# Mensuration-35

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** (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**

**cm ^{2}, 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 = 90^{0}

Therefore, in right triangle, OC^{2} = OD^{2} + DC^{2}

=> (10)^{2} = 5^{2} + DC^{2}

=> DC^{2} = 100^{2} - 5^{2} = 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 cm^{2} = 50 x 1.732 cm^{2}

= 86.6 cm^{2}

(b) It is given that the area of the rhombus is 32 √3 cm^{2}

Therefore, 1/2 x OB x AC = 32 √3

=> 1/2 x r x AC = 32 √3

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

Now, OC^{2} = OD^{2} + DC^{2}

=> r^{2} = (r/2)^{2} + (AC/2)^{2}

=> r^{2} = r^{2} / 4 + AC^{2} / 4

=> AC^{2} / 4 = r^{2} - r^{2} / 4 = 3 r^{2} / 4

Therefore, AC^{2} = 3r^{2} => AC = r √3 ----------- (2)

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

=> r^{2} = 64

=> r = 8 cm