Question

If the length of a building is 2 1 2 times the width and each dimension is increased by 7 ft, then the perimeter is 266 ft. Find the dimensions of the original building in ft.

Answer 1

Answer:

The original dimensions of the building is 95 ft × 38 ft.

Explanation:

Let the original length be 'l' and original width be 'w'.

Given:

Original length (l) = $2\frac{1}{2}\times original\ width$

Original width = 'w'.

So, $l=2\frac{1}{2}w=\frac{5}{2}w$

Now, as per question:

Length and width is increased by 7 ft.

So, new length (l') = $l+7=\frac{5w}{2}+7$

New width (w') = $w+7$

New perimeter (P) = 266 ft

Perimeter is given as:

$P=2(l' +w')\\\\266=2(\frac{5w}{2}+w)\\\\\frac{266}{2}=\frac{5w+2w}{2}\\\\266=7w\\\\w=\frac{266}{7}=38\ ft$

Therefore, original width = 38 ft.

Original length is, $l=\frac{5\times 38}{2}=\frac{190}{2}=95\ ft$

Hence, the original dimensions of the building is 95 ft × 38 ft.