Question

find the area of an isosceles triangle each of whose equal sides is 13 cm and whose base is 24 cm.​

Answer 1

Given:-

• An isosceles triangle whose equal sides is 13 cm and base is 24 cm.

To Find:-

• Find the area of isosceles triangle.

Solution:-

Let the equal sides of the isosceles triangles be a cm and the base be of b cm.

★ According to the problem

• a = 13 cm
• b = 24 cm

$\tt \: Area \: of \: isosceles \: triangle= \frac{1}{4}b \sqrt{4 {a}^{2} - {b}^{2} }$

$\tt \leadsto \frac{1}{4} \times 24 \sqrt{4 {(13)}^{2} - {24}^{2} }$

$\tt \leadsto 6 \sqrt{4\times 169 - 576}$

$\tt \leadsto 6 \sqrt{676 - 576}$

$\tt \leadsto 6 \sqrt{100}$

$\tt \leadsto 6 \times 10$

$\tt \leadsto 60 \: {cm}^{2}$

ㅤ

So the area of the isosceles triangles be $\tt 60 \:cm^2$.

$\rule{300px}{.5ex}$

Additional information:-

★ Some formulas related to isosceles triangle

Base and Height

• $\boxed{\red{\tt \: A = \frac{1}{2} × b × h}}$

All three sides

• $\boxed{\red{\tt \: A = \frac{1}{2} [ \sqrt{( {a}^{2} − \frac{ {b}^{2} }{2} ) × b}]}}$

Length of 2 sides and an angle between them

• $\boxed{\red{\tt \: A = \frac{1}{2} × b × a × sin(α)}}$

Two angles and length between them

• $\boxed{\red{\tt \: A = [a^2×sin( \frac{β}{2})×sin(α)]}}$

Isosceles right triangle

• $\boxed{\red{\tt \: A = \frac{1}{2} × a^2}}$ㅤ

where,

• b = base of the isosceles triangle
• a = measure of equal sides of the isosceles triangle
• α = measure of equal angles of the isosceles triangle
• β = measure of the angle opposite to the base