Question

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

• A. 3

Answer 1

The correct option is A 3
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 $\Delta ABC$ ,
$s=\frac{a+b+c}{2}=\frac{7.5+7+6.5}{2}=\frac{21}{2}=10.5\text{cm}$
$\therefore Areaof\Delta ABC=\sqrt{s(s-a)(s-b)(s-c)}$ [by Heron’s formula]
$=\sqrt{10.5(10.5-7.5)(10.5-7)(10.5-6.5)}$
$=\sqrt{10.5\times 3\times 3.5\times 4}=\sqrt{441}=21{\text{cm}}^2\cdots \left(i\right)$
Now, area of parallelogram BCED
= Base $\times$ height
$=BC\times DF$
$=7\times DF$ . . . . . .(ii)

According to the question,
Area of $\Delta ABC$ = Area of parallelogram BCED
$\Rightarrow 21=7\times DF$ [from Eqs.(i) and (ii)]
$\Rightarrow DF=\frac{21}{7}=3\text{cm}$
Hence, the height of parallelogram is 3 cm.