Question

Area of isosceles triangle where 2 equal sides are 13cm and base is 24
cm ​

Answer 1

$\huge \tt{ \underline{ \underline{Answer}}}$

$Let \: ∆ABC \: is \: an \: isosceles \\ triangle.$

$AB = AC = 13 cm \\ And,$

$BC= 24 cm \\ Therefore,$

$Area \: of \: triangle ABC = \frac{1}{4} ×B \sqrt{4 \times A²-B²}$

$\\ = \frac{1}{4} × 24 × \sqrt{4 × (13)² - (24)²}$

$= [ \frac{1}{2} × 24 × \sqrt{4 × 169 - 24 × 24} ] cm²$

$=> [ \frac{1}{4} × ( 676 - 576)]$

$=> 6 × \sqrt{100}$

$=> 6 × 10$

$=> 60 cm²$

Answer 2

Answer:

156

Step-by-step explanation:

height = 13

base=24

Area=bh/2

13×24/2

=312/2

=156cm Ans.