Jharkhand Board JAC Class 9 Maths Solutions Chapter 10 Circles Ex 10.5 Textbook Exercise Questions and Answers.

## JAC Board Class 9th Maths Solutions Chapter 10 Circles Ex 10.5

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Question 1.

In the 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.

Answer:

Here, ∠AOC = ∠AOB + ∠BOC

∠AOC = 60° + 30°

⇒ ∠AOC = 90°

We know that angle subtended by an arc at centre is double the angle subtended by it at any point on the remaining part of the circle.

∠ADC = \(\frac{1}{2}\) ∠AOC = \(\frac{1}{2}\) × 90° = 45°

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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 nugor arc.

Answer:

Given: Chord AB is equal to the radius of the circle.

In ∆OAB, OA = OB = AB = radius of the circle.

Thus, AOAB is an equilateral triangle.

∠AOB = 60°

also, ∠ACB = \(\frac{1}{2}\) ∠AOB = \(\frac{1}{2}\) × 60° = 30°

ACBD is a cyclic quadrilateral,

∠ACB + ∠ADB = 180° (Opposite angles of cyclic quadrilateral)

⇒ ∠ADB = 180° – 30° = 150°

Thus, angle subtend by the chord at a point on the minor arc and also at a point on the major arc are 150° and 30° respectively.

Question 3.

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

Answer:

Reflex ∠POR = 2 × ∠PQR = 2 × 100° = 200°

∴ ∠POR = 360° – 200° = 160°

In ∆OPR,

OP = OR (radii of the circle)

∠OPR = ∠ORP

(Angles opposite to equal sides)

Now,

∠OPR + ∠ORP +∠POR = 180°

(Sum of the angles in a triangle)

⇒ ∠OPR + ∠OPR + 160° = 180°

⇒ 2∠OPR= 180°- 160°

⇒ ∠OPR =10°

Question 4.

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

Answer:

∠BAC = ∠BDC (Angles in the same segment of the circle)

In ∆ABC,

∠BAC + ∠ABC+ ∠ACB =180° (Sum of the angles of a triangle)

⇒ ∠BAC + 69°+ 31° = 180°

⇒ ∠BAC = 180°- 100°

⇒ ∠BAC = 80°

Thus, ∠BDC = 80°

Question 5.

In Fig, 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.

Answer:

∠BAC = ∠CDE

(Angles in the segment of the circle)

In ∆CDE,

∠CEB = ∠CDE + ∠DCE (Exterior angles of the triangle)

=> 130° = ∠CDE + 20°

⇒ ∠CDE =110°

Thus, ∠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.

Answer:

For chord CD,

∠CBD = ∠CAD (Angles in same segment)

∴ ∠CAD = 70°

∠BAD = ∠BAC + ∠CAD

= 30° + 70° = 100°

∠BCD + ∠BAD = 180° (Opposite angles of a cyclic quadrilateral)

⇒ ∠BCD + 100° = 180°

⇒ ∠BCD = 80°

In ∆ABC, AB = BC (given)

∠BCA = ∠CAB (Angles opposite to equal sides of a triangle)

∴ ∠BCA = 30°

Also, ∠BCD = 80°

∴ ∠BCA + ∠ACD = 80°

⇒ 30° + ∠ACD = 80°

∠ACD = 50°

i.e. ∠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.

Answer:

Given: ABCD is a cyclic quadrilateral, whose diagonals AC and BD are diameters of the circle passing through A, B, C and D.

To prove: ABCD is a rectangle

Proof: In AAOD and ACOB AO = CO (Radii of a circle)

OD = OB (Radii of a circle)

∠AOD = ∠COB (Vertically opposite angles)

∴ ∆AOD ≅ ∆COB (By SAS axiom)

∴ ∠AOD S ∠OCB (By CPCT )

∴ AD || BC Similarly AB || CD Hence quadrilateral ABCD is a parallelogram.

∠ABC = ∠BCD = ∠CDA = ∠DAB = 90°

(Angles in the semi-circle)

Thus, ABCD is a parallelogram with each internal angle as 90°. So, ABCD is a rectangle.

Question 8.

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

Answer:

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

Construction: DM and CN are perpendiculars drawn on AB from D and C respectively.

To prove: ABCD is cyclic trapezium.

Proof: In ∆DAM and ∆CBN,

AD = BC (Given)

∠AMD = ∠BNC = 90° (By construction) DM = CN (Distance between the parallel lines)

∆DAM ≅ ∆CBN by RHS congruence criterion.

Now, ∠A = ∠B by CPCT

Also, ∠B + ∠C = 180° (sum of the co-interior angles)

⇒ ∠A + ∠C = 180°

In trapezium ABCD,

∠A + ∠B + ∠C + ∠D = 360°

=> 180° + ∠B + ∠D = 360°

⇒ ∠B + ∠D = 360° – 180°

= 180°

Thus, ABCD is a cyclic trapezium as sum of the pair of opposite angles is 180°.

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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 Fig). Prove that ∠ACP = ∠QCD.

Answer:

Chords AP and DQ are joined.

For chord AP,

∠PBA = ∠ACP

(Angles in the same segment) …(i)

For chord DQ,

∠DBQ = ∠QCD

(Angles in same segment) …(ii)

ABD and PBQ are line segments intersecting at B.

∠PBA = ∠DBQ

(Vertically opposite angles) …(iii)

By the equations (i), (ii) and (iii),

∠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.

Answer:

Given: Two circles are drawn on the sides AB and AC of the triangle ΔABC as diameters. The circles intersected at D.

Construction: AD is joined.

To prove: D lies on BC. We have to prove that BDC is a straight line.

Proof: ∠ADB = ∠ADC = 90° (Angle in the semi circle)

Now, ∠ADB + ∠ADC = 90° + 90° = 180°

⇒ BDC is straight line.

Thus, D lies on the side BC.

Question 11.

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

Answer:

Given: AC is the common hypotenuse.

∠ABC = ∠ADC = 90°.

To prove: ∠CAD = ∠CBD

Proof: Since, ∠ABC and ∠ADC are 90°.

These angles are in the semi circle. Thus, both the triangles are lying in the semi circle and AC is the diameter of the circle.

⇒ Points A, B, C and D are concyclic.

Thus, CD is the chord.

⇒ ∠CAD = ∠CBD (Angles in the same segment of the circle)

Question 12.

Prove that a cyclic parallelogram is a rectangle.

Answer:

Given: ABCD is a cyclic parallelogram.

To prove: ABCD is rectangle.

Proof: ∠1 + ∠2 = 180° (Opposite angles of a cyclic parallelogram)

also, opposite angles of a parallelogram are equal.

Thus,

∠1 = ∠2

⇒ ∠1 + ∠1 = 180°

⇒ ∠1 = 90°

One of the interior angles of the parallelogram is right angle. Thus, ABCD is a rectangle.