The length of a rectangle is $3x+10$ feet and its width is $x+12$ feet. If the perimeter of the rectangle is 76 feet, how many square feet are in the area of the rectangle?

Step-by-step explanation:

Given the length of a rectangle = 3x+10

Width of the rectangle = x+12

Perimeter of the rectangle = 76ft

For us to get the area of the rectangle we need to get the value of its length and width.

Given the perimeter of a rectangle P = 2(L+W) where;

L is the length and W is the width, on substitution

76 = 2(3x+10+x+12)

76 = 2(4x+22)

76 = 8x+44

8x = 76-44

8x = 32

x = 32/8

x = 4

Substituting x = 4 into the function of the length and width we have;

Length = 3(4)+10

Length = 22feet

Width = 4+12

width = 16feet

Area of the rectangle = Length * Width

Area of the rectangle = 22feet * 16feet

Area of the rectangle = 352ft²

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