Irregular pyramid

Geometry Level 3

Sides of a triangular pyramid are right-angled triangles with right angle vertices at the pyramid vertex. Areas of these sides are $6 \text{ cm}^2$ , $8 \text{ cm}^2$ and $12 \text{ cm}^2$ . Volume of the pyramid (in $\text{ cm}^3$ ) is:

8

12

6

8\sqrt{2}

6\sqrt{2}

1 solution

Aleksa Radovanović
Nov 1, 2015

If we were to place the pyramid on the side, we would get Now we have: $V=\frac{BH}{3} = \frac{\frac{ab}{2}\times c}{3} = \frac{abc}{6}$ Now, we know the areas of these sides as: $\frac{ab}{2}=6 cm^2$ $\frac{bc}{2}=8 cm^2$ $\frac{ac}{2}=12 cm^2$ If we multiply these 3 equations we get: $\frac{a^2b^2c^2}{8} = 576$ From there we get $abc=48\sqrt{2}$ , and the volume is $8\sqrt{2}cm^3$

how did you project the area into the plane, can you explain more vividly?

Sagar Ojha - 5 years, 7 months ago

I didn't understand the question exactly, but you can imagine this pyramid as part of a rectangular cuboid (i.e. to "cut" the rectangular cuboid from one vertex to the opposite side)

Aleksa Radovanović - 5 years, 7 months ago

Irregular pyramid

1 solution

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