Telangana SCERT Class 9 Math Solution Chapter 12 Circles Exercise 12.5

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:


Updated: October 2, 2021 — 4:41 pm

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