Question

The area of the pentagon whose vertices are (4,1), (3,6), (-5,1), (-3,-3) and (-3,0) is
A.30 sq. units
B.60 sq. units
C.120 sq. units
D.None of these

Answer 1

Hint: We are given the coordinates of all the five points, we use the formula of area of pentagon which is
Area of the pentagon $$={\displaystyle \frac{1}{2}}\left|\sum x_n{y}_{n+1}-y_n{x}_{n+1}\right|$$; where n = 1, 2, 3, 4, 5. We substitute the values and find the area.

Complete step-by-step answer:
Pentagon is a plane figure with five straight sides and five angles.
Here, the given vertices are (4,1), (3,6), (-5,1), (-3,-3) and (-3,0).

Now, Let,
$$\left(x_1,y_1\right)$$=(4,1)
$$\left(x_2,y_2\right)$$=(3,6)
$$\left(x_3,y_3\right)$$=(-5,1)
$$\left(x_4,y_4\right)$$=(-3,-3)
$$\left(x_5,y_5\right)$$=(-3,0)
Now, apply the formula of area of a pentagon, which is
$$={\displaystyle \frac{1}{2}}\left|\sum x_n{y}_{n+1}-y_n{x}_{n+1}\right|$$; where n = 1, 2, 3, 4, 5
$$={\displaystyle \frac{1}{2}}\left|\left(x_1{y}_2+x_2{y}_3+x_3{y}_4+x_4{y}_5+x_5{y}_1\right)-\left(y_1{x}_2+y_2{x}_3+y_3{x}_4+y_4{x}_5+y_5{x}_1\right)\right|$$
Putting the values of xi and yi where i = 1, 2, 3, 4, 5 in the formula we get
$$\begin{gathered} A={\displaystyle \frac{1}{2}}\left|\left(4\times 6+3\times 1+\left(-5\right)\times \left(-3\right)+\left(-3\right)\times 0+\left(-3\right)\times 1\right)-\left(1\times 3+6\times \left(-5\right)+1\times \left(-3\right)+\left(-3\right)\times \left(-3\right)+0\times 4\right)\right|sq.units \\ ={\displaystyle \frac{1}{2}}\left|\left(24+3+15+0-3\right)-\left(3-30-3+9+0\right)\right|sq.units \\ ={\displaystyle \frac{1}{2}}\left|39+21\right|sq.units \\ ={\displaystyle \frac{1}{2}}\left|60\right|sq.units \\ =30sq.units \end{gathered}$$
Hence, the area of the pentagon is 30 sq. units .
Hence, the correct option is (A).

Note: 1. A pentagon is any five –sided polygon. Sometimes it is called 5-gon. The sum of the angles in a simple pentagon is $${540}^{\circ }$$.
2.The formula given below is also valid for all polygons ,
$$A={\displaystyle \frac{1}{2}}\left|\left(x_1{y}_2+x_2{y}_3+x_3{y}_4+.....+x_{n-1}{y}_n+x_n{y}_1\right)-\left(y_1{x}_2+y_2{x}_3+y_3{x}_4+.....+y_{n-1}{x}_n+y_n{x}_1\right)\right|$$