Question

The length and width of a rectangular patio are, (x + 8) feet and (x + 6) feet, respectively. If the area of the patio is 160 square feet, what are the dimensions of the patio?
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Answer 1

Answer:

Area of the rectangle(A) is given by:

$A=lw$

where,

l is the length and w is the width of the rectangle.

As per the statement:

The length and width of a rectangular patio are, (x + 8) feet and (x + 6) feet, respectively. If the area of the patio is 160 square feet

⇒A = 160 square feet, l= x+8 feet and w = x+6 feet

then;

$(x+8)(x+6) = 160$

⇒$x^2+6x+8x+48 = 160$

⇒$x^2+14x+48 = 160$

Subtract 48 from both sides we have;

$x^2+14x=112$

Using completing square method:

$x^2+14x+7^2=112+7^2$

⇒$(x+7)^2=112+49$

⇒$(x+7)^2=161$

Simplify:

$x+7 = \pm \sqrt{161}$

Subtract 7 from both sides we have;

$x = -7 \pm \sqrt{161}$

Since, the sides cannot be in negative;

⇒$x = -7+\sqrt{161}$

Length of patio = $x+8 = -7+\sqrt{161}+8 = 1+\sqrt{161}$ feet

Width of patio = $x+6 = -7+\sqrt{161}+6 = -1+\sqrt{161}$ feet

Therefore, the dimensions of the patio are:

$1+\sqrt{161}$ feet and $-1+\sqrt{161}$ feet