Quadratic equation problem

Question

Hi, 

 

The problem is a drawing of a rectangle and a triangle at the top. It looks like a house. Unfortunately I cant draw it so i can just describe it. 

 

The height of the triangle is (x-3). 

The width of the rectangle is x. 

The length of the rectangle is (x+4) 

The total area is 50m²

 

How do i find out what x is?

Answer 1

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Answer 2

Assuming the width of the rectangle is the base, draw and label a diagram.

 

area of rectangle = x(x + 4)

area of triangle = (0.5)(x)(x - 3)

 

50 = x(x + 4) + (0.5)(x)(x - 3)

50 = x² + 4x + 0.5x² - 1.5x

50 = 1.5x² + 2.5x

100 = 3x² + 5x

0 = 3x² + 5x - 100

0 = (3x + 20)(x - 5)

 

x = 5

Answer 3

Set up expressions for the areas of the rectangle and triangle. 

 

The area of the rectangle is x(x+4)

 

**I cant tell if the base of the triangle is the same as the width or the length of the rectangle so ill assume width**

 

 The area of the triangle is .5(x)(x-3)

 

Simplifying we have the area of the rectangle = x² + 4x

and the area of the triangle = .5x² - 1.5x

 

Add them to equal the total area of the figure and set equal to 50m²

 

x² + 4x + .5x² - 1.5x = 50m²

1.5 x² + 2.5x = 50m²

1.5x² +2.5x -50 = 0

 

Use quadratic formula to solve for x.

 

-2.5 +/- √(2.5²-4(1.5)(-50)

------------------------------- = x        

              2(1.5)

 

 **to save space the square root part calculates to: √2.52-4(1.5)(-50) = 17.5

 

since -2.5 - 17.5 is negative we can discard that option so the answer is 

 

-2.5 + 17.5          15

-------------  =  -------- = 5 = x

       3                   3