Question

NO LINKS!!! Part 2: Find the Lateral Area, Total Surface Area, and Volume. Round your answer to two decimal places.​

Answer 1

Answer:

Question 7

Lateral Surface Area

The bases of a triangular prism are the triangles.

Therefore, the Lateral Surface Area (L.A.) is the total surface area excluding the areas of the triangles (bases).

$\implies \sf L.A.=2(10 \times 6)+(3 \times 6)=138\:\:m^2$

Total Surface Area

Area of the isosceles triangle:

$\implies \sf A=\dfrac{1}{2}\times base \times height=\dfrac{1}{2}\cdot3 \cdot \sqrt{10^2-1.5^2}=\dfrac{3\sqrt{391}}{4}\:m^2$

Total surface area:

$\implies \sf T.A.=2\:bases+L.A.=2\left(\dfrac{3\sqrt{391}}{4}\right)+138=167.66\:\:m^2\:(2\:d.p.)$

Volume

$\sf \implies Vol.=area\:of\:base \times height=\left(\dfrac{3\sqrt{391}}{4}\right) \times 6=88.98\:\:m^3\:(2\:d.p.)$

Question 8

Lateral Surface Area

The bases of a hexagonal prism are the pentagons.

Therefore, the Lateral Surface Area (L.A.) is the total surface area excluding the areas of the pentagons (bases).

$\implies \sf L.A.=5(5 \times 6)=150\:\:cm^2$

Total Surface Area

Area of a pentagon:

$\sf A=\dfrac{1}{4}\sqrt{5(5+2\sqrt{5})}a^2$

where a is the side length.

Therefore:

$\implies \sf A=\dfrac{1}{4}\sqrt{5(5+2\sqrt{5})}\cdot 5^2=43.01\:\:cm^2\:(2\:d.p.)$

Total surface area:

$\sf \implies T.A.=2\:bases+L.A.=2(43.01)+150=236.02\:\:cm^2\:(2\:d.p.)$

Volume

$\sf \implies Vol.=area\:of\:base \times height=43.011193... \times 6=258.07\:\:cm^3\:(2\:d.p.)$