ BAC + 69Â° + 31Â° = 180Â°

âˆ BAC + 100Â° = 1800

âˆ BAC = 180Â° = 100Â° = 80Â° ………..(1)

Now, âˆ BDC = âˆ BAC

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

âˆ BDC = 80Â° [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 âˆ BEC = 130Â° and âˆ ECD = 20Â°. Find âˆ BAC.

Solution:

In Î”CDE âˆ CDE + âˆ DCE = âˆ BEC

[Exterior angle property of a triangle]

â‡’ âˆ CDE + 20Â° = 130Â°

â‡’ âˆ CDE = 130Â° – 20Â° = 110Â° ………..(2)

Now, âˆ BAC = âˆ CDE

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

âˆ´ âˆ BAC = âˆ 110Â° [Usmg(1)]

Question 6.

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

Solution:

âˆ CDB = âˆ BAC

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

âˆ CDB = 30Â° ………..(1)

âˆ DBC = 70Â° ………..(2) (Given)

In Î”BCD,

âˆ BCD + âˆ DBC + âˆ CDB = 180Â°

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

â‡’ âˆ BCD + 70Â° + 300 = 180Â° [Using (1) and (2)1

â‡’ âˆ BCD + 100Â° = 180Â°

âˆ BCD = 180Â° – 100Â°

â‡’ âˆ BCD = 80Â° ………(3)

In Î”ABC, AB = BC

âˆ BCA = âˆ BAC

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

â‡’ âˆ BCA = 30Â° ………(4)

âˆ BAC = 30Â° (given]

Now, âˆ BCD = 80Â° (From (3))

= âˆ BCA + âˆ ECD = 80Â°

â‡’ 30Â°+âˆ ECD = 80Â°

â‡’ âˆ ECD = 80 – 30Â°

â‡’ âˆ ECD = 50Â°

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 Î”OAB and Î”OCD,

OA = OC [Radii of a circle]

OB = OD [Radii of a circle]

âˆ AOB = âˆ COD [Vertically opposite angles]

âˆ´ Î”OAB = Î”OCD [SAS rule]

âˆ´ 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

âˆ´ âˆ BAD = \(\frac {1}{2}\)BOD

= \(\frac {1}{2}\) x 180Â° = 90Â°

âˆ´ ABCD is the rectangle. (A parallelogram with an angle 90Â° 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]

âˆ´ Quadrilateral ABED is a parallelogram.

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

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

But AD = BC ………(3) [Given]

From (2) and (3),

BE = BC

âˆ´ âˆ BEC = âˆ BCE ………(4)

[Angles opposite to equal sides]

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

â‡’ âˆ BCE + âˆ BAD = 1800 [From (4) and (1)]

â‡’ Trapezium ABCD is cyclic.

[ If a pair of opposite angles of a quadrilateral is 180Â°, 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

âˆ ACP = âˆ QCD.

Solution:

âˆ ACP = âˆ ABP …………(1)

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

And ., âˆ QCD = âˆ QBD …………(2)

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

But âˆ ABP = âˆ QBD [Vertically opposite angles]

âˆ ACP = âˆ 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 âˆ ABC.

Construction: Join AD.

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

âˆ´ âˆ ADB = 90Â°

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

âˆ ADC = 90Â° ………..(2)

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

âˆ ADB + âˆ ADC = 180Â°

âˆ´ Points B, D, C are collinear.

âˆ´ D lies on BC.

Question 11.

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

Solution:

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

âˆ´ âˆ ABC = 90Â° = âˆ 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.

âˆ´ Points A, B, D and C are concyclic.

Case II – Triangles are on opposite sides, then

âˆ ABC + âˆ ADC = 1800

âˆ´ A, B, C and D are concyclic.

Now, in both cases, DC is a chord

âˆ´âˆ CAD = âˆ CBD

[âˆ´ 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.

âˆ´ âˆ 1 + âˆ 2 = 180Â° ……….(1)

[âˆ´ Opposite angles of a cyclic quadrilateral are supplementary]

âˆ´ ABCD is a parailelograin.

âˆ´ âˆ 1 = âˆ 2 ………(2) [Opp. angles of a || gm]

From (1) and (2),

âˆ 1 = âˆ 2 = 90Â°

âˆ´ || gm ABCD is a rectangle.