Question

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

Answer 1

Area of $△ABC$

$S=\frac{7.5+6.5+7}{2}=10.5$

$A=\sqrt{S(S-A)(S-B)(S-C)}=\sqrt{10.5(10.5-7.5)(10.5-6.5)(10.5-7)}$

$=\sqrt{441}$

$=21c{m}^2$

Area of quadrilateral $BDEC=$ Base $\times$ Height

$=7\times DF$

$7\times DF=21$

$DF=3cm$