MAJOR prism problem

Ana.stasia

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The base of the prism is a right-angled triangle with the surface area of 9 times square root of 3 and one angle of 30 degrees. The surface area of the biggest rectangle is 8cm2. Calculate the volume of the prism.

I tried to calculate the edges of the triangle first. Using sin and the surface area which is 9 times square root of 3.

121329911_2325866090893175_2779208842234239850_n.jpg

I calculated that c was the biggest and therefore I came to a conclusion that the side with c was the biggest and therefore one whose surface area equals 8cm2.
So I calculated H.

121307211_672136767026068_5605020847437871246_n.jpg

The volume i got was 6 times square root of 6. The solution from the book is 12 times square root of 3. Where did I go wrong?
 
if the base area is [MATH]9\sqrt{3}[/MATH] and the area of the largest rectangular face is 8, then the height of the prism is [MATH]\dfrac{2\sqrt{2}}{3}[/MATH].

[MATH]V=Bh = 6\sqrt{6} \, cm^3[/MATH]
 
The base of the prism is a right-angled triangle with the surface area of 9 times square root of 3 and one angle of 30 degrees. The surface area of the biggest rectangle is 8cm2. Calculate the volume of the prism.

I tried to calculate the edges of the triangle first. Using sin and the surface area which is 9 times square root of 3.

View attachment 22228

I calculated that c was the biggest and therefore I came to a conclusion that the side with c was the biggest and therefore one whose surface area equals 8cm2.
So I calculated H.

View attachment 22229

The volume i got was 6 times square root of 6. The solution from the book is 12 times square root of 3. Where did I go wrong?
60-30-90 triangle (base) has area = 32√3

The largest side of the triangle (hypotenuse) is then = 3*2 = 6 ..................... Incorrect

height of the prism = 8/6

volume of the prism = 9 * √3 * 8/6 = 12√3 (units?)
 
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30-60-90 side ratios, [MATH]a:a\sqrt{3}:2a[/MATH]
base area [MATH]\dfrac{a^2\sqrt{3}}{2} = 9\sqrt{3} \implies a = 3\sqrt{2}[/MATH]
hypotenuse, [MATH]2a = 6\sqrt{2}[/MATH]
so, the big question ...

are the sum of the two bases [MATH]9\sqrt{3}[/MATH], or is that value the area of a single base ... ???
 
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30-60-90 side ratios, [MATH]a:a\sqrt{3}:2a[/MATH]
base area [MATH]\dfrac{a^2\sqrt{3}}{2} = 9\sqrt{3} \implies a = 3\sqrt{2}[/MATH]
hypotenuse, [MATH]2a = 6\sqrt{2}[/MATH]
so, the big question ...

are the sum of the two bases [MATH]9\sqrt{3}[/MATH], or is that value the area of a single base ... ???
You are correct. I crossed out my response.

Going to the corner before Jomo squawks about it.......
 
if the base area is [MATH]9\sqrt{3}[/MATH] and the area of the largest rectangular face is 8, then the height of the prism is [MATH]\dfrac{2\sqrt{2}}{3}[/MATH].

[MATH]V=Bh = 6\sqrt{6} \, cm^3[/MATH]

So It would appear the solution from the book isn’t correct.
30-60-90 side ratios, [MATH]a:a\sqrt{3}:2a[/MATH]
base area [MATH]\dfrac{a^2\sqrt{3}}{2} = 9\sqrt{3} \implies a = 3\sqrt{2}[/MATH]
hypotenuse, [MATH]2a = 6\sqrt{2}[/MATH]
so, the big question ...

are the sum of the two bases [MATH]9\sqrt{3}[/MATH], or is that value the area of a single base ... ???

Of the single base.
 
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