Find the area of an iscoceles triangle whose base is 16 cm, and length of each side = 10 cm.

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Solution

We are given the following data:
Base, b = 16 cm
Equal side, a = 10 cm
Using the formula for area of isosceles triangle:
$A r e a = \frac{1}{4} \times b \times \sqrt{4 a^{2} - b^{2}} = \frac{1}{4} \times 16 \times \sqrt{4 \times 10^{2} - 16^{2}} = 4 \times \sqrt{400 - 256} = 4 \times \sqrt{144} = 4 \times 12 = 48 c m^{2}$

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Find the area of an iscoceles triangle whose base is 16 cm, and length of each side = 10 cm.

We are given the following data: Base, b = 16 cm Equal side, a = 10 cm Using the formula for area of isosceles triangle: Area = 14 ×b ×√4a2 − b2 =14 ×16 ×√4×102 − 162 = 4 ×√400 −256 = 4 ×√144= 4 × 12 = 48 cm2