SL 3.1 Solve distance, midpoint, surface-area, volume, and angle problems | International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

Geometry and Trigonometry

A practical guide to choosing and applying geometry methods

This lesson brings together several ways to describe shape and size. A distance is a length, a midpoint marks the halfway position between two points, surface area measures the outside covering of a solid, volume measures the space it occupies, and an angle measures a turn or opening. The problems may look different, but a reliable approach is to identify the quantities given, select a suitable rule, substitute carefully, and interpret the result with appropriate units. You will use coordinate methods, familiar solid formulas, and right-triangle trigonometry.

What you will learn

Calculate the distance and midpoint between two points on a coordinate plane.
Choose and apply suitable surface-area and volume formulas for familiar solids.
Use right-triangle trigonometry to calculate an unknown angle.
Check whether answers, units, and calculator results fit the context.

1. Distance and midpoint on a coordinate plane

A point on a coordinate plane is written as

(x,y)

:

x

gives its horizontal position and

y

its vertical position. For points

A(x_1,y_1)

and

B(x_2,y_2)

, the horizontal change is

x_2-x_1

and the vertical change is

y_2-y_1

. These changes make the legs of a right triangle, so the Pythagorean theorem gives the distance between the points. The result is a length, so use the same units as the coordinates.

The midpoint is halfway between the endpoints in both directions. Average the two

x

-coordinates and average the two

y

-coordinates. This works even when a change is negative: averaging places the midpoint between the values. On a graph, plot both points, draw or imagine the segment, and check that the calculated midpoint lies halfway along it.

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2},\quad M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)

Distance is non-negative; reversing the order of the points does not change it.
A midpoint averages corresponding coordinates, not the distance between the points.
If coordinates represent metres, the distance and midpoint coordinates are also in metres.

2. Surface area and volume of familiar solids

Surface area is the total area of the outside faces or curved surfaces of a solid. Volume is the amount of three-dimensional space inside it. Surface area is measured in square units, while volume is measured in cubic units. Before choosing a formula, identify the solid and check that all dimensions use compatible units.

For a prism, the cross-section stays the same along its length: its volume is the cross-sectional area multiplied by the length. A cylinder follows the same idea, with a circular cross-section. For a cylinder of radius

r

and height

h

, the two circular ends contribute

2\pi r^2

to total surface area, and the curved side contributes

2\pi rh

. If a problem asks for only the curved surface, do not include the ends.

Other common formulas include the volume of a pyramid or cone, which is one-third of the corresponding base area times the perpendicular height. A sphere has surface area

4\pi r^2

and volume

\frac{4}{3}\pi r^3

. Read the wording carefully: an open container, for example, may not have every face included in its surface area.

V_{\text{cylinder}}=\pi r^2h,\quad A_{\text{cylinder}}=2\pi r^2+2\pi rh

Use a perpendicular height for volume formulas, not a sloping edge.
Check whether the question asks for total surface area or only a specified part.
Keep exact answers involving $\pi$ when appropriate; round only as requested.

3. Finding angles with right-triangle trigonometry

In a right triangle, choose one of the two acute angles and name the sides relative to that angle. The hypotenuse is opposite the right angle. The opposite side is across from the chosen angle, and the adjacent side touches it but is not the hypotenuse.

The sine, cosine, and tangent ratios connect an angle to two side lengths. To find an angle when two sides are known, use the matching inverse operation on a calculator: inverse sine, inverse cosine, or inverse tangent. For example, if the known sides are opposite and adjacent, use inverse tangent. Check the calculator is in degree mode when the context or question uses degrees.

Sketching the triangle helps connect the calculation to the situation. Label the known sides, the right angle, and the unknown angle before choosing a ratio. A calculator gives a numerical angle; the side labels and ratio explain why that is the correct calculation.

\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}},\quad\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}},\quad\tan\theta=\frac{\text{opposite}}{\text{adjacent}}

Choose the trigonometric ratio from the sides that are known relative to the angle.
Inverse trigonometric operations return an angle; ordinary sine, cosine, and tangent take an angle as input.
State angle units and round to the accuracy requested.

4. Representation, technology, and checking

A useful solution connects a model to its meaning. Coordinate calculations describe points on a graph; solid formulas turn dimensions into areas or volumes; trigonometric ratios connect a diagram to an angle. Units and a labelled sketch make the interpretation visible, while an exact expression or a rounded decimal communicates the numerical result.

A graphing calculator can plot two coordinate points and help you inspect their separation and midpoint. It can also evaluate a formula or an inverse trigonometric function. Use it to check arithmetic or explore a diagram, not as a substitute for stating the formula and showing which values were substituted. For angles, verify degree or radian mode before interpreting the result.

A final reasonableness check can catch common errors. Distance cannot be negative; a midpoint should lie between the endpoints; surface area has square units; volume has cubic units; and a right-triangle angle should be between

0^\circ

and

90^\circ

. Keep extra digits during intermediate calculations and round at the end.

Write down the relevant rule before substituting values.
Use a labelled diagram or plotted points to support the calculation.
Check units, mode, and requested rounding in the final answer.

Worked example

Distance and midpoint

Find the distance and midpoint of the segment joining

A(1,2)

and

B(7,10)

.

Find coordinate changes
The horizontal and vertical changes form the perpendicular legs of a right triangle.
$7-1=6,\quad 10-2=8$
Calculate the length
Apply the Pythagorean theorem to the two changes. The distance is a length in coordinate units.
$d=\sqrt{6^2+8^2}=\sqrt{100}=10$
Average the coordinates
The midpoint has the average of the two horizontal coordinates and the average of the two vertical coordinates.
$M=\left(\frac{1+7}{2},\frac{2+10}{2}\right)=(4,6)$

Answer: The distance is

10

units and the midpoint is

(4,6)

.

Check: The point

(4,6)

is halfway between the endpoint coordinates. The distance is positive and agrees with a triangle having legs

6

and

8

.

Worked example

Cylinder surface area and volume

A closed cylindrical container has radius

3

cm and height

8

cm. Find its total surface area and volume, giving exact answers in terms of

\pi

.

Select the formulas
The container is closed, so total surface area includes both circular ends and the curved side. Volume is circular base area multiplied by height.
$A=2\pi r^2+2\pi rh,\quad V=\pi r^2h$
Substitute the dimensions
Use $r=3$ cm and $h=8$ cm in both formulas. The area calculation produces square centimetres, and the volume calculation produces cubic centimetres.
$A=2\pi(3)^2+2\pi(3)(8)=66\pi\text{ cm}^2,\quad V=\pi(3)^2(8)=72\pi\text{ cm}^3$

Answer: The total surface area is

66\pi\text{ cm}^2

and the volume is

72\pi\text{ cm}^3

.

Check: The surface-area result includes two ends, contributing

18\pi\text{ cm}^2

, and the curved side, contributing

48\pi\text{ cm}^2

. The volume has cubic units.

Worked example

Angle from two sides

A right triangle has an opposite side of length

7

cm and an adjacent side of length

24

cm relative to angle

\theta

. Find

\theta

to the nearest tenth of a degree.

Choose a ratio
The known sides are opposite and adjacent, so the tangent ratio relates them directly.
$\tan\theta=\frac{7}{24}$
Use the inverse operation
Apply inverse tangent to find the angle. Set the calculator to degrees because the requested answer is in degrees.
$\theta=\tan^{-1}\left(\frac{7}{24}\right)\approx16.3^\circ$

Answer: The angle is approximately

16.3^\circ

.

Check: The opposite side is shorter than the adjacent side, so the angle should be less than

45^\circ

. The calculated angle is consistent with that comparison.

Common mistakes and how to avoid them

Subtracting coordinate values but forgetting to square both changes in the distance calculation.

Correction: Square each coordinate change, add the squares, and then take the square root.

Using only the curved side when a closed cylinder's total surface area is requested.

Correction: Include both circular ends unless the question explicitly asks for curved surface area only.

Using ordinary tangent instead of inverse tangent when the angle is unknown.

Correction: First form the ratio from the known sides, then use the corresponding inverse function to obtain the angle.

Giving a volume in square units or an area in cubic units.

Correction: Use squared units for surface area and cubed units for volume.

Lesson summary

Use coordinate changes and the Pythagorean theorem to find distance; average coordinates to find a midpoint.
Identify the solid and the surface requested before selecting an area or volume formula.
For a right triangle, match the known sides to sine, cosine, or tangent and use an inverse function to find an angle.
Show the method, use a calculator as a check, and report units and rounding clearly.

Check your understanding

Question 1

What is the midpoint of the points

(2,4)

and

(8,10)

?

$(5,7)$
$(6,6)$
$(10,14)$
$(3,5)$

Show answer and explanation

(5,7)

Average the horizontal coordinates and the vertical coordinates:

\left(\frac{2+8}{2},\frac{4+10}{2}\right)=(5,7)

.

Question 2

A cylinder has radius

2

m and height

5

m. What is its volume?

$20\pi\text{ m}^3$
$10\pi\text{ m}^3$
$28\pi\text{ m}^3$
$40\pi\text{ m}^2$

Show answer and explanation

20\pi\text{ m}^3

Use

V=\pi r^2h

:

\pi(2)^2(5)=20\pi\text{ m}^3

.

Question 3

In a right triangle, the opposite side to

\theta

is

5

and the adjacent side is

12

. Which calculation gives

\theta

in degrees?

$\tan^{-1}\left(\frac{5}{12}\right)$
$\tan\left(\frac{5}{12}\right)$
$\sin^{-1}\left(\frac{5}{12}\right)$
$\cos^{-1}\left(\frac{5}{12}\right)$

Show answer and explanation

\tan^{-1}\left(\frac{5}{12}\right)

Opposite and adjacent sides form the tangent ratio, so the angle is

\tan^{-1}(5/12)

, approximately

22.6^\circ

.

Key terms

Midpoint: The point halfway between two endpoints of a segment.
Surface area: The total area covering the outside of a three-dimensional solid.
Volume: The amount of three-dimensional space inside a solid.
Hypotenuse: The longest side of a right triangle, opposite its right angle.
Inverse trigonometric function: A calculator operation used to find an angle from a trigonometric ratio.

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SL 3.1 · Solve distance, midpoint, surface-area, volume, and angle problems

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

A practical guide to choosing and applying geometry methods

What you will learn

1. Distance and midpoint on a coordinate plane

2. Surface area and volume of familiar solids

3. Finding angles with right-triangle trigonometry

4. Representation, technology, and checking

Distance and midpoint

Cylinder surface area and volume

Angle from two sides

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through IB AA SL

About this lesson and its review