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A regular square pyramid has height 12 and slant height 15. If the height of a frustum of this pyramid is 4, find: Surface area of the frustum
Answers
Looking at this in the simplest form, given a slant height of 15, and a vertical height of 12, we seem to have a triangle in proportion to 3:4:5 its base, (proportional to 3) will be in the same proportions to the actual dimension as 4 is to 12, and 5 is to 15 (or 1:3) the base of the right angled triangle formed by the slant height, vertical height and base is, therefore 3 x 3, or 9. 9 is, you must remember, half of the base square’s side, which will be 18 (I’m omitting units, since you have…)
If the base square’s side is 18, and the height is 12, the volume of the whole pyramid is base area times 1/3rd of the height, or 18x18x12/3, or 1296 cubic units, BUT! we’re interested in the frustrum of that pyramid. That can be achieved by subtracting the volume of the pyramid which has been removed from the top of the frustrum. That pyramid is 8 units high, so is 2/3rds of the height of the original. its base sides will be in proportion to the original, so will be 2/3rds of 18 units long, or 12 units. That means the pyramid of which that is the base (12x12) will have a volume of 12x12x8/3 cubic units, or 384 cubic units. Subtract that volume from your original pyramid volume, and you arrive at a volume of 1296–384 , or 912 cubic units. Little “real” trigonometry needed.
You can check that, quite easily, knowing that the volumes of regular solids increase with the cube, so, by multiplying the volume of the “top” pyramid by 1.5^3 (1.5 cubed) because the “whole” pyramid is 1.5 times as tall, we get 384 x 1.5 x 1.5 x 1.5, which turns out, luckily for me, to be 1296!
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