Question

The base of an isoscle triangle is 8 cm finds its area the perimeters of

Answer 1

16√2 is the answer of your

Answer 2

Let ABC be the isosceles triangle, the AD be the altitude

Let AB=AC=x, then BC=32−2x [because parameter =2 (side) + Base]

Since in an issoceles triangle the altitude bisects the base, so

BD = DC = 16 − x

In a triangle ADC, we get

AC² = AD² + CD²

⇒ x² = 8² + (16 - x²)

⇒ x = 10cm

BC = 32 - 2x = 32 - 20 = 12cm

$\bold{Hence, \: required \: area =} \small\blue{\: \frac{1}{2} } \:\small\bold{ AD \times BC} = \small\green {\frac{1}{2} \times 8 \times 12 }= \huge\bold\red{48{cm}^{2} }$