Question

Find the area of an isosceles triangle each of whose equal sides measures 13 cm and whose base measures 20 cm

Answer 1

By heron's formula....
a=13cm,b=13cm,c=20 cm
S=a+b+c/2
= 13+13+20/2=46/2
=23 cm

Area=under root 23 (23-13)(23-13)(23-20)
=under root 23×10×10×3
=10 under root 69 is the answer

Answer 2

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Solution

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$\rm \: here \: each \: equal\: side \: \: a = 13cm \: and \: base = 20cm \\ \\ \therefore \rm \: area \: of \: the \: triangle = \big({ \frac{1}{4} b. \sqrt{4a {}^{2} - b {}^{2} } } \big)cm {}^{2} \\ = \big( \frac{1}{4 } + 20 \times \sqrt{4 \times 169 - 20 \times 20} \big) cm {}^{2} \\ = 83.07 \: cm {}^{2}$