## HM1 – 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**, 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.

## HM2 – 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.

## HM3 – Areas of Shapes

**Answer:** As we need to find a missing side-length rather than the area, we’re going to have to set up an equation and rearrange it to find x. The formula for the area we’ll need here is

\dfrac{1}{2}ab \sin(C),

so, our equation is

\dfrac{1}{2} \times 2.15 \times x \times \sin(26) = 1.47

Simplifying the left-hand-side, we get

1.075 \sin (26) \times x = 1.47

Finally, dividing through by 1.075\sin(26), and putting it into a calculator, we get

x = \dfrac{1.47}{1.075\sin(26)} = 3.12\text{m (2 d.p.)}

## HM4 – 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 [\latex]

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.

## HM5 - Sectors of Circles

**Answer:** The question asks for the total perimeter of the shape. We know that one side is 14mm but the other two are missing. Immediately we can identify that the other straight side is also a radius of the circle and so will also be 14mm long. Then, all that remains is to calculate the arc length and add up our answers.

The angle in this sector is 165 degrees, meaning that the arc length will be equal to \frac{165}{360} of the total circumference. The formula for the circumference is \pi d, or alternatively (and more helpfully in this case), 2\pi r So, we get:

\text{Arc length } = \dfrac{165}{360} \times 2 \pi \times 14 = 40.3\text{mm}

Therefore, \text{total perimeter } = 14 + 14 + 40.3 = 68.3\text{mm}

## HM6 - Circle Theorems 1

**Answer:** When a question like this tells you to show our workings, you __must__ state what circle theorem/geometry fact you use when you use it.

There’s no way for us to immediately find the angle we want, so we’re going to try to find the other angles in the quadrilateral ABCD. The first circle theorem we’re going to use here is: **the angle at the centre is twice the angle at the circumference**. The angle at the centre is 126° , so \text{angle BAD } = 126 \div 2 = 63\degree.

We now know two out of the four angles inside ABCD. To find a third, simply observe that **angles around a point sum to 360**, then we get that the angle at point C (the one inside ABCD) is 360 - 126 = 234\degree. Since the **angles in a quadrilateral sum to 360****°** , if we subtract the ones we know from 360 then find the angle we’re looking for.

\text{Angle ABC } = 360 - 33 - 63 - 234 = 30 \degree.

## HM7 - Circle Theorems 2

**Answer:** Our first circle theorem here will be: **tangents to a circle from the same point are equal**, which in this case tells us that AB and BD are equal in length.

This means that ABD must be an isosceles triangle, and so the two angles at the base must be equal. In this case those two angles are angles BAD and ADB, neither of which know. Let the size of one of these angles be x, then using the fact that **angles in a triangle add to 180**° , we get

x + x + 42 = 180.

Then, subtract 42 from both sides to get

2x = 180 - 42 = 138,

and divide both sides by 2 to get

x = 69\degree.

Now we can use our second circle theorem, this time the **alternate segment theorem**. This tells us that the angle between the tangent and the side of the triangle is equal to the opposite interior angle. Given that angle ADB, which is 69\degree, is the angle between the side of the triangle and the tangent, then the alternate segment theorem immediately gives us that the opposite interior angle, angle AED (the one we’re looking for), is also 69\degree.

## HM8 - 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.

## HM9 - 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.

## HM10 - Translations and Reflections

**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.

## HM11 - 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.

## HM12 - Negative Enlargements

**Answer:** This question has it all going on: the enlargement factor is negative **and** it’s less than one. Don’t panic – we start this in the same way we start any enlargement question: by drawing lines from the corners of the shape to the centre of enlargement.

Now we think about the new lines we’re going to draw. The fact the enlargement factor is negative tells us that A’ ( A’ is just the new position of A) is going to be in the opposite direction than A, relative to the centre of enlargement. The fact that the size of the enlargement factor is 0.5 tells us that the line to OA’ will be **half** the length of the line to OA.

So, we should get something like this (where X is the centre of enlargement).

**Tip**: A good way of plotting these lines is not to think about the **length** exactly, but the height and width of the lines. For example, the line XA goes 4 up and 1 to the right. We know the line XA’ will go half this in the opposite direction: so 2 down and ½ to the left.

## HM13 -Volumes of 3D shapes

**ANSWER: **So, to work out the volume of a prism we must multiply the area of the cross section by the length. In this case, the cross section is a trapezium, and the area of the trapezium is

\text{area of cross section }=\dfrac{1}{2}\times(45+60)\times20=1,050\text{ cm}^2

The length of the prism is 80cm, so we get

\text{volume of prism }=1,050\times80=84,000\text{ cm}^3

## HM14 -Volumes of 3D shapes 2

**Answer:** To work out the volume of the shape, we need to work out the two volume 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)}

## HM15 - Surface Areas of 3D Shapes

**Answer:** We know the whole surface area is 120 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

Now, this must be the area of the curved face, and the formula for the area of the curved face is given to us: \pi r l. So, setting this formula equal to the value we worked out, we get

\pi r l=120-9\pi

Then, to find the slant height, we will divide both sides by \pi r to get

l=\frac{120-9\pi}{3\pi}=9.7 cm (1dp)

## HM16 - 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.

## HM17 - 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.