Find the area of an isosceles triangles with base of length $6 c m$ and length of each equal side $5 c m$ .

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Solution

$\begin{matrix} h = \sqrt{{(5)}^{2} - {(3)}^{2}} = 4 A r e a o f i s o s c e l e s Δ = \frac{b h}{2} b = 6 A r e a = \frac{6 \times 4}{2} = 12 m^{2} \end{matrix}$

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Find the area of an isosceles triangles with base of length 6cm and length of each equal side 5cm.

h=√(5)2−(3)2=4AreaofisoscelesΔ=bh2b=6Area=6×42=12m2

Find the area of an isosceles triangles with base of length $6 c m$ and length of each equal side $5 c m$ .

$\begin{matrix} h = \sqrt{{(5)}^{2} - {(3)}^{2}} = 4 A r e a o f i s o s c e l e s Δ = \frac{b h}{2} b = 6 A r e a = \frac{6 \times 4}{2} = 12 m^{2} \end{matrix}$