Question

9) Find the area of an isosceles triangle each of whose equal sides measures 13 cm and
bases measure 20 cm.​

Answer 1

h=√s²-(b/2)²

=√169-100

=√69=8.30

A =1/2×base×height

=1/2×20×8.3

=83.0

Answer 2

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