Question

Find the area enclosed by the lines x=0, x=2, y=x, and y=3​

Answer 1

Given the lines:

x=0, x=2, y=x, and y=3​

When we plot the lines, they form closed figure ABCD which is a trapezium.

Please refer to the attached image for the graph of lines given.

The coordinates of points are:

A(0,0)

B(0,3)

C(2,3)

D(2,2)

To calculate the area of trapezium, we have the following formula:

$A = \dfrac{1}{2} \times \text{Sum of parallel sides} \times {\text{Distance between parallel lines}}$

Here, as per figure, the parallel sides are AB and CD.

Distance between parallel lines AB and CD is side BC.

Distance between 2 points ($x_1,y_1$) and ($x_2,y_2$) is given by distance formula:

$D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$AB = \sqrt{(0-0)^2+(3-0)^2}\\AB = 3\ units$

$CD = \sqrt{(2-2)^2+(3-2)^2}\\CD = 1\ unit$

$BC = \sqrt{(2-0)^2+(3-3)^2}\\BC = 2\ units$

$A = \dfrac{1}{2} \times (3+1) \times {2}\\\Rightarrow A=\dfrac{1}{2} \times 8\\\Rightarrow A = 4\ sq\ units$

So, area of enclosed figure is 4 sq units.

Answer 2

Answer:

Given the lines:

x=0, x=2, y=x, and y=3​

When we plot the lines, they form closed figure ABCD which is a trapezium.

Please refer to the attached image for the graph of lines given.

The coordinates of points are:

A(0,0)

B(0,3)

C(2,3)

D(2,2)

To calculate the area of trapezium, we have the following formula:

Here, as per figure, the parallel sides are AB and CD.

Distance between parallel lines AB and CD is side BC.

Distance between 2 points () and () is given by distance formula:

So, area of enclosed figure is 4 sq units.