Word problem involving shapes

Question

The length of a rectangle is three feet more than twice it's width. The area of the triangle is 44 square feet. Find the dimensions of the triangle. 

Answer 1

If this is a rectangle only, and the area of the rectangle is 44 ft²:

 

L = 2W +3     length is 3 ft more than twice the width

LW = 44     area is 44 ft²

 

W(2W+3) = 44

2W² + 3W = 44

2W² + 3W - 44 = 0

 

We can factor this by grouping.  The coefficient of the squared

term "2" multiplied by the constant "-44" = 2(-44) = -88.

The factors of -88 that add to 3 are 11(-8).  Break up the 3W

into 11W-8W giving:

 

2W²-8W + 11W - 44 = 0

2W(W-4) + 11(W-4) = 0

(2W+11)(W-4) = 0

 

Either 2W+11 = 0    OR   W-4=0

          2W = -11              W = 4 ft

            W = -11/2

 

Since the width will not be a negative value, we disregard -11/2.

The width of the rectangle is 4 ft.

The length is 2W+3 = 2(4)+3 = 11 ft

 

The rectangle is 4ft by 11ft

 

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If the rectangle is divided into two triangles, each with an

area of 44 ft², then the area of the rectangle is 88 ft² and,

setting up the same with 88 ft instead of 44 as the total:

 

W(2W+3) = 88

2W² + 3W = 88

2W² + 3W - 88 = 0

 

To solve this you need to use the quadratic equation

w = [-b±√(b²-4ac)]/2a   with a=2, b=3, c=-88

 

w = [-3 ±√(3² -4(2)(-88))]/2(2)

= [-3 ±√713]/4

= (-3 ± 26.7)/4

 

Since we can disregard the negative answer we will

go with the answer that gives the positive result:

 

w = (-3+26.7)/4 = 23.7/4 = 5.925 ft

L = 2W + 3 = 2(5.925) + 3 = 14.85 ft

 

You need to verify with your instructor which is the case for this problem.

However, it seems more likely the problem was misprinted and the first

case with the rectangle of 4 ft by 11 ft is probably correct.