# Daily Maths Revision – Week 6 Walkthrough

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, these are the relevant ones this week.

Sectors $\text{area of a sector} = \frac{\theta}{360} \times \pi r^2$

The Area of a Triangle $\text{area} = \frac{1}{2} ab\sin C$

In the 2023 exams, you will be given the area of a triangle but you won’t be given the formula for the area of a sector.

This week, we’re exploring:

### Question 1: Area of a Sector

A sector is simply a fraction of a circle – you can think of it as a pizza slice. To find the area of a sector, we need to find what fraction of the whole circle it is. This means you need to be confident with finding the area of a circle – we covered this in week 2 so it might be worth heading to that blog for a quick refresher.

Let’s find the area of the sector below. We will give our answer to 2 decimal places.

There are two methods – the first follows a simple procedure and the second jumps straight in with a formula. In an exam, either of the methods will get you the marks.

For the first method, we will find the area of the whole circle, then find the area of a 1° slice and finally the area of the sector itself.

So, the area of the sector is 86.42cm2 to 2 decimal places.

If you prefer to use the formula, you need to start by identifying the different variables. Remember, $\theta$ represents the angle of the sector. $\theta = 117^{\text{o}}$ $r = 9.2 \text{cm}$ \begin{aligned} A &= \frac{\theta}{360} \times \pi r^2 \\ &= \frac{117}{360} \times \pi \times 9.2^2 \\ &= 86.42 \text{cm}^2 &&(2\text{d.p.}) \end{aligned}

### Question 2: Simplifying Algebraic Fractions

To simplify a numerical fraction, we look for common factors. Then, we divide the numerator and denominator by the common factors.

We apply the same concept to algebraic fractions. We look for common factors in the numerator and denominator and divide by these factors. To find factors in algebraic expressions, we often have to factorise, although we can sometimes use index laws.

Let’s simplify this fraction first: \begin{aligned} \boldsymbol{\frac{15x^7y^3}{12x^2y}} \end{aligned}

On questions like this, look at the coefficients first then consider each variable (letter) in turn.

15 and 12 have a common factor of 3. $x^7$ and $x^2$ have a common factor of $x^2$. $y^3$ and $y$ have a common factor of $y$.

So, we divide the numerator and denominator by 3, $x^2$ and $y$ (or $3x^2y$). Remember, when we divide indices, we subtract the powers. \begin{aligned} \frac{15x^7y^3}{12x^2y} &= \boldsymbol{\frac{5x^5y^2}{4}} \end{aligned}

Let’s simplify this fraction: \begin{aligned} \boldsymbol{\frac{x^2+8x+15}{2x+10}} \end{aligned}

When we have a fraction with multiple terms, we need to start by factorising the numerator and denominator.

For a reminder on how to factorise quadratics, take a look back at week 2.

To factorise the numerator, we want to find a pair of numbers that multiply together to give 15 and add to give 8. In this case, the numbers we want are 3 and 5. $x^2+8x+15 = (x+3)(x+5)$

For the denominator, we take out the largest common factor (which is 2). $2x+10 = 2(x+5)$

Putting these together gives: \begin{aligned} \frac{x^2+8x+15}{2x+10} &= \frac{(x+3)(x+5)}{2(x+5)} \end{aligned}

Now, we can see that the numerator and denominator have a common factor of $(x + 5)$.so we can divide both by this expression: \begin{aligned} \frac{(x+3)(x+5)}{2(x+5)} &= \boldsymbol{\frac{x+3}{2}} \end{aligned}

Vectors can be added using diagrams or just numerically. Remember, the top part of the vector gives information about how far to the right a vector goes. The lower part of the vector is how far up the vector goes.

Let’s find: $\huge{\boldsymbol{2\begin{pmatrix} 3\\\text{-}1 \end{pmatrix} + \begin{pmatrix} \text{-}7\\\text{-}1 \end{pmatrix}}}$

To do this question numerically, we simply multiply the first vector by two, then add the second vector. We treat the top numbers and bottom numbers completely separately. \begin{aligned} 2\begin{pmatrix} 3\\\text{-}1 \end{pmatrix} + \begin{pmatrix} \text{-}7\\\text{-}1 \end{pmatrix} &= \begin{pmatrix} 6\\\text{-}2 \end{pmatrix} + \begin{pmatrix} \text{-}7\\\text{-}1 \end{pmatrix} \\ &=\begin{pmatrix} 6-7\\\text{-}2-1 \end{pmatrix} \\ &= \boldsymbol{\begin{pmatrix} \text{-}1\\\text{-}3 \end{pmatrix}} \end{aligned}

This method is quite abstract – you might find it easier to draw the vectors out. We’ll start by drawing the vectors on a grid. The first vector is 3 units to the right and 1 unit down:

We need two of the first vector – one after the other. Then, we need the second vector. This will be 7 units to the left and 1 unit down.

The resultant vector (the answer to our addition) is the vector from our starting point to the finish point. It is drawn below in colour.

This is the vector: $\boldsymbol{\begin{pmatrix} \text{-}1\\\text{-}3 \end{pmatrix}}$

### Question 4: The Area of a Triangle

The area of a triangle can be found my multiplying the base by the height and dividing by 2. However, this requires us to know the perpendicular height of the triangle.

If we don’t know this measurement, we can use a slightly different formula to find the area. Consider the triangle below:

Any triangle can be labelled like this. The lowercase letters denote the sides and the uppercase letters denote the angles. Each angle is directly opposite the side it shares a letter with.

Once we’ve labelled a triangle like this, we can find the area using: $\text{area} = \frac{1}{2} ab\sin C$

The angle, $C$, should be enclosed by the two sides, $a$ and $b$.

Find the area of the triangle below. Give your answer correct to 2 decimal places.

Let’s start by labelling the sides and angles. It’s a good idea to label the angle we know as C so we can use the formula in the format it’s given.

Now, we can substitute $a$ = 15.7cm, $b$ = 18.2cm and $C$ = 78° into the formula. \begin{aligned} \text{area} &= \frac{1}{2} ab\sin C \\ &= \frac{1}{2} \times 15.7 \times 18.2 \times \sin (78) \\ &= \boldsymbol{139.7}\text{cm}^2 \text{ (1d.p.)}\end{aligned}

### Question 5: Venn Diagram Notation

Venn diagram notation is all about knowing the symbols. $\boldsymbol{\cup}$ means union and can be read as “or”.

Outside of maths, a union is the act of joining together. In Venn diagrams, it means exactly the same thing: the union of A and B is everything that is in A or in B. It’s the two sets joined together. The green section below shows the set A $\boldsymbol{\cup}$ B: $\boldsymbol{\cap}$ means intersect and can be read as “and”.

Similarly, intersect means two or more thing that lie across each other. In Venn diagrams, it is the elements that lie across both sets. More visually, it is the point where the two sets cross. The green section below shows the set A $\boldsymbol{\cap}$ B:

An apostrophe (‘) after a set means the complement of the set – this is not the same as a compliment! Two things complement each other if they go well together or make a complete group. It’s this second definition that’s relevant here: the complement of a set is everything that isn’t in that set. The green section below is the set B’:

We can combine these 3 symbols to define any part of a Venn diagram.

Shade the region represented by A’ $\boldsymbol{\cup}$ B.

Start by marking each part of the union. In the diagram below, the areas that represent A’ and the areas that represent B are marked as such:

Since we are looking for the union, we shade any region with a label in as these will either be in A’ or in B (or both).

Shade the region represented by A’ $\boldsymbol{\cap}$ B.

In this case, we start by labelling the areas as before. However, we only shade the region that is labelled both A’ and B:

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