Question

The base of an isosceles triangle is $10$ cm and one of its equal sides is $13$ cm. Find its area using Heron's Formula.

Answer 1

Base of the isosceles triangle is $10$ cm

One of its equal side is $13$ cm, so the third side is also $13$ cm

Now, the three sides are $a=10,b=13,c=13$

According to Hero's formula,

Area of a triangle, $A=\sqrt{s(s-a)(s-b)(s-c)}$

where,

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

$s=\frac{10+13+13}{2}=18$

Now,

$A=\sqrt{18(18-10)(18-13)(18-13)}$

$=\sqrt{18\left(8\right)\left(5\right)\left(5\right)}$

$=\sqrt{9\left(16\right)\left(5\right)\left(5\right)}$

$=3\times 4\times 5$

$=60c{m}^2$