The volume of the base is 5 x 2, then multiply by the height, 6, giveives volume = 5 x 2 x 6

b) 2 faces are 5 x 2, 2 faces are 6 x 2 and 2 faces are 5 x 6, giving a total of 20 + 24 + 60 = 104cm²

a) Using pythagoras theorem, h² = 3² + 4²

b) Area = (3 x 4)/2 = 6

c) Volume is corss sectional area x length, so V = 6 x 12 = 72

d) There are 5 faces, 2 triangles on each end each 6, the bottom which is 3 x 12, the back which is 4 x 12 and the sloping face which is 5 x 12, so 6 + 6 + 36 + 48 + 60 = 156

a) Volume = Cross section x length, so 300 = Cross section x 15. Cross section = 300/15

b) Area of truiangle is base x height / 2. If base x 5/2 = 20 then base must be 8

a) Volume of a cylinder = (pi)r²h. in this case pi x 7² x 15 = 2039.07060 which is 2040 to 3sf

b) Surface area of a cylinder = 2 (pi)r² + 2(pi)rh, substitute r = 7 and h = 15 and you get SA = 4926.01728 which rounds to 4930 to 3sf

a) \(V=\frac { 1 }{ 3 } Area\quad of\quad base\quad \times \quad heigth\\ V\quad =\quad \frac { 2\times 4\times 3 }{ 3 } =8{ cm }^{ 3 }\)

b) Consider m to be the midpoint of CD then we have the folllowing triangle

The vertical height of triangle ECD is

\(h\quad =\quad \sqrt { { 3 }^{ 2 }+{ 1 }^{ 2 } } \\ h\quad =\quad 3.162277\\ h\quad =\quad 3.16\quad cm\quad (3sf)\)

c) Consider n to be the midpoint of BC then we have the folllowing triangle

The vertical height of triangle EBC is

\(p\quad =\quad \sqrt { { 3 }^{ 2 }+{ 2 }^{ 2 } } \\ p\quad =\quad 3.605551\\ p\quad =\quad 3.61\quad cm\quad (3sf)\)

d) \(Base\quad area\quad =\quad 2\quad x\quad 4\quad =\quad 8{ cm }^{ 2 }\\ \\ Triangle\quad ECD\quad =\quad triangle\quad EBA\quad \\ Area\quad =\quad \frac { 3.16\times 4 }{ 2 } =\quad 6.32\\ \\ Triangle\quad EBC\quad =\quad triangle\quad EAD\quad \\ Area\quad =\quad \frac { 3.61\times 2 }{ 2 } =\quad 3.61\\ \\ Total\quad surface\quad area\quad =\quad 8\quad +\quad 2\times 6.32\quad +\quad 2\times 3.61\quad =\quad 27.86\quad { cm }^{ 2 }\quad \\ \\ \)

SA = 27.9 (3sf)

a) \(Volume\quad of\quad a\quad cone\quad =\quad \frac { \pi { r }^{ 2 }h }{ 3 } \\ V\quad =\quad \frac { \pi { \quad \times \quad 5 }^{ 2 }\quad \times \quad 12 }{ 3 } \\ V\quad =\quad 314.15926\\ V\quad =\quad 314\quad { cm }^{ 3 }\\ \\ \)

b) The profile of the cone looks like the diagram to the left. the slant height, l is calculated....

\(Slant\quad height\quad of\quad a\quad cone,\quad l\\ l\quad =\quad \sqrt { { 5 }^{ 2 }+{ 12 }^{ 2 } } \\ l\quad =\quad \sqrt { 169 } \\ l\quad =\quad 13\quad cm\\ \\ \\ \)

\(Surface\quad area\quad of\quad a\quad cone\quad =\quad \pi { r }^{ 2 }+\pi rl\\ SA\quad =\quad \pi \quad \times \quad 25\quad +\quad \pi \quad \times \quad 5\quad \times \quad 13\\ SA\quad =\quad 282.7433\\ SA\quad =\quad 283\quad { cm }^{ 2 }\)

\(Volume\quad of\quad pyramid\quad =\quad \frac { 1 }{ 3 } Area\quad of\quad base\quad \times \quad height\quad \\ 90\quad =\quad \frac { 1 }{ 3 } \quad \times \quad 6\quad \times \quad h\\ 270\quad =\quad 6h\\ h\quad =\quad \frac { 270 }{ 6 } \quad =\quad 45cm\\ \\ \\ \)

a) \(Volume\quad of\quad sphere\quad =\quad \frac { 4 }{ 3 } \pi { r }^{ 3 }\\ V\quad =\quad \frac { 4 }{ 3 } \pi \quad \times \quad 10^{ 3 }\\ V\quad =\quad 4188.7902\\ V\quad =\quad 4190\quad { cm }^{ 3 }\quad (3sf)\\ \\ \\ \)

b) \(Surface\quad Area\quad of\quad sphere\quad =\quad 4\pi { r }^{ 2 }\\ SA\quad =\quad 4\pi { \quad \times \quad 10 }^{ 2 }\\ SA\quad =\quad 1256.6370\\ SA\quad =\quad 1260\quad { cm }^{ 2 }\quad (3sf)\\ \\ \\ \)

\(Volume\quad of\quad sphere\quad =\quad \frac { 4 }{ 3 } \pi { r }^{ 3 }\\ { V }_{ sphere }\quad =\quad \frac { 4 }{ 3 } \pi \quad \times \quad 7^{ 3 }\\ { V }_{ sphere }\quad =\quad 1436.7550402\quad (Keep\quad calculator\quad display)\\ \\ { V }_{ sphere }\quad =\quad { V }_{ cone\quad }=\quad \frac { 1 }{ 3 } \pi { r }^{ 2 }h\quad =\quad \frac { 1 }{ 3 } \pi { \quad \times \quad 7 }^{ 2 }h\\ (\times 3)\quad \quad \quad 3\quad \times \quad { V }_{ sphere }\quad =\quad \pi { \quad \times \quad 49 }h\\ (\div 49\pi )\quad \frac { 3\quad \times \quad { V }_{ sphere } }{ 49\pi } \quad =\quad h\\ h\quad =\quad 27.99999999\\ h\quad =\quad 28.0\quad cm\quad (3sf)\\ \\ \)