Question 1207323
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The problem as stated is nonsense, since we don't know what "summit height" of 2 meters is.<br>
It is a good problem if we assume that the 2 meters is supposed to be the length of a side of the square top of the truncated pyramid.  So I'll go ahead with that assumption.<br>
Consider the pyramid before it was truncated.<br>
Use similar triangles and the given dimensions to determine that the height of the original pyramid was 10m.<br>
The slant height of a face of the original pyramid (peak to midpoint of a side of the base) is the hypotenuse of a right triangle with legs 5m and 10m:<br>
{{{h=sqrt(5^2+10^2)=sqrt(125)=5sqrt(5)}}}<br>
The area of each triangular face of the original pyramid is (one-half base times height)<br>
{{{A=(1/2)(10)(5sqrt(5))=25sqrt(5)}}}<br>
The area of each face of the small pyramid that was cut off to form the truncated pyramid is<br>
{{{A+(1/2)(2)(sqrt(5))=sqrt(5)}}}<br>
So the area of each face of the truncated pyramid is<br>
{{{25sqrt(5)-sqrt(5)=24sqrt(5)}}}<br>
The total surface area of the truncated pyramid is the area of the two bases, plus the area of the four faces:<br>
{{{(100+4)+4(24sqrt(5))=104+96sqrt(5)}}}<br>
ANSWER: {{{104+96sqrt(5)}}} square meters<br>