End of chapter exercises

\begin{align*} A_{\text{rectangle}} & = l \times b\\ & = (15)(5)\\ & = \text{75}\text{ cm$^{2}$} \end{align*}

\begin{align*} A_{\text{circle}} & = \pi r^{2}\\ & = (3,1415...)(\text{7})^{2} \\ & = \text{153,93804...} \\ & \approx \text{153,94}\text{ mm$^{2}$} \end{align*}

We first need to find the height:

\begin{align*} h^{2} & = (14)^{2} - (8)^{2}\\ h & = \sqrt{132} \end{align*}

Now we can find the area of the parallelogram. Note that the length of the base is \(b = 8 + 14 = 22\).

\begin{align*} A_{\text{parallelogram}} & = b \times h\\ & = (22)(\sqrt{132})\\ & \approx \text{252,76}\text{ cm$^{2}$} \end{align*}

\begin{align*} A & = bh \\ A & = (-5y)(-3y + 2) \\ A & = 15y^{2} - 10y \end{align*}

\begin{align*} A & = \frac{1}{2}bh \\ A & = \frac{1}{2}(-5y)(-3y + 2) \\ A & = \frac{15y^{2}}{2} - 5y \end{align*}

Cone

We first need to calculate the slant height:

\begin{align*} h_{s}^{2} & = r^{2} + h^{2} \\ & = 3^{2} + 10^{2} \\ h_{s} & = \sqrt{109} \end{align*}

Now we can calculate the surface area:

\begin{align*} \text{Surface area} &= \pi r (r + h_{s}) \\ & = \pi (3)(3 + \sqrt{109})\\ & \approx \text{126,67}\text{ cm$^{2}$} \end{align*}

Square pyramid

We first need to calculate the slant height:

\begin{align*} h_{s}^{2} & = b^{2} + h^{2} \\ & = (\text{7,5})^{2} + 12^{2} \\ h_{s} & = \sqrt{\text{200,25}} \end{align*}

Now we can calculate the surface area:

\begin{align*} \text{Surface area} &= b(b + 2h_{s})\\ & = (\text{15})(\text{15} + 2\sqrt{\text{200,25}}) \\ & \approx \text{437,26}\text{ cm$^{2}$} \end{align*}

Half sphere

For a half sphere we need to divide the surface area of a sphere by 2. We also need to include the area of a circle.

\begin{align*} \text{Surface area} &= \frac{4 \pi r^2}{2} + \pi r^{2}\\ & = 2 \pi (4)^{2} + \pi (4)^{2} \\ & = 48 \pi \\ & \approx \text{150,80}\text{ cm$^{2}$} \end{align*}

The surface area of each of the objects is:

• \(A_{\text{cone}} = \text{126,67}\text{ cm$^{2}$}\)
• \(A_{\text{square pyramid}} = \text{437,26}\text{ cm$^{2}$}\)
• \(A_{\text{half sphere}} = \text{150,80}\text{cm$^{2}$}\)>

Cone

\begin{align*} V &= \frac{1}{3} \pi r^{2} H \\ & = \frac{1}{3} \pi (3)^2 \times 10 \\ & = 30 \pi \\ & \approx \text{94,25}\text{ cm$^{3}$} \end{align*}

Square pyramid:

\begin{align*} V & = \frac{1}{3} b^{2} \times H \\ & = \frac{1}{3} (15)^2 \times 12 \\ & = \text{900}\text{ cm$^{3}$} \end{align*}

Half sphere

The volume of a half sphere is half the volume of a sphere.

\begin{align*} V & = \frac{4}{3}\pi r^{3} \times \frac{1}{2}\\ & = \frac{4}{3} \pi (4)^{3} \times \frac{1}{2} \\ & = 18 \pi\\ & \approx \text{134,04}\text{ cm$^{3}$} \end{align*}

The volume of each of the objects is:

• \(V_{\text{cone}} = \text{94,25}\text{ cm$^{3}$}\)
• \(V_{\text{square pyramid}} = \text{900}\text{ cm$^{3}$}\)
• \(V_{\text{half sphere}}= \text{134,04}\text{ cm$^{3}$}\)

Cylinder

A cylinder is composed of two circles and a rectangle. The breadth of the rectangle is the circumference of the circle.

\begin{align*} \text{Surface area} & = 2 \pi r^2 + 2 \pi r h \\ & = 2 \pi (4)^2 + 2 \pi(4)(10) \\ & = 112 \pi \\ & \approx \text{351,9}\text{ cm$^{2}$} \end{align*}

Triangular prism

A triangular prism is composed of three rectangles and two triangles. We are given the vertical height of the triangles as well as the slant height.

\begin{align*} \text{Surface area} &= 2(\frac{1}{2}b \times h) + 2(H \times h_{s}) + (H \times b)\\ &= 2(\frac{1}{2}(8)(3)) + 2(20 \times 5) + (20 \times 8)\\ &= \text{384}\text{ cm$^{2}$} \end{align*}

Rectangular prism

A rectangular prism is composed of 6 rectangles. We have the dimensions of all the rectangles.

\begin{align*} \text{Surface area} &= 2[(L \times b) + (b \times h) + (L \times h)] \\ &= 2[(5 \times 4) + (4 \times 2) + (5 \times 2)] \\ &= \text{76}\text{ cm$^{2}$} \end{align*}

The surface area of each shape is:

• \(A_{\text{cylinder}} = \text{351,9}\text{ cm$^{2}$}\)
• \(A_{\text{triangular prism}} = \text{384}\text{cm$^{2}$}\)
• \(A_{\text{rectangular prism}}= \text{76}\text{ cm$^{2}$}\)

Cylinder

\begin{align*} V & = \pi r^2 h \\ & = \pi (4)^2(10) \\ & = 160 \pi \\ & = \text{502,7}\text{ cm$^{3}$} \end{align*}

Triangular prism

\begin{align*} V &= \frac{1}{2} \times h \times b \times H\\ &= \frac{1}{2}(3)(8)(20)\\ &= \text{240}\text{ cm$^{3}$} \end{align*}

Rectangular prism

\begin{align*} \text{Volume} &= l \times b \times h\\ &= 5 \times 4 \times 2 \\ &= \text{40}\text{ cm$^{3}$} \end{align*}

The volume of each shape is:

• \(V_{\text{cylinder}} = \text{502,7}\text{ cm$^{3}$}\)
• \(V_{\text{triangular prism}} = \text{240}\text{ cm$^{3}$}\)
• \(V_{\text{rectangular prism}} = \text{40}\text{ cm$^{3}$}\)

Start with the faces of the cube, which are all squares:

\begin{align*} A_{\text{five squares}} & = 5 \times s^{2}\\ & = 5 \times (\text{7})^{2}\\ & = \text{245}\text{ cm$^{2}$} \end{align*}

Next we note that the height of the pyramid is:

\[h_{\text{pyramid}} = \text{22} - \text{7} = \text{15}\]

And we need to calculate the slant height using the theorem of Pythagoras:

\begin{align*} h_{s} & = \sqrt{(\text{15})^{2} + \left(\frac{1}{2}(\text{7})\right)^{2}} \\ & = \sqrt{\left(\text{225} + \text{12,25}\right) } \\ & = \sqrt{\text{237,25}} \end{align*}

Now we can calculate the area of each of the four triangles:

\begin{align*} A_{\text{four triangles}} & = 4 \times \frac{1}{2}bh_{s}\\ & = 4 \times \frac{1}{2}(\text{7})(\sqrt{\text{237,25}})\\ & = 14\sqrt{\text{237,25}} \end{align*}

Finally we can calculate the total surface area:

\begin{align*} A_{\text{total}} & = A_{\text{triangles}} + A_{\text{squares}} \\ & = 14\sqrt{\text{237,25}} + \text{245}\\ & = \text{460,6409...}\text{ cm$^{2}$} \\ & \approx \text{460,64}\text{ cm$^{2}$} \end{align*}

Therefore the surface area is: \(\text{460,64}\) \(\text{cm$^{2}$}\).

Volume of the pyramid:

\begin{align*} V_{\text{pyramid}} & = \frac{1}{3}b^{2}H\\ & = \frac{1}{3}(\text{7})^{2}(\text{15})\\ & = \text{245}\text{ cm$^{3}$} \end{align*}

Volume of the cube:

\begin{align*} V_{\text{cube}} & = l^{3}\\ & = (\text{7})^{3}\\ & = \text{343}\text{ cm$^{3}$} \end{align*}

Total volume:

\begin{align*} V_{\text{total}} & = V_{\text{cube}} + V_{\text{pyramid}}\\ & = \text{343} + \text{245}\\ & = \text{588}\text{ cm$^{3}$} \end{align*}

Therefore the total volume is: \(\text{588}\) \(\text{cm$^{3}$}\).

The shape is a half sphere on top of a right cone. We can calculate the volume of a cone and add this to half the volume of a sphere. The volume is:

\begin{align*} V & = \frac{1}{3}\pi r^{2}h + \frac{1}{2}\left(\frac{4}{3} \pi r^{3}\right) \\ & = \frac{1}{3} \pi (10)^{2}(15) + \frac{2}{3} \pi (10)^{3} \\ & = \frac{\text{3 500}}{\text{3}}\pi \\ & = \text{3 665,19}\text{ cm$^{3}$} \end{align*}

For the surface area we first need to find the slant height:

\begin{align*} h_{s}^{2} & = 10^{2} + 15^{2} \\ & = 325 \\ h_{s} & = 5\sqrt{13} \end{align*}

We have a half sphere on top of a cone. The half sphere covers the circle on top of the cone and so we need to exclude this part from our calculation. For the half sphere we can use half the surface area of a sphere as this does not include the circle at the base of the half sphere.

The surface area is:

\begin{align*} \text{surface area } & = \pi r h_{s} + 2\pi r^{2} \\ & = \pi \left(10\right)\left(5 \sqrt{13}\right) + 2\pi\left(10\right)^{2} \\ & \approx \text{1 194,68}\text{ cm$^{2}$} \end{align*}

Therefore the volume and surface area are \(\text{3 665,19}\) \(\text{cm$^{3}$}\) and \(\text{1 194,67}\) \(\text{cm$^{2}$}\) respectively.

We have a cylinder with two half spheres. We can calculate the volume of a cylinder and add the volume of a sphere to this. The volume is:

\begin{align*} V & = \pi r^{2}h + \frac{4}{3} \pi r^{3} \\ & = \pi (6)^{2}(11) + \frac{4}{3} \pi (6)^{3} \\ & = \text{684} \pi \\ & = \text{2 148,85}\text{ m$^{3}$} \end{align*}

For the surface area of the two half spheres we can use the surface area of a sphere. For the cylinder we need to exclude the area of the two circles from our calculation since these are covered up by the two half spheres. The surface area is:

\begin{align*} \text{surface area } & = 2\pi rh + 4\pi r^{2} \\ & = 2\pi \left(6\right)\left(11\right) + 4\pi\left(6\right)^{2} \\ & = \text{276} \pi \\ & \approx \text{867,08}\text{ m$^{2}$} \end{align*}

Therefore the volume and surface area are: \(\text{2 148,85}\) \(\text{m$^{3}$}\) and \(\text{867,08}\) \(\text{m$^{2}$}\) respectively.

This shape consists of a triangular prism and a rectangular prism. The volume is:

\begin{align*} V & = \frac{1}{2}bhH + lbh \\ & = \frac{1}{2}(12)(9)(4) + (4)(6)(16) \\ & = \text{600}\text{ ft$^{3}$} \end{align*}

For the surface area we need to exclude the base of the triangular prism as well as part of the top of the rectangular prism.

We first need to calculate the slant height for the triangular prism:

\begin{align*} h_{s}^{2} & = 9^{2} + 4^{2} \\ & = 97 \\ h_{s} & = \sqrt{97} \end{align*}

Now we can calculate the surface area of the triangular prism. Remember that we do not need to include the base in our calculation so we only have 2 triangles and 2 rectangles.

\begin{align*} \text{surface area } & = 2\left(\frac{1}{2}bH\right) + 2\left(bh_{s}\right) \\ & = (12)(9) + 2(12)(\sqrt{97}) \\ & = 108 + 24\sqrt{97} \end{align*}

For the rectangular prism we can calculate the full surface area and then subtract the base of triangular prism from this.

\begin{align*} \text{surface area } & = 2(bh) + 2(bl) + 2(hl) - \text{base triangular prism} \\ & = 2(16)(6) + 2(16)(4) + 2(6)(4) - (12)(4) \\ & = 256 \end{align*}

Now we can add the two surface areas together to get the total surface area:

\begin{align*} \text{surface area } & = 256 + (108 + 24\sqrt{97}) \\ & \approx \text{600,37}\text{ ft$^{2}$} \end{align*}

The volume and surface area are \(\text{600}\) \(\text{ft$^{3}$}\) and \(\text{600,37}\) \(\text{ft$^{2}$}\) respectively.

We can draw the following two triangles based on the information in the figure:

These two triangles are similar triangles. They are both right-angled and share a common angle.

Therefore we can use the ratio of the sides to find \(h\):

\begin{align*} \frac{h}{120} & = \frac{45}{60} \quad \text{(similar triangles)} \\ \therefore h & = \frac{(45)(120)}{60} \\ & = \text{90}\text{ cm} \end{align*}

The volume of fuel that has been siphoned out is the total volume of fuel minus the volume of fuel left. The volume of a cone is \(\frac{4}{3}\pi r^{2}H\). From the previous question we have the vertical height for both cones.

\begin{align*} \text{Volume siphoned out } & = \frac{1}{3}\pi R^{2}H_{\text{start}} - \frac{1}{3}\pi r^{2}H_{\text{end}} \\ & = \frac{1}{3}\pi (60)^{2}(120) - \frac{1}{3}\pi (45)^{2}(90) \\ & = \text{144 000} \pi - \text{60 750} \pi \\ & \approx \text{261 537,59}\text{ cm$^{3}$} \\ & \approx \text{261,5}\text{ l} \end{align*}

We are given the diameter of the cylinder. The radius is half the diameter.

\begin{align*} V & = \pi r^2 h \\ & = \pi (10)^2(15) \\ & = \text{15 000} \pi \\ & \approx \text{47 123,89}\text{ cm$^{3}$} \end{align*} \begin{align*} A & = 2 \pi r h + 2 (\pi r^2) \\ & = 2 (\pi (10)(15)) + 2(\pi (10)^{2}) \\ & = 300\pi + 200 \pi \\ & \approx \text{1 570,80}\text{ cm$^{2}$} \end{align*}

Therefore the volume and surface area are \(\text{47 123,89}\) \(\text{cm$^{3}$}\) and \(\text{1 570,80}\) \(\text{cm$^{2}$}\) respectively.

This is a triangular pyramid. We are given the vertical height as well as an angle. Since it is a right-angled triangle we can use trigonometry to help us find the missing length.

We redraw the triangle we are interested in:

Now we can calculate \(x\) (the slant height) and \(y\) (the base):

\begin{align*} \frac{x}{8} & = \tan 30° \\ x & = 8 \tan 30° \\\\ \frac{8}{y} & = \sin 30° \\ \frac{8}{\sin 30°} & = y \end{align*}

Now we know all the lengths we need to know to calculate the volume.

\begin{align*} V & = (\frac{1}{2} bh ) \times H \\ & = (\frac{1}{2} (8)(8\tan{30^{\circ}}) ) \times 5\\ & = 160 \tan 30^{\circ} \\ & \approx \text{92,38} \end{align*}

And the surface area is:

\begin{align*} A &= 2(\frac{1}{2} bh )+ (H \times h_s) + (H \times h_s) + (H \times b) + (H \times h) \\ \\ & = 2(\frac{1}{2} (8)(8\tan{30^{\circ}}) + (5 \times \frac{8}{\sin{30^{\circ}}}) \\ & + (5 \times 8\tan{30^{\circ}}) + (5 \times 8) \\ & = 64 \tan{30^{\circ}} + \frac{40}{\sin{30^{\circ}}} + 40\tan{30^{\circ}} + 40 \\ & \approx \text{180,04} \end{align*}

The volume and surface area are: \(\text{92,38}\) and \(\text{180,04}\) respectively.

Let: \(L = 9\), \(B = 8\), \(H = 15\), \(l = 2\), \(b = 2\) and \(h = 15\).

We can view this shape as three rectangular prisms. Two of the three prisms are exactly the same. The volume is therefore:

\begin{align*} V &= 2(lbh) + LBH \\ &= 2((2)(2)(15)) + ((8)(9)(15))\\ &= 120 + \text{1 080} \\ &= \text{1 200} \end{align*}

For the surface area we have several different rectangles. Each of the smaller prisms has 5 exposed rectangles. The larger rectangular prism has 4 rectangles that are not covered up by the smaller prisms. The remaining two rectangles are partly covered up by the smaller prisms and so can be considered as 4 separate rectangles.

We will start by finding the surface area of one of the smaller prisms:

\begin{align*} A_{\text{smaller prism}} & = 2(bl) + 2(bh) + lh \\ & = 2((2)(2)) + 2((2)(15)) + (2)(15) \\ & = 98 \end{align*}

For the larger prism we get:

\begin{align*} A_{\text{larger prism}} & = 2(BH) + 2(BL) + 4(Hx) \\ & = 2((8)(15)) + 2((9)(15)) + 4((15)(3)) \\ & = 690 \end{align*}

Therefore the total surface area is:

\begin{align*} A & = 98 + 98 + 690 \\ & = 886 \end{align*}

The volume and surface area are \(\text{1 200}\) and 886 respectively.

We first need to find the vertical height (\(H\)):

\begin{align*} H &= \sqrt{(20)^2 - (12)^2}\\ & = 16 \end{align*} \begin{align*} V & = \frac{1}{3} \pi r^2 H\\ & = \frac{1}{3} \pi (12)^2 (16)\\ & = 768 \pi \\ & \approx \text{2 412,743} \end{align*}

We first find the vertical height:

\begin{align*} H & = \sqrt{(10)^{2} - (6)^{2}} \\ & = 8 \end{align*}

We also need to find the length of the side of the square. To do this we note that triangle \(ABC\) is a right-angled isosceles triangle. So we can find the length of the side of the square using the theorem of Pythagoras:

\begin{align*} AC^{2} & = AB^{2} + BC^{2} \\ 12^{2} & = 2(AB^{2}) \\ 77 & = AB^{2} \\ \therefore AB & = \sqrt{77} \end{align*}

Now we can find the volume:

\begin{align*} V &= \frac{1}{3} \pi b^2 H \\ & = \frac{1}{3} \pi (77) (8) \\ & = \frac{616}{3} \pi \\ & \approx \text{645,07}\text{ cm$^{3}$} \end{align*}

Since \(D\) is the centre of the circle \(BD = DC\) (they are both radii of the arc).

\(AB \parallel EC\) and \(BD\) joins \(AB\) and \(EC\), therefore \(AB = ED = 4\text{ units}\).

We also know that \(EC = 8\text{ units}\) and since \(EC = ED + DC\), \(DC = 4 \text{ units}\). Therefore \(BD\) is 4 units.

We have just calculated that \(BD = 4\). We also know that \(AB = ED = 4\) and so \(ABDE\) is a square (\(AB \parallel EC\)). This means that we have the area of a square plus one quarter the area of a circle.

The total area is:

\begin{align*} A & = AB^2 + \frac{1}{4} \pi r^2\\ & = (4)^2 + \frac{1}{4} \pi (4)^2\\ & = 16 + 4 \pi \\ & = \text{28,57} \text{ units}^2 \end{align*}

The area of the shaded piece is the area of the base. For the volume we know that we can calculate the volume by multiplying the area of the base and the height.

\begin{align*} V &= \text{area of base} \times \text{height} \\ &= (16 + 4 \pi) \times (10) \\ & = \text{285,664} \text{ units}^3 \end{align*}

You can also calculate the volume using the volume of a rectangular prism and one quarter of the volume of a cylinder.

1. \(\triangle UIJ\) is a right-angled triangle. We can use trigonometry to help us find \(UI\). In this case we will use the cosine ratio as we have the hypotenuse (\(JI\)) and are looking for the adjacent side (\(UI\)).

   \begin{align*} \cos{\text{33,577}^{\circ}} &= \frac{UI}{JI} \\ UI & = 18 \cos{\text{33,577}^{\circ}} \\ &= \text{15}\text{ cm} \end{align*}

2. \(\triangle UIJ\) is a right-angled triangle. We can use trigonometry to help us find \(JU\). In this case we will use the sine ratio as we have the hypotenuse (\(JI\)) and are looking for the opposite side (\(JU\)).

   \begin{align*} \sin{\text{33,577}^{\circ}} &= \frac{JU}{JI} \\ JU & = 18 \sin{\text{33,577}^{\circ}} \\ &= \text{10}\text{ cm} \end{align*}
3. 

   We first find \(h\):

   \begin{align*} h & = \sqrt{15^2 - \text{3,5}^2} \\ & = \text{14,586} \end{align*}

   Now we can find the area:

   \begin{align*} A &= \frac{1}{2} bh \\ &= \frac{1}{2} (\text{3,5})(\text{14,586}) \\ &= \text{25,526}\text{ cm$^{2}$} \end{align*}
4. \begin{align*} V &= \frac{1}{3} \times \frac{1}{2} (bh) \times JU \\ & = \frac{1}{3} \times (\text{25,526}) (\text{9,950}) \\ V & = \text{84,661}\text{ cm$^{3}$} \end{align*}

To find the volume of the juice we need to multiply the total volume of the container by the percentage of juice in the container.

\begin{align*} V_j &= V \times (1 - \text{0,1185})\\ & = \text{0,8815} (\text{84,661}) \\ V_j & = \text{74,626}\text{ cm$^{3}$} \end{align*}

\begin{align*} A & = b^2 \\ & = (2a)^2 \\ & = 4a^2 \end{align*}

\begin{align*} BC &= 2a \\ \\ AF & = \sqrt{a^2 + h^2} \\ & = \sqrt{(\text{0,5}BC)^2 + (OF)^2} \\ & = \sqrt{(\text{377,895})^2 + (\text{481,4})^2} \\ & = \text{612,00538}\text{ feet} \end{align*}

\begin{align*} \frac{s}{a} & = \frac{\text{612,00538}}{\text{377,895}} \\ & = \text{1,620} \end{align*}

\begin{align*} V &= \frac{1}{3} \pi b^2 H\\ & = \frac{1}{3} \pi (2a)^2 h \\ & = \frac{1}{3} \pi (\text{755,79})^2 (\text{481,4}) \\ & \approx \text{91 661 532,5}\text{ feet$^{3}$} \end{align*} \begin{align*} A &= b(b + 2h_{s})\\ &= 2a(2a + 2(s)) \\ &= \text{755,79}(\text{755,79} + \text{1 224,0176}) \\ & \approx \text{1 551 425,432}\text{ feet$^{2}$} \end{align*}

The volume and surface area are: \(\text{91 661 532,5}\) \(\text{feet$^{3}$}\) and \(\text{1 551 425,432}\) \(\text{feet$^{2}$}\) respectively.