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Mathematics

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labwork [276]2 years ago

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1. The prism-shaped roof has equilateral triangular bases. Create an equation that models the height of one of the roof's triang

Mandarinka [93]

1. Check picture 1. Let the one side of the triangle be a, drop one perpendicular, CD. Then triangle ADB is a right triangle, with hypothenuse a and one side equal to 1/2a. By the Pythegorean theorem, as shown in the picture, the height is $\frac{ \sqrt{3} }{2} a$

2. if a a=25 ft, then the height is $\frac{ \sqrt{3} }{2} a=\frac{ \sqrt{3} }{2} *25= \frac{1.732}{2}*25= 21. 7$ (ft)

3. consider picture 2. Let the length of the roof be l feet.

one side of the prism (the roof) is a rectangle with dimensions a and l, so the area of one side is a*l
the lateral Area of the roof is 3a*l

the area of the equilateral surfaces is $2*( \frac{1}{2} *a* \frac{ \sqrt{3} }{2}a )=\frac{ \sqrt{3} }{2} a^{2}$

so the total area of the roof is $\frac{ \sqrt{3} }{2} a^{2} +3al$

4. The total area was the 2 triangular surfaces + the 3 equal lateral rectangular surfaces. Now instead of 3 lateral triangular surfaces, we have 2.
So the total area found previously will be decreased by al

5. so the area now is $\frac{ \sqrt{3} }{2} a^{2} +2al$

6. now a=25 and l=2a=50

Area= $\frac{ \sqrt{3} }{2} a^{2} +2al=\frac{ \sqrt{3} }{2} * 25^{2}+2*25*50=25 ^{2} (\frac{ \sqrt{3} }{2} +4)=625*4.866$

=3041.3 (ft squared)