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

If it’s a formula you need, here are the ones that are relevant this week. You’re not given these in the exam, so you need to learn them.

**Density**

This week, we’re exploring:

### Question 1: Rationalising Surds

Rationalising a surd means finding an **equivalent fraction **that has an **integer** as the denominator. In other words, you want to manipulate the fraction so that any surds are on the top.

**Let’s simplify:**

To eliminate the surd in the denominator, we multiply the denominator by its conjugate. This is the same expression but with the sign in the middle changed. In this case, we multiply by . As we need to find an equivalent fraction, we need to multiply both the numerator and denominator by this.

Let’s see the effect of this on the denominator:

As you can see, the surd has been eliminated. This will always be the case when we use the conjugate of the denominator.

Now, let’s consider the fraction as a whole:

That’s it! You have no surds on the bottom of the fraction, but you haven’t changed the value of the number.

### Question 2: Fractional and Negative Indices

Let’s start with the general rules and then we will look at how we **apply** that rule:

This looks a bit complicated but what it means is slightly simpler. If we have a number to the power of one half, we take the square root of that number; if it’s to the power of one third, we take the cube root and so on. For negative indices, you find the **reciprocal **(the reciprocal of is ). Here are some examples:

Sometimes you will have to evaluate a mix of negative and fractional indices, similar to the final example. The key is to take it one step at a time. This way, you are less likely to make mistakes and will get marks for any working.

**Let’s find the value of**:

First, we will consider the negative part of the power. Remember, this means we find the reciprocal.

Now, we take the square root of the fraction – simply square root the numerator and the denominator.

Finally, we cube the fraction.

### Question 3: Congruent Triangles

Two shapes are **congruent** if they are the same size and shape. When working with triangles, we can prove two (or more) triangles are congruent using one of four rules:

- SSS – all of their sides are equal.
- SAS – two sides and an
**enclosed**angle are equal. - ASA – two angles and a side are equal.
- RHS – the hypotenuse and another side are equal in a right-angle triangle.

**Show that the triangles ABE and CDE are congruent.**

First, we can see that ED = EB as they are both labelled as 12.2cm.

Next, we can know that AEB = DEC as they are vertically opposite angles, so they are equal.

Finally, we know AB and DC are parallel. This means that ABE and EDC are **alternate** angles which means they are equal.

So, we have a side (ED = EB) and two angles (ABE = EDC and AEB = DEC) so, by **ASA, the triangles are congruent.**

There are other methods that show the two triangles are congruent. This is often the case but you only need to give one complete answer to get the marks.

### Question 4: Density

The density of an object is the mass per unit volume; usually g/cm^{3} or kg/m^{3}. You aren’t given the formula for density, so you need to learn it.

You can also write this as a formula triangle:

**Let’s find the mass of an object of volume 30cm ^{3} that has a density of 0.8g/cm^{3}.**

In this case, we want to find the mass so we cover this up in the formula and use the other two variables. Since density and volume are on the same level, we multiply them together.

mass = density × volume

= 0.8 × 30

= **24g**

Make sure you include units in your answer – there is often a mark for this.

If the question asked for the density, we would use density = mass ÷ volume. If the question asked for the volume, we would use volume = mass ÷ density.

### Question 5: Circle Theorems

These are the two circle theorems that are relevant this week.

**Find the size of the angle marked .**

This looks like the first circle theorem but is actually the second one. The point O is the centre of the circle so the angle COB is double the angle CAB:

= 21 × 2

= 42°

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