## FM1 – Types of Angle and Angle Facts

**Answer: **The angles in the diagram go all the way around a point, so they must add up to 360\degree. One of the angles doesn’t have a measurement, but it is marked with a square which means it’s a right angle ( 90\degree )

We can write the following equation:

63+57+62+68+90+z=360

Then we just simplify it by adding together the number on the left hand side:

340+z=360

Then subtract 340 from both sides of the equation to give us the answer:

z=20\degree

## FM2 – Angles in Parallel Lines

**Answer:** Firstly, because angles BEF and EHJ are **corresponding angles**, we get

\text{angle EHJ } = 39\degree.

Next, because angles EDH and DHG are **alternate angles, **we get

\text{angle DHG } = 76\degree.

Then, because angles DHG, DHE, and EHJ are angles on a straight line and **angles on a straight line add to **180\degree, we get

\text{angle DHE } = 180 - 76 - 39 = 65\degree

Finally, because angle DHE and angle x are **vertically opposite** **angles**, we get

x = 65\degree.

## FM3 – Areas of Shapes

**Answer:** This is a bit of a weird-looking trapezium, but it is a trapezium nonetheless. To work out the area, we will, as the question states, need to find the perpendicular height. The word ‘perpendicular’ is key here, because it means that we can draw one line and form a right-angled triangle, as seen below. Because the top of a trapezium is always parallel to the base, we can find the base of the small right-angled triangle by subtracting the length of the top from the length of the base to get 3cm.

Then, this is where Pythagoras comes in. The hypotenuse of the right-angled triangle is 5cm and the base is 3cm, so we get the other side (the perpendicular height of the trapezium) as such:

\text{Perpendicular height } = \sqrt{5^2 - 3^2} = \sqrt{16} = 4\text{cm}

Now we know the perpendicular height, we can calculate the area.

\text{Area } = \dfrac{1}{2}(a + b)h = \dfrac{1}{2}(5 + 8) \times 4 = 26\text{cm}^2

## FM4 – Interior & Exterior Angles

**Answer:** This is a 4-sided shape, which we now know has interior angles that add up to 180\times 2=360. So, if we know all the interior angles other than x, then we can find x.

Currently we don’t know them all, however we do have an exterior angle, and we know that exterior angles form a straight line with their associated interior angles, we get the interior angle at D to be 180 - 121 = 59\degree.

Now we know all 4 interior angles, we get that

x = 360 - 84 - 100 - 59 = 117\degree.

You will find algebra involved more in questions on this topic, but as long as you know what the interior and exterior angles add up to then you can write the statement “these angles add up to ___” as an equation which you can then solve.

## FM5 – Parts of the Circle

**Answer:**The straight line passing through A and B intersects the circumference twice and doesn’t pass through the centre, so it’s a **chord.** The straight line BC touches the circumference of the circle just once, so it’s a **tangent**. There are two curved lines joining A and B: the ‘big part’ of the circumference and the small part of the circumference. These are both **arcs**.

## FM6 – Area and Circumference of a Circle

**Question: **A circle has an area of exactly 25{\pi} m^2 . Calculate the **exact** radius and circumference of the circle.

**Answer: **Note the use of the word “exact” in this question. The area given is “exact” because it’s in terms of pi. If we try to write pi (or any multiple of pi) as a decimal, it would go on forever (we say that pi is an “irrational” number). So if we actually wanted to write it as a decimal, we’d have to round it, which would make it inaccurate.

This means that when the question asks us for an “exact” answer, we have to make sure we’re not rounding anything. So we will need to express our answer in terms of pi where appropriate.

For the first part of the question we want to find the radius of the circle. We know the formula that relates radius and area: A=\pi r^2. We need to rearrange this to make r the subject. First we divide both sides of the equation by pi:

\frac{A}{\pi} =r^2

Then we square root both sides to give us r:

\sqrt(\frac{A}{\pi})=r

Substituting the given area for A:

\sqrt(\frac{25\pi}{\pi})=r

The pi’s cancel out:

\sqrt(25)=r

So we have r=5

Next, to find the circumference we just use the formula c=2\pi r

c=2\times \pi \times 5=10\pi

We could convert this answer into a decimal, which would be 31.4 to 1 d.p. But the question asks for an exact answer, so we can’t round it. The only way to represent the answer exactly is to leave it in terms of pi.

## FM7 – Perimeters of Shapes

**Answer:** We can see that the side lengths of the rectangle are 4cm and 6cm… however it’s not immediately clear how to get the length of the curved edge joining A and B!

We can work this out using the formula for the circumference of a circle, c=\pi d. We can see the diameter of the circle is 4cm, the same as the length of the shortest edge of the rectangle. The circumference of a circle with a diameter of 4cm would be 4\pi. However we only have *half* a circumference here, so the length of the curved edge is 2\pi. This means the total perimeter of the shape is:

6+4+6+2\pi =22.3 (1 d.p.)

## FM8 – Congruence

**Answer:** Let’s check each shape individually.

Shape B: it has two angles in common with A, but the side is a different length.

Shape C: this has two angles and a side-length in common with A, but to pass the ASA test the side-length needs to be between the two angles, which in C’s case it isn’t.

Shape D: this does what shape C didn’t – all the numbers match, and the side we know is between the two angles which means that shape D is congruent to A by the **ASA** criteria.

The real value in being able to spot when two triangles are congruent like this is that we suddenly know that all the other angles and side-lengths must also be the same. This is useful in making quick leaps towards solving bigger problems, for example in circle theorems, so keep the definition of congruence as well as the 4 tests for congruent triangles in mind when solving all kinds of geometry problems.

## FM9 – Similar Shapes

**Answer:** Firstly, we will determine the scale factor that relates the side-lengths by dividing the side-length of the bigger shape by that of the smaller shape: SF=28\div7=4.

Now, if the scale factor for the side-lengths is 4, then that means that the scale factor for the areas is: SF_A=4^2=16.

Therefore, to find the area of the smaller shape, we need to divide the area of the bigger shape by the area scale factor: 16. Doing so, we get

\text{Area of A }=320\div16=20\text{ cm}^2.

## FM10 – Solving Quadratics by Factorisation

**Answer:** Firstly, we must draw the line y=1 onto the graph. Then, you can either choose to use tracing paper or, if you’re confident without it, just go right into the reflection.

If you’re using tracing paper, you should firstly trace over the shape and the mirror line. Then, flip over the tracing paper, and line up perfectly the mirror line on the page with the one on the tracing paper such that the trace of the shape is on the opposite side of the line to the original shape.

Then, the trace of the shape is the result of the reflection. Draw that shape onto the original axes, mark it with a C and you should get the resulting picture below.

## FM11 – Rotations and Enlargements

**Answer:** We need to draw lines from the point (0, 1) to all corners of this shape. Then, since this is scale factor 3 enlargement, we need to extend these lines until they are 3 times longer. For example, the line from (0, 1) to A goes 1 space to the right and 1 up. So, once we’ve extended it, the resulting line should go 3 spaces to right and 3 spaces up.

Then, once all these lines have been drawn, their ends will be the corners of the enlarged shape. Joining these corners up, we get the completed shape, as seen below.

## FM12 – Edges, Faces and Vertices

**Answer:** There’s no real “how-to” here – we just need to count them up! Here we’ve been given a diagram, but it always helps to draw one if not so you don’t forget any “hidden” features. There are 5 faces (2 triangles and 3 rectangles), 9 edges, and 6 vertices.

## FM13 – Projections, Plans and Elevations

**Answer:** As we can see, the shape is one block high and is the shape of a cross, or letter x. As a result, the plan is going to be the shape of a cross, and both elevations will only be one block high. Furthermore, we can see that in looking both from the front and from the side, the shape in question is 3 blocks wide, so the resulting elevations will both be 3 blocks wide and 1 block high. All 3 projections are as seen below.

## FM14 -Volume of a Prism

**Answer:** A cylinder is a “circular prism”. To find its volume, like with any other prism, we just need to multiply the cross-sectional area by the length. Because the cross-section of a cylinder is a circle, we can calculate its area using the formula for the area of a circle: A=\pi r^2.

So in this case, the cross-sectional area is: A=\pi 3^2=28.274…cm2.

Next we just multiply this by the length to get the volume:

V=28.274…\times 8=226.19cm^3 (2 d.p.)

## FM15 -Volume of a Cone and Sphere

**Question: **The shape shown was made by attaching a cone to the top of a cylinder. The base of the cylinder has radius 4mm, the height of the cylinder portion is 3mm, and the height of the cone portion is 5.5mm. The formula for the \text{volume of a cone}=\pi r^2 \dfrac{h}{3}. Work out the volume of the whole shape to 1dp.

**Answer:** To work out the volume of the shape, we need to work out the two volumes separately.

Firstly, the cylinder is a type of prism, so we know we need to multiply the area of the cross section by its length. Here, the cross section is a circle with radius 4mm, and the length of the cylinder is 3mm, so we get

\text{volume of cylinder}=\pi\times4^2\times3=48\pi

We’ll worry about the rounding at the end. Next, we have to work out the volume of the cone part, and fortunately the formula for this is given in exams!

h is the height of the cone, which we know to be 5.5, and r is the radius of the cone, which is the same as that of the base: 4mm. Therefore, we get

\text{volume of cone }=\dfrac{1}{3}\pi\times4^2\times5.5=\dfrac{88}{3}\pi

Then, the volume of the shape is the sum of these two answers:

\text{volume of whole shape }=48\pi+\dfrac{88}{3}\pi=242.9498...=242.9\text{mm}^3\text{ (1dp)}

## FM16 – Surface Areas of 3D Shapes

**Answer:** We know the whole surface area is 120{\mathrm{cm}}^2 and we also know the radius. To work out the slant height, we need to first work out what the curved surface area is. In other words, we need to subtract the surface area of the base of the cone (since that’s the only other face) from 120 to get the curved surface area. The base is a circle, so its area is

\pi×3^2=9\pi cm^2

Subtracting this from the total we have

120-9\pi=91.72566…

This is written as a decimal to give an idea of how big it is, but when you put it into your calculator you should either use the ANS key to store the value, or you should directly type in 120-9\pi. If you don’t, it could affect your final answer.

Now, this must the area of the curved face, and the formula for the area of the curved face is given to us: \pi rl\ = 91.72566 l (since we know r=3). So, setting this formula equal to the value we worked out, we get

l=\dfrac{91.72566}{3\pi}

Then, to find the slant height, we will put this into our calculator

l=\dfrac{91.72566}{3\pi}=9.7323

## FM17 – Loci and Constructions

**Answer:** For the fountain to be at least 3m away from his house along CD, we need to only consider the area to the left of the straight line which is parallel to CD and 3cm away from it.

Then, the locus of points which are 1.5m away from the tree at E will be a circle of radius 1.5cm – for the fountain to be at least 1.5m away, it must be outside this circle.

So, the locus of points where he could place the fountain is to the left of the (blue) line 3m away from the house, and outside the (green) circle which is 1.5m away from the tree. The correct region is shaded red on the picture below.

## FM18 – Bearings

**Answer:** We can’t simply measure the angle, since the picture is not drawn accurately. Instead, we will use the fact that the two North lines are parallel to one another.

Firstly, recognise that we can find the other angle around the point B by subtracting 295 from 360.

360 - 295 = 65 \degree

Then, because the two north lines are parallel, we can say that the bearing of B from A and the 65\degree angle we just found are **interior **(sometimes co-interior, or allied, depending on what your teacher likes). From our facts about angles in parallel lines, we know the two angles (marked with red below) must add to 180.

So, we get:

\text{Bearing of B from A } = 180 - 65 = 115\degree.