Question

find the area of an isosceles triangle each of whose equal sides are 13 cm dot and base is 24 cm​

Answer 1

Step-by-step explanation:

area of isosceles triangle=1/2×base(a^-b^/4)

=1/2×24(13^-13^/4)

=12(169-169/4)

=12(676/4-169/4)

=12(507/4)

=12×126.75

=1521cm^

Answer 2

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