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Question

If the area of the rectangle is 9x212x32, then find out the perimeter of the rectangle.
[2 marks]

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Solution

Given, the area of the rectangle =9x212x32

Now, by using the identity, (a+b)×(ab)=a2b2

9x212x32 can be written as:
=(3x)22×3x×2+436=(3x2)262=(3x2+6)×(3x26)=(3x+4)×(3x8)
[1 mark]

Now,
Length =3x+4; Breadth =3x8

Perimeter =2×(length+breadth)=2×(3x+4+3x8)=2×(6x4)
[1 mark]

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