A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram?

Let ABC be a triangle with sides AB = 26 cm, BC = 28 cm, CA = 30 cm



Now, we know that,

\(s = \frac{AB + BC + CA}{2}\)
\(\therefore \) \(s = \frac{26 + 28 + 30}{2} = \frac{84}{2} = 42cm\)

Now, Area of triangle
= \(\sqrt{42(42 - 26)(42 - 28)(42 - 30)}\)
(Since, Heron's formula [area = \(\sqrt{s(s - a)(s - b)(s - c)}\)])

= \(\sqrt{42 × 16 × 14 × 12}\)
= \(\sqrt{7 × 6 × 4 × 4 × 2 × 7 × 2 × 6}\)
= \(7 × 6 × 4 × 2 {cm}^2\)
= \(336 {cm}^2\) ...(i)

We know that,

Area of parallelogram = Base × Height ...(ii)

We also have,
Area of parallelogram = Area of \(\triangle{ABC}\) ...(Given)

Thus, from (i) and (ii),

\(\Rightarrow \) Base × Height = 336
\(\Rightarrow \) 28 × Height = 336 ...(Given)
\(\Rightarrow \) Height = 12cm

Therefore, the height of the parallelogram is 12cm.