If you’re struggling with this week’s questions, this walkthrough should help you get the skills you need.

This week, we’re exploring:

- Similar Solids
- Pythagoras’ Theorem in 3D
- Capture Recapture
- Changing the Subject of a Formula
- Expanding Double Brackets

### Question 1: Similar Solids

Two shapes are similar if one is an enlargement of the other. The key to solving problems involving similar shapes is finding the scale factor.

**The two prisms below are similar. The surface area of the smaller prism is 243cm ^{2}. Find the surface area of the larger prism.**

First, we compare the two lengths we’re given. By dividing the larger length by the smaller length, we can find the scale factor for length.

Scale factor for length =

=

So, if we multiply any of the lengths on the smaller shape by this scale factor, we will get the corresponding length on the larger shape.

The area has been enlarged in **two dimensions**, so we need to **square** the scale factor for length to find the scale factor for area.

Scale factor for area =

=

Surface area of the larger shape = 243 ×

= **432cm ^{2}**

**Using the same shapes, find the volume of the smaller prism given that the volume of the larger prism is 320cm ^{3}.**

The volume has been enlarged in **three dimensions**, so we need to **cube** the scale factor for length to find the scale factor for volume.

Scale factor for volume =

=

Volume of the smaller shape = 320 ÷

= **135cm ^{3}**

This is the connection between the scale factors for length, area (including surface area) and volume:

Scale factor for length =

Scale factor for area =

Scale factor for volume =

### Question 2: Pythagoras’ Theorem in 3D

When using Pythagoras’ theorem to solve problems in 3D, the most difficult part is **visualising** the problem. It can help to draw out the 2D triangles you are working with, as well as labelling the 3D shape. Colour coding can also help.

**Find the length of AH in the cuboid below.**

We will start by finding the length AD. We will use triangle ABD as shown below:

This right-angled triangle has shorter sides of 21cm and 12cm and we are looking for the hypotenuse.

AD^{2} = AB^{2} + BD^{2}

AD^{2} = 21^{2} + 12^{2}

AD^{2} = 585

AD =

Now, we can use this length with triangle ADH to find the length of AH.

This right-angled triangle has shorter sides of cm and 16cm and we are looking for the hypotenuse.

AH^{2} = AD^{2} + DH^{2}

AH^{2} = ()^{2} + 16^{2}

AH^{2} = 841

AH = 29cm

The length AH is **29cm**.

### Question 3: Capture Recapture

Capture recapture is a statistics related question that involves **equivalent fractions**.

Essentially, a sample of a population is taken and marked somehow. This sample is then released into the population and, after some time, a second sample is taken. The size of the population is estimated by the number of marked items in the second sample.

It works on a few assumptions:

- The size of the population is fixed – it can’t increase or decrease significantly between the two samples.
- The population is contained. For example, this method wouldn’t work with wild birds as they are likely to migrate in the time between the two samples being taken.
- The first sample has enough time to integrate with the population before the second sample is taken.

**Cole catches 10 rabbits from a warren. He tags them and releases them back into the warren. The next day, Cole catches 20 rabbits from the warren. 4 of them are tagged. Estimate the number of rabbits in the warren.**

Let’s call the whole population . 10 of the population are tagged. We assume the same proportion of tagged rabbits are in the second sample. We can write these proportions as two fractions and set them equal.

Now, we need to solve this equation for .

There are approximately **50 rabbits** in the warren.

### Question 4: Changing the Subject of a Formula

To change the subject of a formula, we need to make sure we keep both sides of the formula **balanced**. Just like when you are solving equations, whatever you do to one side, you must do to the other.

**Let’s make the subject of the formula below.**

At first glance, it may look like is already the subject of this formula. However, there is a second occurrence of on the right-hand side of the formula. For to be the subject, it needs to be in the form = … with no other terms on the other side of the formula.

We start by multiplying by the denominator of the fraction.

Next, we need to **factorise** the out of the expression on the left hand side.

Finally, we divide by the expression in the backets:

### Question 5: Expanding Double Brackets

There are a lot of different methods to expand double brackets. We will look at the **grid method **as this is the simplest method.

**Expand and simplify:**

We start by drawing a grid. Write one of the brackets across the top and the other down the side – it doesn’t matter which way round.

Now, we simply multiply the term in each row by each column. Make sure you are careful with the negatives.

Don’t forget to read even more of our blogs **here** and you can find our main Daily Maths Revision Page here! You can also **subscribe to Beyond** for access to thousands of secondary teaching resources. You can **sign up for a free account here** and take a look around **at our free resources** before you subscribe too.