Question

In the given figure, a ∆ABC has been given in which AB = 7.5 cm, AC = 6.5 cm and BC = 7 cm. On base BC, a parallelogram DBCE of the same area as that of ∆ABC is constructed. Find the height DL of the parallelogram.

Answer 1

In ∆ABC,
The sides of the triangle are of length 7.5 cm, 6.5 cm and 7 cm.
∴ Semi-perimeter of the triangle is
$s=\frac{7.5+6.5+7}{2}=\frac{21}{2}=10.5\mathrm{cm}$

∴ By Heron's formula,
$$\begin{gathered} \mathrm{Area}\mathrm{of}∆ABC=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{10.5\left(10.5-7.5\right)\left(10.5-6.5\right)\left(10.5-7\right)} \\ =\sqrt{10.5\left(3\right)\left(4\right)\left(3.5\right)} \\ =21{\mathrm{cm}}^2...\left(2\right) \end{gathered}$$

Now,
Area of parallelogram DBCE = Area of ∆ABC
= 21 cm²

Also,
Area of parallelogram DBCE = base × height
$$\begin{gathered} \Rightarrow 21=BC\times DL \\ \Rightarrow 21=7\times DL \\ \Rightarrow DL=\frac{21}{7}=3\mathrm{cm} \end{gathered}$$

Hence, the height DL of the parallelogram is 3 cm.