Question

A rectangle is twice as long as it is wide. If the length and width are both increased by 5 cm, the resulting rectangle has an area of 50cm^2. Find the dimensions of the original rectangle to the nearest hundredth.

Answer 1

Considering the sides of the rectangle are:

w = width

l = lenght

Since it is twice as long as it is wide:

l = 2w

The area of a rectangle (A) is the product of the width and length.

Then,

A = lw

Knowing that: If the length and width are both increased by 5 cm, the resulting rectangle has an area of 50cm^2, then:

$50=(l+5)*(w+5)$

Substituting l by 2w:

$50=(2w+5)*(w+5)$

Solving the multiplications:

$\begin{gathered} 50=2w*w+2w*5+5*w+5*5 \\ 50=2w^2+10w+5w+25 \\ 50=2w^2+15w+25 \end{gathered}$

Subtracting 50 from both sides:

$\begin{gathered} 50-50=2w^2+15w+25-50 \\ 2w^2+15w-25=0 \end{gathered}$

To find w, use the quadratic formula.

For a quadratic equation ax² + bx + c = 0, the quadratic formula is:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

In this question:

x = w

a = 2

b = 15

c = -25

Substituting the values:

$\begin{gathered} w=\frac{-15\pm\sqrt{(-15)^2-4*2*(-25)}}{2*2} \\ w=\frac{-15\pm\sqrt{225+200}}{4} \\ w=\frac{-15\pm20.62}{4} \\ w_1=\frac{-15-20.62}{4}=\frac{-35.62}{4}=-8.90 \\ w_2=\frac{-15+20.62}{4}=\frac{5.62}{4}=1.40 \end{gathered}$

Since the measure must be posite, width = 1.40 cm.

Also, l = 2w.

Then, l = 2*1.40

l = 2.80 cm.

Answer:

width = 1.40 cm

length = 2.80 cm.