Question

find area of an isosceles triangle having unequal sides is 12 cm, each of the equal sides is 24 cm also find its altitude corresponding to unequal side

Answer 1

perimeter= 24+24+12=60
semi perimeter=30
area=
$\sqrt{s \times (s - a) \times (s - b) \times (s - c)}$
=
$\sqrt{30 \times6 \times 6 \times 18}$
=139.43


the altitude will be
area=1/2*b*h
139.43*2/12=23.24

Answer 2

Answer:

$\text{The area and altitude is }139.43 cm^2\text{ and }23.2cm$

Step-by-step explanation:

Given the isosceles triangle having unequal sides is 12 cm, each of the equal sides is 24 cm.

we have to find the area of an isosceles triangle and  altitude corresponding to unequal side.

$s=\frac{a+b+c}{2}$

$s=\frac{24+24+12}{2}=\frac{60}{2}=30$

Area of triangle can be calculated as

$=\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{30(30-24)(30-24)(30-12)}$

$=\sqrt{30\times 6\times 6\times 18}=36\sqrt{15}=139.427 cm^2$

$\text{Area of triangle=}\frac{1}{2}\times base\times altitude$

$139.427=\frac{1}{2}\times 12\times altitude$

$altitude=\frac{2\times 139.427}{12}=23.2379000772\sim 23.2 cm$