Question

ABCDE is a polygon whose vertices are $A(-1,0)$, $B(4,0)$, $C(4,4)$, $D(0,7)$ and $E(-6,2)$. Find the area of the polygon.

• A. $84$ sq. units
• B. $64$ sq. units
• C. $44$ sq. units
• D. $24$ sq. units

Answer 1

The correct option is C $44$ sq. units

Given-

The vertices of a polygon ABCDE are

$A(-1,0)=A(x_1,y_1),B(4,0)=B(x_2,y_2),C(4,4)=C(x_3,y_3),D(0,7)=D(x_4,y_{41})$

and $E(-6,2)=E(x_5,y_5).$

To find out-

$A\left(\square ABCDE\right)$

Solution-

We join AD & AC.

Then we get $\Delta ABC,\Delta ACD$ & $\Delta ADE.$

So, $A\left(\square ABCDE\right)=A\left(\Delta ABC\right)+A\left(\Delta ACD\right)+A\left(\Delta ADE\right).$

We shall apply the formula

$A\left(\Delta \right)=\frac{1}{2}\{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\}.$

So $A\left(\Delta ABC\right)=\frac{1}{2}\{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\}$

$=\frac{1}{2}\{-1(0-4)+4(4-0)+4(0-0)\}units=10sq.units,$

$A\left(\Delta ACD\right)=\frac{1}{2}\{x_1(y_3-y_4)+x_3(y_4-y_1)+x_4(y_1-y_3)\}$

$=\frac{1}{2}\{-1(4-7)+4(7-0)+0(0-4)\}sq.units=\frac{31}{2}$ sq.units

$A\left(\Delta ADE\right)=\frac{1}{2}\{x_1(y_4-y_5)+x_4(y_5-y_1)+x_5(y_1-y_4)\}$

$=\frac{1}{2}\{-1(7-2)+0(-6+1)-6(0-7)\}$ sq.units

$=\frac{37}{2}$ sq.units.

$\therefore A\left(\square ABCDE\right)=(10+\frac{31}{2}+\frac{37}{2})sq.units=44$ sq.units.