Question

In the given, $\Delta ABC$ has sides $AB=7.5cm,AC=6.5cm$ 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. $7cm$
• B. $5cm$
• C. $3cm$
• D. $2cm$

Answer 1

The correct option is C $3cm$

In triangle ABC, we have $a=7.5cm,b=7cmandc=6.5cm$
So, Semiperimeter $s=\frac{7.5+7+6.5}{2}=\frac{21}{2}cm=10.5cm$
$\therefore$ Area of $\Delta ABC=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{10.5(10.5-7.5)(10.5-7)(10.5-6.5)}$
$=\sqrt{10.5\times 3\times 3.5\times 4}$
$=\sqrt{441}=21{cm}^2$
Given, area of triangle $=$ area of parallelogram $=21{cm}^2$
or, Area of parallelogram $=21{cm}^2$
$\Rightarrow \text{base}\times \text{height}=21\Rightarrow BC\times DF=21\Rightarrow 7\times DF=21\Rightarrow DF=3cm$
Hence, the height of the parallelogram is $3cm.$