This page defines the relationship between the square side length and square area via a coordinate graph. Take the square area, square side as the coordinates of a point. And plot those points on the graph and read the unknown values from the graph easily. You can get the solved example questions on how to draw a graph of area vs side of a square in the following sections.
Relation Between Square Side Length & Area
The square area is defined as the product of the length of each side with itself. Its formula is given as side_length².
So, the relationship between the square side and the area is a square graph.
Square Area A = side² = s².
Solved Example Questions
Example 1.
Draw a graph of area vs side of a square. From the graph, find the value of the area, when the side length of the square = 3.
Solution:
Square Area A = side² = s².
For different values of s, we get the corresponding value of A.
When s = 0, A = 0² = 0
When s = 1, A = 1² = 1
When s = 2, A = 2² = 4
| s | 0 | 1 | 2 |
|---|---|---|---|
| A | 0 | 1 | 4 |
Thus, we have the points O (0, 0), A (1, 1), B (2, 4).
Plot these points on a graph paper and join them successively to obtain the required graph given below.
Reading off from the graph of area vs. side of a square:
On the x-axis, take the point L at s = 3.
Draw LP ⊥ x-axis, meeting the given graph at P.
Clearly, PL = 9 units.
Therefore, s = 3 ⇒ A = 9.
Thus, when s = 3 units, then A = 9 sq. units
Example 2.
Draw a graph for the following.
Side of the square (in cm) 1, 2, 3, 4 and Area (in cm) 1, 4, 9, 16 Is it a linear graph?
Solution:
Area of the square A = side² = s².
Draw these square side, area on a table.
| s | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| A | 1 | 4 | 9 | 16 |
Take the side of the square on the x-axis, area on the y-axis.
Plot the points P (1, 1), Q (2, 4), R (3, 9), s (4, 16) on the graph paper.
From the graph, we can say that square area vs side does not form a linear graph. It forms a square graph.
Example 3.
(a). Consider the relation between the area and the side of a square, given by A = s². Draw a graph of the above function.
(b). From the graph, find the value of A, when s = 2.5, 3.5.
Solution:
Given that,
Square Area A = s².
For different values of s, we get the corresponding value of A.
s = 0 ⇒ A = 0² = 0
s = 0.5 ⇒ A = 0.5² = 0.25
s = 1 ⇒ A = 1² = 1
s = 1.5 ⇒ A = 1.5² = 2.25
| s | 0 | 0.5 | 1 | 1.5 |
|---|---|---|---|---|
| A | 0 | 0.25 | 1 | 2.25 |
Thus, we get the points O (0, 0), A (0.5, 0.25), B (1, 1), C 1.5, 2.25)
Plot these points on a graph paper and join them successively to obtain the required graph given below.
(b). Reading off from the graph of area vs. side of a square:
On the x-axis, take the point L at s = 2.5.
Draw LP ⊥ x-axis, meeting the given graph at P.
Clearly, PL = 6.25 cm².
Therefore, s = 2.5 ⇒ A = 6.25.
Thus, when s = 2.5 cm, then A = 6.25 cm²
On the x-axis, take the point M at s = 3.5.
Draw MQ ⊥ x-axis, meeting the given graph at Q.
Clearly, MQ = 12.25 cm².
Therefore, s = 3.5 ⇒ A = 12.25 cm²
Thus, when s = 3.5 cm, then A = 12.25 cm²