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# Ex 10.5 NCERT Solutions- Circles Class 9 Notes | EduRev

## Mathematics (Maths) Class 9

Created by: Full Circle

## Class 9 : Ex 10.5 NCERT Solutions- Circles Class 9 Notes | EduRev

The document Ex 10.5 NCERT Solutions- Circles Class 9 Notes | EduRev is a part of the Class 9 Course Mathematics (Maths) Class 9.
All you need of Class 9 at this link: Class 9

Question 1. In the figure, A, B and C are three points on a circle with centre O such that ∠ BOC = 30º 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 = 60º and ∠BOC = 30º
∵ ∠AOB + ∠BOC = ∠AOC
∴ ∠AOC = 60º + 30º = 90º
Now, the arc ABC subtends ∠AOC = 90º
at the centre and ∠ADC at a point D on the circle other than the arc ABC.
⇒ ∠ADC = (1/2)(90º) = 45º

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
⇒ ΔAOB is an equilateral triangle.
Since,  each angle of an equilateral = 60º.
⇒ ∠AOB = 60º Since, the arc ACB makes reflex ∠AOB = 360º ∠ 60º = 300º at the centre of the circle and ∠ABC at a point on the minor arc of the circle.
∴ ∠ACB = (1/2)           [reflex ∠AOB]
= (1/2)[300º] = 150º
= (1/2)x [60º] = 30°

Thus, the angle subtended by the chord on the minor arc = 150º and on the major arc = 30º.

Question 3. In the figure, ∠ PQR = 100º, 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 = 100º
∴ reflex ∠POR = 2 x 100º = 200º
Since, ∠POR + reflex ∠POR = 360º
∴ ∠POR + 200º = 360º
⇒ ∠POR = 360º ∠ 200º
⇒ ∠POR = 160º
Since, OP = OR               [Radii of the same circle]
∴ In ΔPOR,∠OPR = ∠ORP               [Angles opposite to equal sides of a triangle are equal]
Also, ∠OPR + ∠ORP  + ∠POR = 180º               [Sum of the angles of a triangle = 180°]
⇒ ∠OPR + ∠OPR + 160º = 180º               [∵ ∠OPR = ∠ORP]
⇒ 2∠OPR = 180º ∠ 160º = 20º
⇒ ∠OPR = (200/2)= 10º

Question 4. In the figure, ∠ ABC = 69º, ∠ ACB = 31º, find ∠ BDC.
Solution:
We have, in ΔABC, ∠ABC = 69º and ∠ACB = 31º But ∠ABC + ∠ACB + ∠BAC = 180º
∴ 69º + 31º + ∠BAC = 180º
⇒ ∠BAC = 180º ∠ 69º ∠ 31º = 80º
Since, angles in the same segment are equal.
∴ ∠BDC = ∠BAC
⇒ ∠BDC = 80º

Question 5. In the 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 = 20º.
Find ∠BAC.
Solution:
In D CDE, Exterior ∠BEC = {Sum of interior opposite angles}               [∵ BD is a straight line.] 130º = ∠EDC + ∠ECD 130º
= ∠EDC + 20º
⇒ ∠EDC = 130º ∠ 20º = 110º
⇒ ∠BDC = 110º
Since, angles in the same segment are equal.
∴ ∠BAC = ∠BDC
⇒ ∠BAC = 110º

Question 6. ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70º, ∠BAC is 30º, find ∠BCD. Further, if AB = BC, find ∠ECD.
Solution
: ∵ Angles in the same segment of a circle are equal.
∴ ∠BAC = ∠BDC
⇒ 30º = ∠BDC Also ∠DBC = 70º               [Given] ∴ In ΔBCD, we have ∠BCD + ∠DBC + ∠CDB = 180º               [∵ Sum of angles of a triangle is 180º]
⇒ ∠BCD + 70º + 30º = 180º
⇒ ∠BCD = 180º ∠ 70º ∠ 30º = 80º
Now, in ΔABC,
∵ AB = BC               [Given]
∴ ∠BCA = ∠BAC               [Angles opposite to equal sides of a triangle are equal]
⇒ ∠BCA = 30º               [∵ ∠BAC = 30º]
Now, ∠BCA + ∠ECD = ∠BCD
⇒ 30º + ∠ECD = 80º
⇒ ∠ECD = 80º ∠ 30º = 50º

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:
∵ AC and BD are diameters.
∴ AC = BD               [∵ All diameters of a circle are equal]               …(1)
Since a diameter divides a circle into equal parts. ∴ ∠BAD = 90º               [∵ Angle formed in a semicircle is 90º]
Similarly, ∠ABC = 90º ∠BCD = 90º and ∠CDA = 90º
Now, in right ΔABC and right ΔBAD, AC = BD              [From (1)]
AB = AB               [Common]
∴ ΔABC ≌ ΔBAD               [RHS criterion]
Similarly, AB = CD Thus, the cyclic quadrilateral ABCD is such that its opposite sides are equal and each of its angle is 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 of a parallelogram are equal. and AD = BE               [Opposite sides of a parallelogram]               …(2)
But AD = BC               [Given]               …(3)
∴ From (2) and (3), we have BE = BC
⇒ ∠BEC = ∠BCE               [Angles opposite to equal sides of a triangle D are equal]               …(4)
Now, ∠BED + ∠BEC = 180º               [Linear pairs]
⇒ ∠BAD + ∠BCE = 180º               [Using (1) and (4)]
i.e. A pair of opposite angles of quadrilateral ABCD is 180º.
∴ ABCD is cyclic.
⇒ The trapezium ABCD is a 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 the figure). Prove that ∠ACP = ∠QCD.
Solution:
Since angles in the same segment of a circle are equal. ∴ ∠ACP = ∠ABP …(1)
Similarly, ∠QCD = ∠QBD               …(2)
Since ∠ABP = ∠QBD               [Vertically opposite angles are equal]
∴ From (1) and (2), 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 a DABC, and two circles described with diameters as AB and AC respectively. They intersect in a point D.
Let us join A and D.
∵ AB is a diameter.
∴ ∠ADB is an angle formed in a semicircle. 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 triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.
Solution:
We have right ΔABC and right ΔADC such that they are having AC as their common hypotenuse.
∵ AC is a hypotenuse.
∵ ∠ADC = 90º = ∠ABC
∴ Both the triangles are in the same semicircle. ⇒ A, B, C and  D are concyclic.
Let us join B and D.
∵ 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 = 180º
⇒ ∠A + ∠C = 180º               …(1)
But ∠A = ∠C               …(2) [∵ Opposite angles of parallelogram are equal]
From (1) and (2), we have ∠A= ∠C = 90º
Similarly, ∠B= ∠D = 90º
⇒ Each angle of the parallelogram ABCD is of 90º.
Thus, ABCD is a rectangle.

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