Given the above question. What is the height of the truncated piece. I tried creating similar triangles, one with a base equal to the height of the 5cm equilateral triangle and the other having a base equal to the height of the 8cm equilateral triangle.
From there I worked out the height to be equal to 3.6cm but it doesn't feel right. Could someone let me know if got the correct answer or if my intuition was wrong?
(a) Let the height of the truncated piece be $h$, then using proportion
$ \dfrac{ 6 }{6 + h} = \dfrac{ 5 }{ 8 } $
From which $ 30 + 5 h = 48 $ and $ h = \dfrac{18}{5} = 3.6 $
(b) Volume of top is
$ \dfrac{1}{3} (6) \left(\dfrac{\sqrt{3}}{4}\right)(5)^2 = \dfrac{25 \sqrt{3}}{2} $
(c) Volume of original pyramid is
$ \left( \dfrac{8}{5} \right)^3 \left( \dfrac{25 \sqrt{3}}{2} \right) = \dfrac{256 \sqrt{3}}{5} $