The area of a rectangle is 45 ft2. If the length of the long side of the rectangle is 80% greater than the short side of the rectangle, what is the length of the longer side?

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carlosego carlosego · Answer 1 · 2018-10-05T01:11:33+02:00

To solve this problem we propose the following equation:

1) [tex]b*h =45\\[/tex] This is the equation of the area of a rectangle.

Where a and b are the sides of the rectangle

If b is the longest side, then

b is 80% longer than h.

2) [tex]b= h+0.8h\\[/tex]

Now we have 2 equations and two unknowns.

Therefore we solve the following equations

[tex]45=b*\frac{b}{1+0.8}\\[/tex]

[tex]b^2=45(1+0.8)[/tex]

[tex]b =\sqrt{45(1+0.8)}\\[/tex]

b = 9 feet

The length of the longest side is 9 feet.

lisboa lisboa · Answer 2 · 2018-10-05T01:14:31+02:00

The area of a rectangle is 45 ft^2

If the length of the long side of the rectangle is 80% greater than the short side of the rectangle.

Let the shortest side be the width (x)

Length of rectangle L = 80% greater than x

So length = 0.8 x + x

Length = 1.8 x

Area of the rectangle = Length * width

45 = 1.8x * x

[tex]45= 1.8x^2[/tex]

Divide by 1.8 from both sides'

[tex]25= x^2[/tex]

Take square root on both sides

x=5

so width of the rectangle = 5 feet

Length of the rectangle (longer side) = 1.8x = 1.8 * 5 = 9 feet