Question

Find the area of an isosceles triangle whose base is 24 cm and one of its equal sides is 13 cm

Answer 1

Answer:

$\boxed{\bold{60 \: cm^2}}$

Step-by-step explanation:

Here, a = b = 13 cm and c = 24 cm.

Semiperimeter (s) = (13+13+24)/2 = 50/2 = 25 cm

Thus, Area of triangle by heron's formula

$= \sqrt{25(25-24)(25-13)(25-13)}\\\\= \sqrt{5 \times 5 \times 1 \times 12 \times 12}\\\\= 5 \times 12\\\\= \boxed{\bold{60 \: cm^2}}$

Answer 2

${\green {\boxed {\mathtt {✓verified\:answer}}}}$

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$\rm\boxed{ \: here \: each \: equal\: side \: \: a = 13cm \: and \: base = 24cm }\\ \\ \therefore \rm \: area \: of \: the \: triangle = \big({ \frac{1}{4} b. \sqrt{4a {}^{2} - b {}^{2} } } \big)cm {}^{2} \\ = \big( \frac{1}{4 } + 24 \times \sqrt{4 \times 169 - 24 \times 24} \big) cm {}^{2} \\ = 60 \: cm {}^{2}$