Question

a rectangle is 8cm longer then it is wide, and it’s area is 153cm^2. Find the dimensions of the rectangle.

Answer 1

Answer:

Width of 9cm and length of 17cm.

Step-by-step explanation:

To calculate an area of a rectangle, use the formula A=l*w. We know the length is 8cm longer than the width. So l = 8 + w. So the area is A= (8+w)*w.

We also know the area is 153. Substitute this value for A and solve.

$153 = (8+w)(w)\\153 = 8w+w^2\\w^2+8w-153 = 0$

To solve the quadratic, use the quadratic formula:

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

Here a=1, b=8 and c=-153.

$w=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\w=\frac{-8+/-\sqrt{8^2-4(1)(-153)} }{2(1)} \\w=\frac{-8+/-\sqrt{64+612} }{2} \\w=\frac{-8+/-\sqrt{676} }{2} \\w=\frac{-8+/-26 }{2} \\w=\frac{-8+26 }{2}=\frac{18}{2}=9 \\and\\w=\frac{-8-26 }{2} =\frac{-34}{2}=-17$

Since w=9 or w=-17, substitute this value for w in l= 8+w to find l.

l = 8+9 = 17

or

l=8+-17 = -9

Since length cannot be positive, it must be l=17 and w=9.