Question

()) A rectangular piece of paper has a perimeter of 22 inches and an area of 28 square

Answer 1

Given:

Perimeter of a rectangular paper = 22 inches.

Area of the rectangular paper = 28 square inches.

To find:

The dimensions of the rectangular paper.

Solution:

Let l be the length and w be the width of the rectangular paper.

Perimeter of a rectangle is:

$P=2(l+w)$

Perimeter of a rectangular paper is 22 inches.

$2(l+w)=22$

$l+w=\dfrac{22}{2}$

$l=11-w$                      ...(i)

Area of a rectangle is:

$A=lw$

Area of the rectangular paper is 28 square inches.

$28=lw$

Using (i), we get

$28=(11-w)w$

$28=11w-w^2$

$w^2-11w+28=0$

Splitting the middle term, we get

$w^2-7w-4w+28=0$

$w(w-7)-4(w-7)=0$

$(w-7)(w-4)=0$

Using zero product property, we get

$(w-7)=0\text{ and }(w-4)=0$

$w=7\text{ and }w=4$

If $w=7$, then by using (i)

$l=11-7$

$l=4$

If $w=4$, then by using (i)

$l=11-4$

$l=7$

Therefore, the dimensions of the paper are either $7\times 4$ or $4\times 7$.