# Word problem on quadratic unit review! TEST TODAY

User Generated

Nyrkfgre

Mathematics

### Question Description

This is the word problem , how do I do it?

A rectangle has a length 6 more than its width.  If the width is decreased by 2 and the length decreased by 4, the resulting rectangle has an area of 21 square units.  What is the length of the original rectangle?  What is the ratio of the original rectangle's area to the new rectangle's area?  what is the perimeter of the new rectangle?  Show all your work.

Can someone please show me how you did this?  I'm lost.

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Let the width of the original rectangle be x

Length of the original rectangle is 6 more than its width.

So, length of the original rectangle = x + 6

Width is decreased by 2 and the length decreased by 4.

Width of the new rectangle = x - 2

Length of the new rectangle = x + 6 - 4 = x + 2

Area of the new rectangle = length * width= (x + 2)(x - 2) = $x^2-4$

But, it is given that area of new rectangle = 21 square units

Therefore,

$x^2-4=21\\ \\ x^2=21+4\\ \\ x^2=25\\ \\ x=\sqrt{25}\\ \\ x=5$

So, width of the original rectangle = x = 5 units

Length of the original rectangle = x + 6 = 5 + 6 = 11 units

Area of the original rectangle = length * width = 11 * 5 = 55 square units

$\frac{Area\hspace{5}of\hspace{5}original\hspace{5}rectangle}{Area\hspace{5}of\hspace{5}new\hspace{5}rectangle}=\frac{55}{21}$

Length of new rectangle = x + 2 = 5 + 2 = 7

Width of new rectangle = x - 2 = 5 - 2 = 3

Perimeter of new rectangle = 2(length + width) = 2(7 + 3) = 2*10 = 20 units

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fulnzanve3 (279)
Carnegie Mellon University
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