Posted by **Britt ** on Tuesday, December 20, 2011 at 6:08pm.

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 Heronâ€™s formula. Justify and explain your reasoning.

- trigonometry -
**Steve**, Tuesday, December 20, 2011 at 6:43pm
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

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