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 dimensi

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.

in progress 0
Orla Orla 4 years 2021-09-05T04:17:19+00:00 1 Answers 12 views 0

Answers ( )

  1. minhkhang
    0
    2021-09-05T04:18:54+00:00
    Reply

    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.

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )