Question 56957
The figure shows the cross section of a house consisting of a square surmounted by an isosceles triangle.  If the combined area of the square and the triangle is 1020 square feet, find the height of the triangle. one side of the triangle is 17 feet.

Unfortunately there is no figure here to show which side of the triangle is 17 ft. I will assume that it's not the base of the triangle. The base is one side of the square also.
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Let the base & one side of the square = x, 
the other two sides of the triangle = 17
Let the height of the triangle = h
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Area of the square + area of the triangle = 1020 sq ft
x^2 + .5*x*h = 1020
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Try to find h in terms of x, and then solve for x
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A right triangle is formed from .5x, h, and 17 ft would be the hypotenuse:
Using pythagoras, solve for h:
(.5x)^2 + h^2 = 17^2
h^2 = 17^2 - (.5x)^2
h^2 = 289 - .25x^2
h = SqRt(289-.25x^2)
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Substitute SqRT(289-.25x^2) for h in the equation: x^2 + .5*x*h = 1020
x^2 + .5x[SqRt(289-.25x^2)] = 1020
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.5x(SqRt(289-.25x^2) = 1020 - x^2
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Square both sides:
.25x^2 *(289-.25x^2) = 1040400 - 2040x^2 + x^4
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72.25x^2 - .0625x^4 = 1040400 - 2040x^2 + x^4
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72.25x^2 + 2040x^2 - .0625x^4 - x^4 = 1040400
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2112.25x^2 - 1.0625x^4 = 1040100
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Arrange as a quadratic equation, solve for x^2
-1.0625x^4  + 2112.25x^2 - 1040100 = 0
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Solve this using the your favorite method and you get:
x^2 = 898.5; x = 29.97 ~ 30
x^2 = 1089.5: x = 33.00
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Obviously x = 30 is the value we need, since 33^2 is more than the total area
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Find the area of the square: 30^2 = 900 sSq
Area of the triangle 1020- 900 = 120

Find h: x=30 and area = 120
 .5(30)h = 120
15h = 120
h = 120/15
h = 8 ft high
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