Geometry problems

Alrighty..

1. For this one, the diagonal of the base ABCD is \(\sqrt{12^2+12^2}\) which is \(12\sqrt2\). Dividing that by 2 gives \(6\sqrt2\).

Now, to find the height, we use the pythagorean theroem.

\(10^2={6\sqrt2}^2+b^2\)

\(100 = 72+b^2\)

\(b^2 = 28\)

\(b = 2\sqrt7\)

by the way, b is the height.

now, we get the area of the pyramid is \(\frac{12\cdot12\cdot2\sqrt7}{3}\)

which gives us \(96\sqrt7\)

2. 

Let ABCDEFGH be right rectangular prism. The total surface area of the prism 30. Also, the sum of all the edges of the prism is 44. Find the length of the diagonal joining one corner of the prism to the opposite corner.

h = height

w = width

l = length

4 ( h + w + l)  =  44

h + w +  l =  11                   (1)

2 (lw + lh + hw) = 30          (2)

Square both sides of (1)

h^2 + 2 h l + 2 h w + l^2 + 2 l w + w^2  = 121       rearrange as

h^2 + l^2 + w^2 +  2 (lw + lh + hw)  = 121  (3)

Sub (2)  into (3)

h^2 + l^2 + w^2  +  30 =  121

h^2 + l^2 + w^2  = 91

sqrt  [ h^2 + l^2 + w^2 ] = sqrt (91)  ≈  9.54  =   length of the diagonal joining one corner of the prism to the opposite corner

CORRECTED

Yaey  I got it right! Currently reading number 3.

Alrighty...

3. For this one, I will calculate the white parts and subtract from the volume. You can try and do it the other way around.

The pyramid ABCF has an area that is half of half of the rectangular prism. You can see the divisions, FC and triangle ACF.

Same thing for the other white parts, so this sums to 3/4 of the total volume, therefore the shaded area is 1/4 of the prism. The volume is 60.