Question

Find the area of a parallelogram. Also find the length of the altitude from vertex $A$ on the side of $DC$

Answer 1

Step : Find the area of parallelogram

Find the area $△DBC$

Given : Sides of $△DBC$ is $12$ cm, $17$ cm and $25$ cm

Let $a=12\text{cm},b=17\text{cm},c=25\text{cm}$

$\text{Semi-perimeter}\left(s\right)=\frac{a+b+c}{2}$

$=\frac{12+17+25}{2}$

$=27\text{cm}$




$\therefore$ Area of $△DBC=\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{27(27-12)(27-17)(27-25)}$

$=\sqrt{27\times 15\times 10\times 2}$

$=\sqrt{9\times 3\times 3\times 5\times 5\times 2\times 2}$

$=3\times 3\times 5\times 2$

$=90$

$\Rightarrow$ Area of $△DBC=90{\text{cm}}^2$

Find the area of parallelogram

As, a diagonal of a parallelogram divides it into two congruent triangles.

Thus, $\text{Ar(ABCD)}=2\times \text{Ar}\left(△DBC\right)$

$=2\times 90$

$=180{\text{cm}}^2$


Step 2 : Find the altitude

$\text{Ar}\left(\square ABCD\right)=180{\text{cm}}^2,\text{Base (DC)}=12\text{cm}$

$\text{Height (AE)}=?$

we know,

$A\left(\square ABCD\right)=\text{Base}\times \text{Height}$

$\text{Height}=\frac{A\left(\square ABCD\right)}{\text{Base}}$

$=\frac{180}{12}$

$=15\text{cm}$

Hence, area of parallelogram is $180{\text{cm}}^2$ and the length of the altitude from vertex $A$ on the side of $DC$ is $15$ cm.