flvsstudents24
Answered

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what are the dimensions of a rectangle with an area of 45 square inches and a perimeter of 28 inches?

Sagot :

reinhard10158

Answer:

9 in × 5 in

Step-by-step explanation:

the area of a rectangle :

length × width

and this is 45 in² in our case.

the perimeter of a rectangle is

2×length + 2×width

and this is 28 in in our case.

l×w = 45

l = 45/w

2×l + 2×w = 2×(45/w) + 2w = 28

45/w + w = 14

45 + w² = 14w

w² - 14w + 45 = 0

the general solution to such a squared equation is

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

x = w

a = 1

b = -14

c = 45

w = (14 ± sqrt((-14)² - 4×1×45))/(2×1) r

= (14 ± sqrt(196 - 180))/2 = (14 ± sqrt(16))/2 =

= (14 ± 4)/2 = 7 ± 2

w1 = 7 + 2 = 9 in

w2 = 7 - 2 = 5 in

l1 = 45/w1 = 45/9 = 5 in

l2 = 45/w2 = 45/5 = 9 in.

so, we see, length and width are interchangeable. it is only important to make one 9 in and the other 5 in.

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