A triangle and a parallelogram have the same base and the same area. I

1.	A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 13 cm, 14 cm and 15 cm and the parallelogram stands on the base 14 cm, find the height of the parallelogram.
Answer» Let a, b and c are the sides of triangle and s is t he semi - perimeter, then its area is given by: A = \(\sqrt{s(s-a)(s-b)(s-c)}\)where s = \(\frac{a+b+c}2\)[Heron’s Formula] s = \(\frac{a+b+c}2\) = \(\frac{13+14+15}2\) = 21 A = \(\sqrt{21(21-13)(21-14)(21-15)}\) A = \(\sqrt{21\times8\times7\times6}\) = 84 cm² Therefore area of ∆ = \(\frac{1}2\)(Base x Altitude) 84 × 2 = 14× Altitude Altitude = 12 cm

A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 13 cm, 14 cm and 15 cm and the parallelogram stands on the base 14 cm, find the height of the parallelogram.

Answer»

Let a, b and c are the sides of triangle and s is t

he semi - perimeter, then its area is given by:

A = \(\sqrt{s(s-a)(s-b)(s-c)}\)where s = \(\frac{a+b+c}2\)[Heron’s Formula]

s = \(\frac{a+b+c}2\) = \(\frac{13+14+15}2\) = 21

A = \(\sqrt{21(21-13)(21-14)(21-15)}\)

A = \(\sqrt{21\times8\times7\times6}\) = 84 cm²

Therefore area of ∆ = \(\frac{1}2\)(Base x Altitude)

84 × 2 = 14× Altitude

Altitude = 12 cm