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What are the dimensions of a rectangle with an area of 48 square centimeters and a perimeter of 28 centimeters?

1 Answer

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Explanation:

the area of a rectangle is

length × width

the perimeter of a rectangle is

2×length + 2×width

so,

length×width = 48 cm²

2×length + 2×width = 28 cm

length + width = 14 cm

length = 14 - width

we use this in the first equation :

(14 - width) × width = 48

14×width - width² = 48

-width² + 14×width - 48 = 0

a quadratic equation

ax² + bx + c = 0

has the general solution

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

x = width

a = -1

b = 14

c = -48

width = (-14 ± sqrt(14² - 4×-1×-48))/(2×-1) =

= (-14 ± sqrt(196 - 192))/-2 =

= (-14 ± sqrt(4))/-2 =

= (-14 ± 2)/-2 =

= 7 ± 1

width1 = 7 + 1 = 8 cm

width2 = 7 - 1 = 6 cm

length1 = 14 - width1 = 14 - 8 = 6 cm

length2 = 14 - width2 = 14 - 6 = 8 cm

so, we see, one of them must be 8 cm, and the other 6 cm.

let's length be the longer side, so

length = 8 cm

width = 6 cm

answered
User Gchtr
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