Question

the area of an isosceles triangle base 2cm and the length of one of the equal sides 4cm ​

Answer 1

Correct Question:

find the area of the isosceles triangle with base 2cm and length of one of the equal side 4cm

Given:

base of the isosceles triangle = 2cm

length of one of the side = 4cm

To Find:

area of the isosceles triangle

Solution:

We know that:

${\boxed{\sf{\purple{in\: isosceles\: triangle,\:two\:sides\:are\:equal}}}}$

So,the Sides are:

• a = 2cm

• b = 4cm

• c = 4cm

therefore,

$\large\sf{ Side\: = \frac{a + b + c}{2}}$

$\large\implies\sf{ \frac{2 + 4 + 4}{2}}$

$\large\implies\sf{ \frac{10}{2}}$

$\large\implies\sf\underline\orange{5}$

Now,area of the isosceles triangle:

${\boxed{\sf{\color{red}{area \: = \sqrt{s(s - a)(s - b)(s - c)}}}}}$

$\sf \: area \: = \sqrt{5 \times (5 - 2)(5 - 4)(5 - 4)}$

$\sf \: area \: = \sqrt{5 \times 3 \times 1 \times 1}$

$\sf \: area \: = 1\sqrt{15}$

$\sf \: area \: = 15 {cm}^{2}$

Hence:

the area of the isosceles triangle is 15cm²

Answer 2

Answer:

Given:-

• base of the isosceles triangle = 2cm

• length of one of the side = 4cm

To Find:-

• area of the isosceles triangle

Solution:-

$\begin{gathered} \sf \: s=\frac{4 + 4 + 2}{2} =5\end{gathered}$

$Δ= \sf\sqrt{s(s - a)(s - b)(s - b)}$

$\sf \: \sf \: area \: = \sqrt{5 \times (5 - 2)(5 - 4)(5 - 4)}$