Question

PLZ HELP ILL GIVE 40 points

All of the edges of a regular triangular pyramid have length 12. A frustum of the pyramid has height half of that of the pyramid. What is the: volume and surface area of the frustum

Answer 1

Step-by-step explanation:

Given : Volume =118.2 cu.m

Upper base edge =9 m, lower base edge =3 m

We know that,

Volume =3h(A1+A2+A1A2) ....... (1), where A1,A2 are area of upper and lower bases.

A1=43×92=35.074

A2=43×32=3.897

From (1) we get,

118.2=3h(35.074+3.897+35.074×3.897)

⟹118.2=3h(38.971+11.6911)

⟹118.2×3=h(50.6621)

Answer 2

Given the value=118.2 cu.m

Upper base edge =9m, lower base

edge=3m,

we know that,

$valume = \frac{h}{3}( a_{1} + a_{2} \sqrt{ a_{1} a_{2}}...... |1|$

Where a1,a2 are area of upper and lower bases,

$a_{1} = \sqrt{ \frac{3}{4} } \times {9}^{2} = 35.74 \\ a_{2} = \sqrt{ \frac{3}{4} } \times {3}^{2} = 38.97$

from (1) we get,

$118.2 = \frac{h}{3}(35.074 + 38.97 + \sqrt{35.074 \times 38.97} \\ 118.2 = \frac{h}{2}(38.971 + 11.6911) \\ 118.2 =h (50.6621) \\ h = 7m$

hence the Basse are 7m for from each other.