Question

Find the area of a parallelogram given in Figure. Also find the length of the altitude from vertex A on the side DC. (in $c{m}^2$ )

Answer 1

Hint: To find the area of a parallelogram $ABCD=\text{Area of }\mathrm{\Delta }\text{ABD+Area of }\mathrm{\Delta }BCD$ . First determine the area of a triangle using Heron's formula $\sqrt{s(s-a)(s-b)(s-c)}$ . Also determine the area of parallelogram by using the formula $Base\times Height$ to find the length of the altitude from vertex A on the side DC.

Complete step-by-step answer:

We know that the diagonal of a parallelogram divides it into two congruent triangles.
So, area of a parallelogram $ABCD=\text{Area of }\mathrm{\Delta }\text{ABD+Area of }\mathrm{\Delta }BCD$ .
Here, we will find the area of $\mathrm{\Delta }BCD$
According to Heron’s formula the area of a triangle with sides a, b, & c is given as –
$\text{Area}=\sqrt{s(s-a)(s-b)(s-c)}$ .
Where,
$s={\displaystyle \frac{(a+b+c)}{2}}$ .
Here, we have from the figure –

$a=CD=12cm,b=BC=17cm,\text{ c=BD=25cm}\text{.}$
So, we will get the value of s as –
$\begin{array}{rl}& s={\displaystyle \frac{(12+17+25)}{2}}\\ & ={\displaystyle \frac{54}{2}}\\ & =27\end{array}$
$\therefore s=27$
Now, we will put the value of a, b, c and s in the formula $\text{Area}=\sqrt{s(s-a)(s-b)(s-c)}$ to find $\mathrm{\Delta }BCD$.
$\text{Area}=\sqrt{27\times (27-12)(27-17)(27-25)}$
$\begin{array}{rl}& =\sqrt{27\times 15\times 10\times 2}\\ & =\sqrt{8100}\\ & =90\end{array}$
$\therefore \text{Area of }\mathrm{\Delta }BCD=90c{m}^2$
Since ABCD is a parallelogram,
$\text{Area of }\mathrm{\Delta }BCD\text{=Area of }\mathrm{\Delta }ABD$
Now, we will find the area of parallelogram.
Area of a parallelogram $ABCD=\text{Area of }\mathrm{\Delta }\text{ABD+Area of }\mathrm{\Delta }BCD$ .
$\begin{array}{rl}& ABCD=90+90\\ & =180c{m}^2\end{array}$
$\therefore$ Area of a parallelogram $ABCD=180c{m}^2$
Next we will find the altitude of a parallelogram be h.
Let us consider the altitude of a parallelogram to be ‘h’.
Also, area of a parallelogram $$ABCD=\text{Base}\times \text{Height}$$
We know that the altitude of a parallelogram is also known as Height of a parallelogram.
Therefore,
Height of altitude from vertex A on side CD of the parallelogram $={\displaystyle \frac{\text{Area of parallelogram ABCD}}{Base}}$ .
$$\begin{array}{rl}& ={\displaystyle \frac{180}{12}}\\ & =15cm\end{array}$$
Hence, the area of a parallelogram for the given figure is $180c{m}^2$ and the altitude from vertex A on side CD is 15 cm.

Note: We have taken the area of both triangles as the same here, but if a student doesn't know that, they can find it separately and then add them together. The mistake of finding an area of parallelogram as $$17cm\times 12cm$$ should be avoided, it is not correct at all. A parallelogram is a simple quadrilateral with two pairs of parallel sides. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure. Opposite sides of a parallelogram are parallel and so will never intersect.