(OSTP is the front of the cube incase you view it differently)
A marble base for a statue is formed from a cube of side 7m. Is is placed relative to coordinates axes as shown in the diagram.
A B and C divide ST UT and PT respectuvely, in ration 3 : 4.
a) Find the coordinates of A, B, C
I've done this part:
A(3,0,7) B(7,4,7) C(7,0,3)
b) If the corner of the cube is cut off along the plane defined by ABC find the surface area of the block.
The initial surface area (hopefully) should be: 7 x7 x 6 = 294
I tried finding the area of the 3 triangles (ACT) (ABT) (BTC) and subtract 294 by this but its wrong and don't know what to try next.
Please help, thanks.
Answer = 283.86m^2
You're that close to the correct answer . . .
The surface area of the original cube is: .6 × 7² = 294 m².
When that corner is sliced off, we lose three isosceles right triangles
. . Each has 4m legs and an area of 8 m².
Then we gain an equilateral triangle with side 4√2 m.
. . Its area is: 8√3 m².
The final surface area is: .294 - 3(8) + 8√3 .≈ .283.86