If the sides of the triangular base are a, b and c and the height is h, the area is 2sh + 2 sqrt (s(s-a)(s-b)(s-c)) where s = (a +b +c)/2 [I. e. s is the semiperimeter of the base] and sqrt means"the square root of "
The answer depends on the information that you do have. Suppose you know all the edge lengths: the three sides of the triangle are a, b and c and the length of the prism is d. Let s = (a + b + c)/2 Then the area of the triangular cross section is sqrt[s*(s-a)*(s-b)*(s-c)] square units. So, surface area = 2*sqrt[s*(s-a)*(s-b)*(s-c)] + d*(a+b+c) square units. Volume = sqrt[s*(s-a)*(s-b)*(s-c)]*d cubic units.
One possible form is sqrt{s*(s-a)*(s-b)*(s-c)} square units where the lengths of the three sides of the triangle are a, b and c units and s = (a+b+c)/2.
Kebobs (as in shish kebobs).
First add the lengths of the three sides, a,b,c and divide by two (= half the perimeter). Call that S. Area = square root of (S*(S-a)*(S-b)*(S-c)) First the subtractions, then the multiplications and finally the square root.
If the sides of the triangular base are a, b and c and the height is h, the area is 2sh + 2 sqrt (s(s-a)(s-b)(s-c)) where s = (a +b +c)/2 [I. e. s is the semiperimeter of the base] and sqrt means"the square root of "
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