Surafce Area And Volume Of Solid Shapes - SS1 Mathematics Lesson Note

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the total surface area of a cube=6l2

Where l is the length of the edge of the cube

the voulme of a cube= l3

Where l is the length of the edge of the cube

the total surface area of a cuboid=2(lb+lh+bh)

Where l is the length of the cuboid, b is the breadth (width) and h is the height

the volume of a cuboid=lbh

Where l is the length of the cuboid, b is the breadth (width) and h is the height

the total surface area of a prism=sum of area of its sides+2A

Where A is area of the triangular base

the volume of a prism=Ah

Where h is the height (length) of the prism

the total surface area of a cylinder=area of the curved surface+[area of the circular bases]

=2πrh+2πr2 (closed at both ends)

=2πrh+πr2 (open at one end)

=2πrh (open at both ends)

the volume of a cylinder= πr2h

the total surface area of a cone=area of the curved surface+[area of the circular base]

=πrs+πr2

Where s is the slant height

s=h2+r2

the volume of a cone=13πr2h

the total surface area of a pyramid=sum of area of all faces

the volume of a pyramid= 13Ah

Where A is the area of the base, h is the height of the pyramid

Example 1 Find the total surface area and volume of a cube of side \(4cm\).

Solution \(the\ total\ surface\ area\ of\ a\ cube = 6l^{2} = 6{(4)}^{2} = 6(16) = 96{cm}^{2}\)

\(the\ volume\ of\ the\ cube = \ l^{3} = \ 4^{3} = 64{cm}^{3}\)

Example 2 If the total surface area of a cuboid is \(158{cm}^{2}\), its length is \(8cm\) and its height is \(3cm\). Calculate the breadth of the cuboid and its volume.

Solution \(total\ surface\ area\ of\ a\ cuboid = 2(lb + lh + bh)\)

\(158 = 2(8b + 24 + 3b)\)

\(158 = 2(11b + 24)\)

\(158 = 22b + 48\)

\(22b = 158 - 48\)

\(22b = 110\)

\(b = \frac{110}{22} = \ \frac{10}{2} = 5cm\)

\(volume\ of\ a\ cuboid = lbh = 8 \times 5 \times 3 = 120{cm}^{3}\)

Example: Calculate the total surface area and volume of the prism below

Solution: \(the\ total\ surface\ area\ of\ a\ prism = sum\ of\ area\ of\ its\ sides + 2A\)

\(the\ total\ surface\ area\ of\ a\ prism = \lbrack(8 \times 5) + (8 \times 3) + (8 \times 4)\rbrack + 2(\frac{1}{2} \times 3 \times 4)\)

\(the\ total\ surface\ area\ of\ a\ prism = \lbrack 40 + 24 + 32\rbrack + 12)\)

\(the\ total\ surface\ area\ of\ a\ prism = 108{cm}^{2}\)

\(the\ volume\ of\ a\ prism = Ah\)

\(the\ volume\ of\ a\ prism = (\frac{1}{2} \times 3 \times 4)8\ \)

\[the\ volume\ of\ a\ prism = 48{cm}^{3}\]

Example: Find total surface area of a closed cylindrical oil drum of height \(160cm\) and base radius \(30cm\). What volume of oil in liters can fill the drum when full? (\(\pi = \frac{22}{7}\ and\ 1000{cm}^{3} = 1\ l\))

Solution \(the\ total\ surface\ area\ of\ a\ cylinder = 2\pi rh + 2\pi r^{2}\)

\(the\ total\ surface\ area\ of\ a\ cylinder = 2(\frac{22}{7})(30)(160) + 2(\frac{22}{7}){(30)}^{2}\)

\(the\ total\ surface\ area\ of\ a\ cylinder = 30171.4 + 5657.1 = \ 35,828.5{cm}^{2}\)

\(the\ volume\ of\ a\ cylinder = \ {\pi r}^{2}h\)

\(the\ volume\ of\ a\ cylinder = \ {\left( \frac{22}{7} \right)(30)}^{2}(160)\)

\(the\ volume\ of\ a\ cylinder = \ 452,571.4{cm}^{3}\)

\(1000{cm}^{3} = 1l\)

\(452,571.4{cm}^{3} = xl\)

\(x = \frac{452,571.4}{1000} = 452.5714\ l\ \cong 452.6\ l\)

Example: Calculate the total surface area and volume of the cone below:

Solution \(slant\ height,\ s = \sqrt{h^{2} + r^{2}} = \ \sqrt{25^{2} + 10^{2}} = \ \sqrt{625 + 100} = \ \sqrt{725} = 26.9cm\)

\(total\ surface\ area\ = \pi rs + \pi r^{2}\)

\(total\ surface\ area\ = (\frac{22}{7})(10)(26.9) + (\frac{22}{7}){(10)}^{2}\)

\(= (\frac{22}{7})(10)(26.9) + (\frac{22}{7}){(10)}^{2}\)

\(= 845.4 + 314.3 = \ 1,159.7{cm}^{2}\)

\(the\ volume\ of\ a\ cone = \ \frac{1}{3}{\pi r}^{2}h\)

\(the\ volume\ of\ a\ cone = \ \frac{1}{3}{\left( \frac{22}{7} \right)(10)}^{2}(25)\)

t\(he\ volume\ of\ a\ cone = \ 2,619.05{cm}^{3}\)

Example: Calculate the total surface area and volume of the pyramid below

Solution

\(the\ height\ of\ each\ triangular\ face = \ \sqrt{3^{2} + 1^{2}} = \sqrt{9 + 1} = \sqrt{10} = 3.2cm\), note the height of the faces and the prism are different

\[the\ total\ surface\ area\ of\ a\ pyramid = sum\ of\ area\ of\ all\ faces\]

\(the\ total\ surface\ area\ of\ a\ pyramid = 4\left\lbrack \frac{1}{2}(2)(3.2) \right\rbrack + 2^{2}\)

\(the\ total\ surface\ area\ of\ a\ pyramid = 12.8 + 4 = 16.8{cm}^{2}\ \)

\(the\ volume\ of\ the\ pyramid = \ \frac{1}{3}Ah\)

\(the\ volume\ of\ the\ pyramid = \ \frac{1}{3}\left( 2^{2} \right)(3) = {4cm}^{3}\)