**NCERT Solutions for Class 9
Maths Chapter 10 Circles Ex 10.5**

NCERT Solutions for Class
9 Maths Chapter 10 Circles Ex 10.5 are the part
of NCERT Solutions for Class 9 Maths. Here you can find the NCERT Solutions for
Chapter 10 Circles Ex 10.5.

**Ex 10.5 Class 9 Maths Question 1.****
**In figure, A, B and C are three
points on a circle with centre O such that ∠BOC = 30° and ∠AOB = 60°. If D is a point on the circle other
than the arc ABC, find ∠ ADC.

**Solution:**

Given a circle with centre O, such that ∠AOB = 60° and ∠BOC = 30°

∵ ∠AOB + ∠BOC = ∠AOC

∴ ∠AOC = 60° + 30° = 90°

Since, the angle subtended by an arc at the circle is half the angle subtended by it at the centre.

∴ ∠ADC = ½ (∠AOC) = ½ (90°) = 45°

**Ex 10.5 Class 9 Maths 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.

⇒ ∆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.

Therefore,
∠ACB = ½ [reflex ∠AOB]

= ½ [300°] = 150°

Hence,
the angle subtended by the chord on the minor arc = 150°

Similarly, ∠ADB = ½ [∠AOB] = ½ × 60° = 30°

Hence, the angle subtended by the chord on the
major arc = 30°

**Ex 10.5 Class 9 Maths 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 on the circumference.

∴ reflex ∠POR = 2∠PQR

But ∠PQR = 100°

∴ reflex ∠POR = 2 × 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°

**Ex 10.5 Class 9 Maths 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°

**Ex 10.5 Class 9 Maths 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:
**∠BEC = ∠EDC + ∠ECD

[Exterior angle is equal to the sum of the opposite interior angles]

⇒ 130° = ∠EDC + 20°

⇒ ∠EDC = 130° – 20° = 110°

⇒ ∠BDC = 110°

Since, angles in the same segment are equal.

∴ ∠BAC = ∠BDC

⇒ ∠BAC = 110°

**Ex 10.5 Class 9 Maths 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°

We have ∠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° [∵ ∠BAC = 30°]

Now, ∠BCA + ∠ECD = ∠BCD

⇒ 30° + ∠ECD = 80°

⇒ ∠ECD = 80° – 30° = 50°

**Ex 10.5 Class 9 Maths 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°

AC = BD [From (i)]

AB = BA [Common side]

∠ABC = ∠BAD [Each equal to 90°]

∴ ∆ABC ≅ ∆BAD [By RHS congruence rule]

⇒ BC = AD [C.P.C.T.]

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.

**Ex 10.5 Class 9 Maths 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.

∴ ∠BAD = ∠BED ….…(i)

and AD = BE ….…(ii)

But AD = BC [Given] ….…(iii)

⇒ ∠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.

**Ex 10.5 Class 9 Maths 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.

Since, ∠ABP = ∠QBD ….…(iii) [Vertically opposite angles]

∴ From (i), (ii) and (iii), we have

∠ACP = ∠QCD

**Ex 10.5 Class 9 Maths 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.

∴ ∠ADB is an angle formed in a semicircle.

⇒ ∠ADB = 90° ….…(i)

Similarly, ∠ADC = 90° ..….(ii)

Adding (i) and (ii), we have

∠ADB + ∠ADC = 90° + 90° = 180°

i.e., B, D and C are collinear points.

⇒ BC is a straight line. Thus, D lies on BC.

**Ex 10.5 Class 9 Maths 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 = ∠ABC = 90°

∴ Both the triangles are in semi-circle.

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

∴ ∠CAD and ∠CBD are formed in the same segment.

⇒ ∠CAD = ∠CBD

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

∴ ∠CAD and ∠CBD are formed in the same segment.

⇒ ∠CAD = ∠CBD

**Ex 10.5 Class 9 Maths 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°

⇒ Each angle of the parallelogram ABCD is 90°.

Thus, ABCD is a rectangle.

**NCERT Solutions for Maths Class 10**