Question

find the area of an isosceles with two equal angles of 45 degree each and equal sides of length 20cm

Answer 1

Area of this tria..gle is 200m2

Answer 2

Answer:

$\text{Area of triangle is }200 cm^2$          

Step-by-step explanation:

Given an isosceles triangle of two equal sides of length 20 cm and base angles of 45°

we have to find the area of triangle.

In ΔABD

$\sin45^{\circ}=\frac{AD}{AB}=\frac{AD}{20}$

⇒ $\frac{1}{\sqrt2}\times 20=AD$

$AD= 10\sqrt2$

$\cos45^{\circ}=\frac{BD}{AB}=\frac{BD}{20}$

⇒ $\frac{1}{\sqrt2}\times 20=BD$

$BD= 10\sqrt2$

Since an isosceles triangle has two equal sides, the height of triangle divides the base in two equal parts.

∴ $BC=2(BD)=20\sqrt2$

$\text{Area of ABC }=\frac{1}{2}\times base\times height$

$=\frac{1}{2}\times 20\sqrt2\times 10\sqrt2=200 cm^2$

$\text{Area of triangle is }200 cm^2$