How do you find the area of the rectangle that has the sides of \[x + 7\] and \[x + 5\] ?

Hint:In this question, we need to find the area of the rectangle whose sides are \[x + 7\] and \[x + 5\] . We can use the area of the rectangle formula when the sides of the rectangle are given. The area of the rectangle is always represented in square units. The area of the rectangle is given by the product of its length and breadth. First let us consider \[(x + 7)\] as \[l\] and \[(x + 5)\] as \[b\] . Then we need to substitute the value of \[l\] and \[b\] in the area formula . Using this we can find the area of the rectangle.

Formula used :
The area of the rectangle is
\[A = l \times b\]
Where \[l\] is the length of the rectangle and \[b\] is the breadth of the rectangle.


Complete step by step solution:

Given the sides of the rectangle are \[x + 7\] and \[x + 5\]
Here we need to find the area of the rectangle.
Let us consider \[l = (x + 7)\] and \[b = (x + 5)\]
We know that the area of the rectangle is given by the product of its adjacent sides .
\[A = l \times b\]
By substituting the values of \[l\] and \[b\],
We get,
\[\Rightarrow \ A = (x + 7)\ (x + 5)\]
On multiplying the both the terms ,
We get,
\[\Rightarrow \ A = x^{2} + 7x + 5x + 35\]
On adding the like terms,
We get,
\[A = x^{2} + 12x + 35\]
Thus we get the area of the rectangle is \[x^{2} + 12x + 35\] square units.
The area of the rectangle is \[x^{2} + 12x + 35\] square units.

Note:
A rectangle is nothing but a two dimensional shape having four sides and four corners. The area of the rectangle is given by the product of its adjacent sides whereas the perimeter of a rectangle is the sum of the length of its four sides. While solving these questions, we should know all the formulas , otherwise the question could not be solved because the use of the formula is the only way to solve it. When the units of the given measurements are not mentioned in the question , then we can just consider them as units and after calculating the area, we need to mention square units after the area value.