Question

In the following figure, ∆ABC has sides AB = 7.5 cm, AC = 6.5 cm and BC = 7 cm. On base BC a parallelogram DBCE of same area as that of ∆ABC is constructed. Find the height DF of the parallelogram.

Answer 1

Now, first determine the area of ∆ABC.

The sides of a triangle are AB = a = 7.5 cm, BC = b = 7 cm and CA = c = 6.5 cm

Now, semi-perimeter of a triangle,

`s = (a + b + c)/2`

= `(7.5 + 7 + 6.5)/2`

= `21/2`

= 10.5 cm

∴ Area of ΔABC = `sqrt(s(s - a)(s - b)(s - c))`   ...[By Heron’s formula]

= `sqrt(10.5(10.5 - 7.5)(10.5 - 7)(10.5 - 6.5))`

= `sqrt(10.5 xx 3 xx 3.5 xx 4)`

= `sqrt(441)`

= 21 cm²   ...(i)

Now, area of parallelogram BCED = Base × Height

= BC × DF

= 7 × DF  ...(ii)

According to the question,

Area of ∆ABC = Area of parallelogram BCED

⇒ 21 = 7 × DF  ...[From equations (i) and (ii)]

⇒ DF = `21/4` = 3 cm

Hence, the height of parallelogram is 3 cm.