It is known that the area of the largest equilateral triangular
section of a cube is 140sq cm. What is the side length of the cube?
The distances between the centres of two adjacent faces of another
cube is 8cms. What is the side length of this cube? Another cube
has an edge length of 12cm. At each vertex a tetrahedron with three
mutually perpendicular edges of length 4cm is sliced away. What is
the surface area and volume of the remaining solid?
What is the volume of the solid formed by rotating this right
angled triangle about the hypotenuse?
Imagine two identical cylindrical pipes meeting at right angles and think about the shape of the space which belongs to both pipes. Early Chinese mathematicians call this shape the mouhefanggai.
Several solutions arrived this month but some of you fell into the trap
of assuming that the sides of the frustrum would converge to a point but
they do not so the solution needed you do dissect the mound in the
middle as suggested here by Michael Levitin of St. Margaret's Episcopal
School San Juan Capistrano, California, USA. I like your advice to the
builder Michael, how about a spreadhsheet?
The answer is 23.154 cubic meters.
First we figure out the volume of the whole rectangular prism. Then we
figure out the volume of the frustrum of the pyramid. Then we simply
subtract the volume of the frustrum from the volume of the rectangular
The volume of the rectangular prism is the area of the base times the
height. So that is 6 x 5 x 12 = 36 cubic meters.
To calculate volume of the frustrum:
Put the frustum on a cutting board with the bigger base down. Take a
huge sharp knife and make four straight cuts along the edges of the top
square. Then we will have nine pieces.
And they will be:
One rectangular prism with base sides of 3.8 m by 2.8 m and with height
of 0.9 m. Then the volume will be 3.8 x 2.8 x 0.9 = 9.576 cubic meters.
We will also have four triangular prisms from the sides of frustrum that
we will put one on top of the other to form a triangular prism with the
right triangle base with legs of 0.5 m and 0.9 m and area of (0.5 x
0.9)/2 = 0.225 square meters. The height of the triangular prism is the
perimeter of the frustrum top base and is 2 x (3.8 + 2.8) = 13.2 m. So
the volume of the triangular prism is the area of the base times height
and equal to 0.225 x 13.2 = 2.97 cubic meters.
We will combine four corners of the frustrum and make a pyramid with a
square base with sides of 1 m and a height of 0.9 m. Then the area of
the base is 1 square meter and the volume is 1 x 0.9 / 3 = 0.3 cubic
Then we find the volume of the frustrum as the sum of the volume of the
figures. 9.576 + 2.97 + 0.3 = 12.846 cubic meters. Then we subtract the
volume of the frustrum from the volume of the rectangular prism 36 -
12.846 = 23.154 cubic meters.
Suggestion to the builders: Either do what we have done or make a
computer program that does it.