Telangana SCERT Solution Class IX (9) Math Chapter 12 Circles Exercise 12.5
(1) (i) Solution:
From the adjacent figure sides opposite to ∠x & ∠y are equal
∠x = ∠y [∵opposite angles of equal sides a triangle are equal]
∴ x + y + 30o = 180o [sum of interior angles of triangle is 180o]
Or, x + x + 30o = 180o [∵x = y]
Or, 2x = 180o – 30o
Or, x = 150/2 = 75o
∴ x = y = 75o
∴ x = 75o, y = 75o
(ii) From the adjacent figure
y + 85o = 180o
y = 180o – 85o = 95o
x + 110o = 180o
Or, x = 180o – 110o
= 70o
∴ x = 70o
y = 95o
Since sum of opposite site angles of a cycle quadrilateral is 180o
(iii) From the adjacent figure,
x = 90o [∵the triangle is a right angles triangle at x]
y + x + 50o = 180o [∵sum of interior angles of a is 180o]
Or, y + 90o + 50o = 180o
Or, y = 180o – 140o
= 40o
x = 90o, y = 40o
(2) Given that the vertices A, B, C of a quadrilateral ABCD lie on a circle. Also ∠A + ∠C = 180°, then prove that the vertex D also lie on the same circle
Solution:
Given,
That In quad ABCD, AB lie on an circle
∠A + ∠C = 180o
∴ ∠A & ∠C are opposite angles of quad ABCD
Also, ∠A + ∠C = 180o therefore they are supplementary
∴ Quad ABCD is a cyclic quadrilateral
∴ All the vertices A, B, C, D lie on the circle
(3) If a parallelogram is cyclic, then prove that it is a rectangle
Solution:
Given, ABCD is a cyclic parallelogram
∴ ∠A = ∠C [opposite angles of a ∥gm are equal]
Also,
∠A + ∠C = 180o [∵ opposite angles of cyclic quadrilateral are supplementary]
Or, ∠A + ∠A = 180o [<A = <C]
Or, ∠A = 180o
Or, ∠A = 90o = ___
∴ ||gm ABCD is a rectangle since opposite angles are 90o
(4) Prove that a cyclic rhombus is a square.
Solution: