Question

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

Answer 1

Answer:

Heya !!!

Let ∆ABC is an isosceles triangle.

In which,

AB = AC = 13 cm

And,

BC= 24 cm

Therefore,

Area of triangle ABC = 1/4 ×B under root4A²-B²

=> 1/4 × 24 × under root 4 × (13)² - (24)²

=>[ 1/2 × 24 × under root 4 × 169 - 24 × 24] cm²

=> [ 1/4 × ( 676 - 576)]

=> 6 × rot 100

=> 6 × 10

=> 60 cm²

★ ★ ★ HOPE IT WILL HELP YOU ★ ★ ★

Answer 2

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

$</p><p>{\bf{\blue{\underline{☆☆Given:}}}}$

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