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).
[tex]\implies \sf L.A.=2(10 \times 6)+(3 \times 6)=138\:\:m^2[/tex]
Total Surface Area
Area of the isosceles triangle:
[tex]\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[/tex]
Total surface area:
[tex]\implies \sf T.A.=2\:bases+L.A.=2\left(\dfrac{3\sqrt{391}}{4}\right)+138=167.66\:\:m^2\:(2\:d.p.)[/tex]
Volume
[tex]\sf \implies Vol.=area\:of\:base \times height=\left(\dfrac{3\sqrt{391}}{4}\right) \times 6=88.98\:\:m^3\:(2\:d.p.)[/tex]
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).
[tex]\implies \sf L.A.=5(5 \times 6)=150\:\:cm^2[/tex]
Total Surface Area
Area of a pentagon:
[tex]\sf A=\dfrac{1}{4}\sqrt{5(5+2\sqrt{5})}a^2[/tex]
where a is the side length.
Therefore:
[tex]\implies \sf A=\dfrac{1}{4}\sqrt{5(5+2\sqrt{5})}\cdot 5^2=43.01\:\:cm^2\:(2\:d.p.)[/tex]
Total surface area:
[tex]\sf \implies T.A.=2\:bases+L.A.=2(43.01)+150=236.02\:\:cm^2\:(2\:d.p.)[/tex]
Volume
[tex]\sf \implies Vol.=area\:of\:base \times height=43.011193... \times 6=258.07\:\:cm^3\:(2\:d.p.)[/tex]
