If the slant height is 5 and the actual height it 3, then it is 4 m from the center to the middle of any of the squate sides. A side view would be two 3-4-5 right triangles, back to back.
The base length is therefore 8 m. Each trangular side has an area of
Aside = (1/2)8*5 = 20 m^2. The total lateral area (4 sides) would therefore be 80 m^2.
I think you mistake was assuming that there are three triangular sides. There are four
Find the lateral area of a regular pyramid whose base is a square, whose slant height is 5 m, whose height is 3 m.
I think that the answer is 60. I subtracted 3 squared from 5 squared and I got 4. I multipled 4 by 2 and I got 8. I made a formula 3(1/2(8 x 5)) and I got 60 as my lateral area. Is this right?
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