# CLASS 9 MATH NCERT SOLUTION FOR CHAPTER – 10 CIRCLES EX – 10.5

## Circles

Question 1.
In figure A,B and C are three points on a circle with centre 0 such that ∠BOC = 30° and ∠ AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ ADC.

Solution:
We have a circle with centre O, such that
∠AOB = 60° and ∠BOC = 30°
∵∠AOB + ∠BOC = ∠AOC
∴ ∠AOC = 60° + 30° = 90°
The angle subtended by an arc at the circle is half the angle subtended by it at the centre.
∴ ∠ ADC = 12 (∠AOC) = 12(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:
We have a circle having a chord AB equal to radius of the circle.
∴ AO = BO = AB
⇒ ∆AOB is an equilateral triangle.
Since, each angle of an equilateral triangle is 60°.
⇒ ∠AOB = 60°
Since, the arc ACB makes reflex ∠AOB = 360° – 60° = 300° at the centre of the circle and ∠ACB at a point on the minor arc of the circle.

Hence, the angle subtended by the chord on the minor arc = 150°.
Similarly, ∠ADB = 12 [∠AOB] = 12 x 60° = 30°
Hence, the angle subtended by the chord on the major arc = 30°

Question 3.
In figure, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.

Solution:
The angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point pn the circumference.
∴ reflex ∠POR = 2∠PQR
But ∠PQR = 100°
∴ reflex ∠POR = 2 x 100° = 200°
Since, ∠POR + reflex ∠POR = 360°
⇒ ∠POR = 360° – 200°
⇒ ∠POR = 160°
Since, OP = OR [Radii of the same circle]
∴ In ∆POR, ∠OPR = ∠ORP
[Angles opposite to equal sides of a triangle are equal]
Also, ∠OPR + ∠ORP + ∠POR = 180°
[Sum of the angles of a triangle is 180°]
⇒ ∠OPR + ∠ORP + 160° = 180°
⇒ 2∠OPR = 180° -160° = 20° [∠OPR = ∠ORP]
⇒ ∠OPR = 20∘2 = 10°

Question 4.
In figure, ∠ABC = 69°,∠ACB = 31°, find ∠BDC.

Solution:
In ∆ABC, ∠ABC + ∠ACB + ∠BAC = 180°
⇒ 69° + 31° + ∠BAC = 180°
⇒ ∠BAC = 180° – 100° = 80°
Since, angles in the same segment are equal.
∴∠BDC = ∠BAC ⇒ ∠BDC = 80°

Question 5.
In figure, A, B and C are four points on a circle. AC and BD intersect at a point E such that ∠ BEC = 130° and ∠ ECD = 20°. Find ∠BAC.

Solution:

∠BEC = ∠EDC + ∠ECD
[Sum of interior opposite angles is equal to exterior angle]
⇒ 130° = ∠EDC + 20°
⇒ ∠EDC = 130° – 20° = 110°
⇒ ∠BDC = 110°
Since, angles in the same segment are equal.
∴ ∠BAC = ∠BDC
⇒ ∠BAC = 110°

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:
Since angles in the same segment of a circle are equal.
∴ ∠BAC = ∠BDC
⇒ ∠BDC = 30°

lso, ∠DBC = 70° [Given]
In ∆BCD, we have
∠BCD + ∠DBC + ∠CDB = 180° [Sum of angles of a triangle is 180°]
⇒ ∠BCD + 70° + 30° = 180°
⇒ ∠BCD = 180° -100° = 80°
Now, in ∆ABC,
AB = BC [Given]
∴ ∠BCA = ∠BAC [Angles opposite to equal sides of a triangle are equal]
⇒ ∠BCA = 30° [∵ ∠B AC = 30°]
Now, ∠BCA + ∠BCD = ∠BCD
⇒ 30° + ∠ECD = 80°
⇒ ∠BCD = 80° – 30° = 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:
Since AC and BD are diameters.
⇒ AC = BD …(i) [All diameters of a circle are equal]
Also, ∠BAD = 90° [Angle formed in a semicircle is 90°]
Similarly, ∠ABC = 90°, ∠BCD = 90°
and ∠CDA = 90°

Now, in ∆ABC and ∆BAD, we have
AC = BD [From (i)]
AB = BA [Common hypotenuse]
∠ABC = ∠BAD [Each equal to 90°]
∴ ∆ABC ≅ ∆BAD [By RHS congruence criteria]
Similarly, AB = DC
Thus, the cyclic quadrilateral ABCD is such that its opposite sides are equal and each of its angle is a right angle.
∴ ABCD is a rectangle.

Question 8.
If the non – parallel sides of a trapezium are equal, prove that it is cyclic.

Solution:
We have a trapezium ABCD such that AB ॥ CD and AD = BC.
Let us draw BE ॥ AD such that ABED is a parallelogram.
∵ The opposite angles and opposite sides of a parallelogram are equal.
But AD = BC [Given] …(iii)

∴ From (ii) and (iii), we have BE = BC
⇒ ∠BCE = ∠BEC … (iv) [Angles opposite to equal sides of a triangle are equal]
Now, ∠BED + ∠BEC = 180° [Linear pair]
⇒ ∠BAD + ∠BCE = 180° [Using (i) and (iv)]
i.e., A pair of opposite angles of a quadrilateral ABCD is 180°.
∴ ABCD is cyclic.
⇒ The trapezium ABCD 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:
Since, angles in the same segment of a circle are equal.
∴ ∠ACP = ∠ABP …(i)
Similarly, ∠QCD = ∠QBD …(ii)
Since, ∠ABP = ∠QBD …(iii) [Vertically opposite angles]
∴ From (i), (ii) and (iii), we have
∠ACP = ∠QCD

Question 10.
If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.

Solution:
We have ∆ABC, and two circles described with diameter as AB and AC respectively. They intersect at a point D, other than A.
Let us join A and D.

∵ AB is a diameter.
∴∠ADB is an angle formed in a semicircle.
Adding (i) and (ii), we have
i. e., B, D and C are collinear points.
⇒ BC is a straight line. Thus, D lies on BC.

Question 11.
ABC and ADC are two right angled triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.

Solution:
We have ∆ABC and ∆ADC such that they are having AC as their common hypotenuse and ∠ADC = 90° = ∠ABC
∴ Both the triangles are in semi-circle.

Case – I: If both the triangles are in the same semi-circle.

⇒ A, B, C and D are concyclic.
Join BD.
DC is a chord.
∴ ∠CAD and ∠CBD are formed in the same segment.

Case – II : If both the triangles are not in the same semi-circle.

⇒ A,B,C and D are concyclic. Join BD. DC is a chord.
∴ ∠CAD and ∠CBD are formed in the same segment.

Question 12.
Prove that a cyclic parallelogram is a rectangle.

Solution:
We have a cyclic parallelogram ABCD. Since, ABCD is a cyclic quadrilateral.
∴ Sum of its opposite angles is 180°.
⇒ ∠A + ∠C = 180° …(i)
But ∠A = ∠C …(ii)
[Opposite angles of a parallelogram are equal]
From (i) and (ii), we have
∠A = ∠C = 90°

Similarly,
∠B = ∠D = 90°
⇒ Each angle of the parallelogram ABCD is 90°.
Thus, ABCD is a rectangle.