Question

find the area of isosceles triangle whose base is 16 cm and one of these equal side it is 10 cm​

Answer 1

Answer:

Base = 16cm

equal sides = 10cm

Area of isosceles triangle =

$\frac{b}{4} \times \sqrt{4a {}^{2} } - b {}^{2} \\ = \frac{16}{4} \times \sqrt{4 \times 10 { }^{2} } - 16 {}^{2} \\ = 4 \times \sqrt{4 \times 100 - 256} \\ = 4 \times \sqrt{400 - 256} \\ = 4 \times \sqrt{144} \\ = 4 \times 12 \\ = 48cm {}^{2}$

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