Question

Plot the points A(1,4.5),B(-1,0),C(1,-4.5) and D(3,0) on a graph paper. Name the figure obtained.Find the area of figure ABCD.​

Answer 1

Given:

A = (1 , 4.5)

B = (-1 , 0)

C = (1 , 4.5)

D = (3 , 0)

To be found:

Area of the figure formed from these vertices.

Solution:

The given vertices are plotted on a graph and attached.

From the graph, it can be seen that the figure formed from joining the vertices is a "Rhombus".

Rhombus: Rhombus is a closed polygon with four equal sides. But, unlike square, not all the angles are equal and 90°.

Now, the area of the Rhombus formed, is to be found out.

Let the area of the rhombus be demoted by "A".

Formula to be used:

Area of rhombus is given by:

$A = (1/2).d_1 . d_2$......(1)

where, d₁ and d₂ are the diagonals of the rhombus formed.

Let d₁ be the diagonal joined the vertices A and C.

Let d₂ be the diagonal joining the vertices B and D.

Before calculating the area, the lengths of the diagonals should be calculated.

For that, a diagonal can be considered as the line joining two points.

Hence, here the length of the diagonals can be calculated as:

Length of the diagonal$= \sqrt{((x_2 - x_1)^{2} + ((y_2 - y_1)^ {2}}$......(2)

Now, using the equation (2), length of the diagonal AC (d₁) can be calculated as:

$d_1 = \sqrt{(1-1)^{2} + (-4.5+(-4.5))^{2}} \\d_1 = \sqrt{(0)^{2} + (-9)^{2}}\\ d_1 = \sqrt{81}\\ d_1 = 9 units$

Now, using the above formula, length of the diagonal BD (d₂) can be calculated as:

$d_2 = \sqrt{(3-(-1))^{2}+ (0-0)^{2}} \\d_2 = \sqrt{(3+1)^{2}+0^{2}} \\d_2 = \sqrt{4^{2}} \\d_2 = 4 units$

Now, from equation (1) the area of the rhombus can be calculated as:

$A= (1/2)* 9 * 4$

⇒$A= 9* 2$

⇒$A = 18 square. units$

∴ The required area of the rhombus formed in graph is 18 square units.