If the area of the rectangle is $9 x^{2} - 12 x - 32$ , then find out the perimeter of the rectangle.
[4 Marks]

Open in App

Solution

Given, the area of the rectangle $= 9 x^{2} - 12 x - 32$

Now, we know that, $(a + b) \times (a - b) = a^{2} - b^{2}$

$9 x^{2} - 12 x - 32$ can be written as:
$= (3 x)^{2} - 2 \times 3 x \times 2 + 4 - 36 = (3 x - 2)^{2} - 6^{2} = (3 x - 2 + 6) \times (3 x - 2 - 6) = (3 x + 4) \times (3 x - 8)$
[1.5 Marks]

Since the area of a rectangle is Length $\times$ Breadth,
Therefore, from the above simplification, we can say,
Length $= 3 x + 4$ ; and
Breadth $= 3 x - 8$
[1 Mark]

Perimeter of a rectangle $= 2 \times (l e n g t h + b r e a d t h) = 2 \times (3 x + 4 + 3 x - 8) = 2 \times (6 x - 4) = 12 x - 8$
[1.5 Marks]

Suggest Corrections

Similar questions

9 x^{2} - 12 x - 32

9 x^{2} - 12 x - 32

[4]

9 x^{2} - 12 x - 32

9 x^{2} - 12 x - 32

Join BYJU'S Learning Program