Question

find the area of an isosceles triangle having base 2 centimetre and the length of one of the equal sides 4 cm

Answer 1

s=2+4+4/2=5 using herons formula
area=
$\sqrt{5(5 - 2)(5 - 4)(5 - 4)}$
=
$\sqrt{5 \times 3 \times 1 \times 1 = \sqrt{15} }$
$ar. = \sqrt{15cm {}^{2} }$

Answer 2

Given that,

Base of an isosceles triangle is 2 cm

The length of one of the equal sides is 4 cm.

To find,

The area of an isosceles triangle.

Solution,

Let a = 2 cm, b = 4 cm and c = 4 cm

Using Heron's formula as:

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

s is semi perimeter,

$s=\dfrac{a+b+c}{2}\\\\s=\dfrac{2+4+4}{2}\\\\s=5\ cm$

Now using the above formula as :

$A=\sqrt{5\times (5-2)(5-4)(5-4)}\\\\A=\sqrt{5\times 3\times 1\times 1} \\\\A=\sqrt{15} \ cm^2$

So, the area of an isosceles triangle is $\sqrt{15} \ cm^2$.