Question 67

What is the area of a rhombus (in cm$$^2$$) whose side is 10 cm and the smaller diagonal is 12 cm ?

Solution

Rhombus ABCD, AB= 10cm , So BC = 10cm

Diagonal AC = 12cm

So, area( tri ABC)

= $$\sqrt{ s(s-a)(s-b)(s-c) }$$ , where s is semi perimeter and a,b,c are sides of the triangle

= s = 32/2 = 16

Area = $$\sqrt{16 *(16-10)(16-10)(16-12)}$$

=> area = $$\sqrt{16*6*6*4}$$

=> area = $$\sqrt{2^2*2^2*2^2*3^2*2^2}$$

=> area = 2*2*2*3*2 = 48 $$cm^2$$

Area of Rhombus is 48*2 = 96 $$cm^2$$

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