Circles
Chapter 10
Exercise 10.5
EX 10.5 QUESTION 1.
In Fig. 10.36, 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:
∠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 =(½)∠AOC
EX 10.5 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:
In ∆OAB
OA = AB [Given]
OA = OB [radii of the circle]
∴ OA = OB = AB
∴ ∠OAB = 60° [Each angle of an equilateral triangle is 60°]
⇒ ∠AOB = 2∠ADB [ The angle subtended by an arc at the centre is double the angle subtended by it at any ]
⇒ ∠ADB = ½∠AOB ⇒ ∠ADB = ½ x 60°= 30°
ABCD is a cyclic quadrilateral
∠ACB + ∠ADB = 180° [the sum of either pair of opposite angles of a cyclic quadrilateral is 180°]
∠ACB + 30° = 180°
⇒ ∠ACB 180° – 30° = 150°
EX 10.5 QUESTION 3.
In Fig. 10.37, 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 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]
⇒ 2∠OPR = 20° = ∠OPR = 10°
EX 10.5 QUESTION 4.
In Fig. 10.38, 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 QUESTION 5.
In Fig. 10.39, 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 [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°
EX 10.5 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°
∠DBC = 70° [Given]
In ∆BCD,
∠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°
EX 10.5 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 is the diameter of the circle.
⇒ ∠ADC =90° and ∠ABC = 90° …(i) [angle in semicircle is at 90°]
BD is the diameter of the circle.
Similarly, ∠BAD = 90° and ∠BCD = 90° …(ii) [angle in semicircle is at 90°]
and ∠CDA = 90°
from equation (i and (ii)
∠ADC = ∠ABC = ∠BAD = ∠BCD = 90°
∴ ABCD is a rectangle.
EX 10.5 QUESTION 8.
If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
Solution:
In trapezium ABCD such that AB ॥ CD and AD = BC.
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)
∴ From (ii) and (iii), 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.
EX 10.5 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. 10.40). Prove that ∠ ACP = ∠ QCD.
Solution:
∠ACP = ∠ABP …(i) [angles in the same segment of a circle are equal.]
∠ABP = ∠QBD …(ii) [Vertically opposite angles]
∠QCD = ∠QBD …(iii)
∴ From (i), (ii) and (iii), we have
∠ACP = ∠QCD
EX 10.5 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:
In ∆ABC, and two circles described with the diameter as AB and AC respectively. They intersect at a point D, other than A.
Join A and D.
∵ AB is a diameter.
∴∠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 QUESTION 11.
ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠ CAD = ∠CBD.
Solution:
Given: Triangle ABC and ADC are two triangles on a common base AC.
To prove : ∠CAD = ∠CBD
Proof: ∆ABC and ∆ADC are on common base BC and ∠BAC = ∠BDC.
Points A, B, C and D lie on the same circle.
[If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment, the four points lie on a circle.]
∴ ∠CAD = ∠CBD
EX 10.5 QUESTION 12.
Prove that a cyclic parallelogram is a rectangle.
Solution:
Given: ABCD is a cyclic quadrilateral.
⇒ ∠A + ∠C = 180° …(i) [∴ Sum of its opposite angles is 180°.]
But ∠A = ∠C …(ii) [Opposite angles of a parallelogram are equal]
From (i) and (ii),
∠A = ∠C = 90°
Similarly,
∠B = ∠D = 90°
⇒ Each angle of the parallelogram ABCD is 90°.
∴ ABCD is a rectangle.