Chapter 1.8, Problem 21E

To determine

To calculate: The area and the type of the quadrilateral formed by ABCD along with its figure.

Answer to Problem 21E

The area of the quadrilateral so formed as trapezium is $\text{9 sq}\text{. units}$ and its figure is,

  

Explanation of Solution

Given information:

The four vertices of the quadrilateral $\text{A}\left(1,0\right),\text{B}\left(5,0\right),\text{C}\left(4,3\right)\text{ and D}\left(2,3\right)$ .

Formula used:

Area of a trapezium is half into the product of the sum of its parallel sides and its height. So, if a and b are the parallel sides of the trapezium and h is its height, area of trapezium is calculated as,

  $\text{A}=\frac{1}{2}\cdot \left(a+b\right)\cdot h$

Calculation:

Consider the given four vertices of the quadrilateral $\text{A}\left(1,0\right),\text{B}\left(5,0\right),\text{C}\left(4,3\right)\text{ and D}\left(2,3\right)$ .

Observe that all the points lie in the first quadrant of the coordinate plane, so, plot the points in first quadrant of the coordinate plane and join the vertices to form the quadrilateral as,

  

In the above figure, it is observed that the quadrilateral so formed has one pair of parallel sides, which is the characteristic of the trapezium, hence, the quadrilateral so formed is a trapezium.

Recall that the area of a trapezium is half into the product of the sum of its parallel sides and its height. So, if a and b are the parallel sides of the trapezium and h is its height, area of trapezium is calculated as,

  $\text{A}=\frac{1}{2}\cdot \left(a+b\right)\cdot h$

Here, $a=2,b=4\text{ and }h=3$ , so, area of the trapezium so formed will be calculated as,

  $\begin{array}{c}\text{A}=\frac{1}{2}\cdot \left(a+b\right)\cdot h\\ =\frac{1}{2}\cdot \left(2+4\right)\cdot 3\\ =\frac{1}{2}\cdot 6\cdot 3\\ =9\text{ sq}\text{. units}\end{array}$

Thus, area of the quadrilateral so formed as trapezium is $9\text{ sq}\text{. units}$ and its figure is,