Question

Find the area of an isosceles triangle ,each of whose equal sides are 13 cm and base 24 cm

Answer 1

a= 13 b=13 c =24

s=( a+ b+c)/2

=( 13+13 +24) /2 = 25


area= √ ( s ( s-a) ( s-b) ( s-c) )

= √ ( 25( 25- 13) (25 - 13) ( 25 - 24 )

= √ ( 25* 12*12 * 1)

= 12 cm^2


therefore the area is 12 cm^2

Answer 2

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