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An architect wants to draw a rectangle with a diagonal of 25inches. The length of the rectangle is to be 10 inches more than twice the width. What dimensions should she make the rectangle?

1 Answer

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Explanation:

the length of the diagonal is calculated out of the length and width by using Pythagoras (after all, one length, one width and the diagonal are creating a right-angled triangle).

25² = length² + width²

length = 2×width + 10

we are using the second in the first equation :

25² = (2×width + 10)² + width²

625 = 4×width² + 40×width + 100 + width²

525 = 5×width² + 40×width

105 = width² + 8×width

width² + 8×width - 105 = 0

a quadratic equation

ax² + bx + c = 0

has the general solutions

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

x = width

a = 1

b = 8

c = -105

width = (-8 ± sqrt(8² - 4×1×-105))/(2×1) =

= (-8 ± sqrt(64 + 420))/2 =

= (-8 ± sqrt(484))/2 =

= (-8 ± 22)/2 = -4 ± 11

width1 = -4 + 11 = 7

width2 = -4 - 11 = -15

a negative size of the length of a side of a shadow does not make any sense, so, our valid solution is

width = 7 in.

and then

length = 2×width + 10 = 2×7 + 10 = 14 + 10 = 24 in.

User Firoze Lafeer
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