Question

A solid object consists of a 4*4*4 cube with a 3*3*3 cube sticking out. Three corners of the 3*3*3 cube lie on the edges of the 4*4*4 cube.The same distance along each edge. What is the combined volume of this object?

Answer 1

Answer on Question #57601 – Math – Geometry

Question

A solid object consists of a $4*4*4$ cube with a $3*3*3$ cube sticking out. Three corners of the $3*3*3$ cube lie on the edges of the $4*4*4$ cube. The same distance along each edge. What is the combined volume of this object?

Solution

The combined volume of this object is $V(4*4*4 \text{ cube}) + V(3*3*3 \text{ cube}) - 2*V \text{ (SABC)}.$

$$V(4*4*4 \text{ cube}) = 64$$ $$V(3*3*3 \text{ cube}) = 27$$$$V = \frac{1}{3} Ah$$ $$SC = SB = 3 \text{ (as edges of the } 3*3*3 \text{ cube)}$$ $$\angle CSB = 90{}^\circ$$ $$CB = 3\sqrt{2}$$

Find the same way $AC = DC = AB = 3\sqrt{2}$

ABC is equilateral triangle.

$$A = \frac {a ^ {2} \sqrt {3}}{4} = \frac {(3 \sqrt {2}) ^ {2} \sqrt {3}}{4} = \frac {9 \sqrt {3}}{2}$$

CD is the radius of the circle circumscribed about the triangle:

$$R = \frac {a \sqrt {3}}{3} = \frac {3 \sqrt {2} \sqrt {3}}{3} = \sqrt {6}$$

SDC is right triangle:

$$S C ^ {2} = S D ^ {2} + D C ^ {2}$$

SD=h

$$h = \sqrt {(3) ^ {2} - \sqrt {6} ^ {2}} = \sqrt {3}$$ $$V (S A B C) = \frac {9 \sqrt {3}}{6} \sqrt {3} = 4. 5$$

V (4*4*4 cube) + V (3*3*3 cube) - 2*V (SABC)=64+27-2*4.5=64+18=82.

Answer: 82.

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