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The length of a rectangle is 10 m greater than the width, and the area is 600 sq. m. Find the dimensions

User Dkimot
by
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1 Answer

10 votes

Explanation:

l = length

w = width

l = w + 10

l×w = 600 m²

now, using the identity of the first equation in the second equation

(w + 10)×w = 600

w² + 10w = 600

w² + 10w - 600 = 0

the general solution to such a squared equation is

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

x = w

a = 1

b = 10

c = -600

w = (-10 ± sqrt(10² - 4×1×-600))/(2×1) =

= (-10 ± sqrt(100 + 2400))/2 = (-10 ± sqrt(2500))/2 =

= (-10 ± 50)/2 = -5 ± 25

w1 = -5 + 25 = 20 m

w2 = -5 - 25 = -30

the negative solution does not make any sense for side lengths. so, the only valid solution is w = 20 m.

l = w + 10 = 20 + 10 = 30 m

the length is 30 m.

the width is 20 m.

User Geshode
by
9.2k points

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