# Mensuration-35

(a)OABC is a rhombus whose three vertices A, B and C lie on a circle with center O. If the radius of the circle is 10 cm,
find the area of the rhombus.

(b) If in the rhombus OABC, three of the vertices lie on a circle with center O and the area of the rhombus is 32 √3

cm2, find the radius of the circle.

Solution: (a) OABC is a rhombus; Radius = 10 cm; OA = OB = OC = 10 cm

Since, the diagonals of a rhombus bisects each other at right angles, therefore,

OD = 1/2 x OB = 1/2 x 10 cm = 5 cm

and angle ODC = 900

Therefore, in right triangle, OC2 = OD2 + DC2

=> (10)2 = 52 + DC2

=> DC2 = 1002 - 52 = 100 - 25 = 75

=> DC = √75 = √(25 x 3) = 5 √3 cm

Therefore, Diagonal = AC = 2 DC = 2 x 5 √3 = 10 √3 cm

Therefore, Area of the rhombus = 1/2 x Product of diagonals = 1/2 x AC x OB

= 1/2 x 10 √3 x 10 cm2 = 50 x 1.732 cm2

= 86.6 cm2

(b) It is given that the area of the rhombus is 32 √3 cm2

Therefore, 1/2 x OB x AC = 32 √3

=> 1/2 x r x AC = 32 √3

=> AC = 64 √3 / r ---------------------------------------------------- (1)

Now, OC2 = OD2 + DC2

=> r2 = (r/2)2 + (AC/2)2

=> r2 = r2 / 4 + AC2 / 4

=> AC2 / 4 = r2 - r2 / 4 = 3 r2 / 4

Therefore, AC2 = 3r2 => AC = r √3 ----------- (2)

From (1) and (2), it follows that 64 √3 / r = r √3

=> r2 = 64

=> r = 8 cm