Asked by lulu
Suppose you want to build a rectangular sandbox where the width is 5 feet less than the length and the diagonal is 5 feet longer than the length. What are the dimensions of the sandbox?
Answers
Answered by
oobleck
If the length is x, then you must have
(x-5)^2 + x^2 = (x+5)^2
You can work through the math, but just think of the simple
3-4-5 right triangle. Scale it up by a factor of 5, and you have
15-20-25
(x-5)^2 + x^2 = (x+5)^2
You can work through the math, but just think of the simple
3-4-5 right triangle. Scale it up by a factor of 5, and you have
15-20-25
Answered by
henry2,
Length = L.
Width = L-5 ft.
Diag. = L+5 ft.
L^2 + (L-5)^2 = (L+5)^2.
L^2 + L^2 - 10L +25 = L^2 + 10L + 25,
L^2 - 20L = 0,
Divide both sides by L:
L - 20 = 0,
L = 20 ft.
Width = L-5 = 15 ft.
Width = L-5 ft.
Diag. = L+5 ft.
L^2 + (L-5)^2 = (L+5)^2.
L^2 + L^2 - 10L +25 = L^2 + 10L + 25,
L^2 - 20L = 0,
Divide both sides by L:
L - 20 = 0,
L = 20 ft.
Width = L-5 = 15 ft.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.