Chapter 11.1, Problem 33PPS

To determine

To Find: To find the area of the given figure.

Answer to Problem 33PPS

The answer is 36square units.

Explanation of Solution

Given information: Given coordinates are A(4,7), B(2,1), C(8,1) and D(10,7).

Formula used: Area of triangle $=\frac{1}{2}\times \left|y_1(x_2-x_1)+y_2(x_3-x_1)+y_3(x_1-x_2)\right|$

Calculation:

Join AC which divides the given quadrilateral in two triangles namely ABC and ADC.

So, Area of the given quadrilateral is sum of the areas of both the triangles.

Thus,

  $\text{Area}\text{of}▱\text{ABCD}=\text{Area}\text{of}\Delta \text{ABC}+\text{Area}\text{of}\Delta \text{ACD}$

Where,

  $\begin{array}{l}\text{Area}\text{of}\Delta \text{ABC}\\ =\frac{1}{2}\times \left|y_1(x_2-x_3)+y_2(x_3-x_1)+y_3(x_1-x_2)\right|\\ =\frac{1}{2}\times \left|7(2-8)+1(8-4)+1(4-2)\right|\\ =\frac{1}{2}\times \left|-42+4+2\right|\\ =\frac{1}{2}\times \left|-36\right|\\ =\frac{1}{2}\times 36\\ =18\end{array}$

Similarly,

  $\begin{array}{l}\text{Area}\text{of}\Delta \text{ACD}\\ =\frac{1}{2}\times \left|y_1(x_2-x_3)+y_2(x_3-x_1)+y_3(x_1-x_2)\right|\\ =\frac{1}{2}\times \left|7(8-10)+1(10-4)+7(4-8)\right|\\ =\frac{1}{2}\times \left|-14+6-28\right|\\ =\frac{1}{2}\times \left|-36\right|\\ =\frac{1}{2}\times 36\\ =18\end{array}$

Hence, the $\text{Area}\text{of}▱\text{ABCD}=\text{Area}\text{of}\Delta \text{ABC}+\text{Area}\text{of}\Delta \text{ACD}$

  $\begin{array}{l}=18+18\\ =36\end{array}$