Asked by Britt
A sheet metal worker constructed a triangular metal flashing with sides of 6 ft., 5 ft., and 3 ft. If he charges $2 per square foot for this type of work, then what should he charge? Do not use Heron’s formula. Justify and explain your reasoning.
Answers
Answered by
Steve
Draw a diagram. you have an almost-isosceles triangle, with base 3, sides 5,6.
Drop an altitude h to the base side 3. It will divide the base into two sections, of length x and 3-x.
Let x be the distance from side 5 to the base of h.
x^2 + h^2 = 25
(3-x)^2 + h^2 = 36
25 - x^2 = 36 - (3-x)^2
25 - x^2 = 36 - 9 + 6x - x^2
25 = 27 + 6x
6x = -2
x = -1/3
1/9 + h^2 = 25
h = 4.988
So, the area is 1/2 * 3 * 4.988 = 7.482
That's the area, so you can figure the price.
check: Heron's says a^2 = 7*1*2*4 = 56, so a = 7.483
Drop an altitude h to the base side 3. It will divide the base into two sections, of length x and 3-x.
Let x be the distance from side 5 to the base of h.
x^2 + h^2 = 25
(3-x)^2 + h^2 = 36
25 - x^2 = 36 - (3-x)^2
25 - x^2 = 36 - 9 + 6x - x^2
25 = 27 + 6x
6x = -2
x = -1/3
1/9 + h^2 = 25
h = 4.988
So, the area is 1/2 * 3 * 4.988 = 7.482
That's the area, so you can figure the price.
check: Heron's says a^2 = 7*1*2*4 = 56, so a = 7.483
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