# GSEB Solutions Class 9 Maths Chapter 10 Circles Ex 10.5

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°

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°

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´ 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
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.
Proof: AB || DE [Given]
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.

Proof: Circle described on AB as diameter intersects BC in D.
[Angle in a semicircle Similarly, the circle described on AC as diameter passes through D.]
Now, adding (1) and (2) we get
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´ 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.