Question

The length of a rectangle is 6 more than twice the width. if the area is 40 cm^2, find the length and breadth of the rectangle

Answer 1

Answer: 3.217 & 12.434

Step-by-step explanation:

If we use w to represent the width, the length will be 6 more than 2 times w.

Hence, the length is $2w+6$.

The area of a rectangle would be its length times its width, so let's make an equation to represent it's area.

$A=w(2w+6)$

We can also substitute 40 in for A as it's given in the question.

$40 = w(2w+6)$

Distributing w by multiplying it by both terms in the parentheses, we get

$40 = 2w^2+6w$

We can make the equation simpler by dividing both sides by 2.

$20 = w^2+3w$

Subtracting both sides by 20 will make the left-hand side 0.

$0=w^2+3w-20$

Now that we have put this quadratic equation into standard form (ax²+bx+c), we can find its solutions using the quadratic formula.

For reference, the quadratic formula is

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

In this case, a is 1, b is 3, and c is -20.

Substituting, we get

$w=\frac{-3\pm\sqrt{3^2-4(1)(-20)}}{2(1)}$

$w= \frac{-3\pm\sqrt{9+80}}{2}$

$w=\frac{-3+\sqrt{89}}{2}\hspace{0.1cm}or\hspace{0.1cm}\frac{-3-\sqrt{89}}{2}$

Since the second solution results in a negative number, it cannot be the length of w.

$w=\frac{-3+\sqrt{89}}{2}\approx3.217$

The width/breadth of the rectangle is 3.217 cm.

To calculate the length, let's substitute the width into the expression for the length:

$l=2(3.217)+6$

$l=12.434$

The length of this rectangle is 12.434 cm.