This week, we’re exploring:
Question 1: Cones
You are given the formula for both the curved surface area and volume of a cone. So, finding these values is mostly about substitution. Just make sure you are using the radius for \(r\), the slant height for \(l\) and the perpendicular height for \(h\).
\(\text{volume} = \frac{1}{3}\pi r^2h \)
\(\text{curved surface area} = \pi rl \)
Let’s find the volume and total surface area of this cone. We will give both answers in terms of 
The slant height is 13cm and the perpendicular height is 12cm. 10cm is the diameter but we want the radius, so we simply divide it by 2.
\( \begin{aligned} r &= 10 \div 2 \\ &= 5\text{cm} \\ h &= 12\text{cm} \\ l &= 13\text{cm} \end{aligned} \)
Volume
\( \begin{aligned} V &= \frac{1}{3}\pi r^2h \\ &= \frac{1}{3}\times pi \times 5^2 \times 12 \\ &= \boldsymbol{100\pi}\text{ cm}^2 \end{aligned} \)
Surface Area
We want to find the total surface area but the formula we are given only calculates the curved surface area. To find the total surface area, we need to add the base of the cone – a simple circle with the same radius as the cone.
\(\begin{aligned} \text{curved surface area} &= \pi r l \\ &= \pi \times 5 \times 13 \\ &= 65\pi \text{ cm}^2 \end{aligned} \)
\(\begin{aligned} \text{base of cone} &= \pi r^2 \\ &= \pi \times 5^2 \\ &= 25\pi \text { cm}^2\end{aligned} \)
\( \begin{aligned} \text{total surface area} &= 65\pi + 25\pi \\&= \boldsymbol{90\pi} \text{ cm}^2 \end{aligned} \)
Question 2: Multiplying Algebraic Fractions
Multiplying algebraic fractions is exactly the same process as multiplying numerical fractions. We simply multiply the numerators together, multiply the denominators together and then simplify.
Write \(\frac{x + 3}{x + 7} \times\frac{x – 3}{2x + 1}\) in the form \(\frac{x^2 – m}{ax^2 + bx + c}\) where 𝒂, 𝒃, 𝒄 and 𝒎 are integers to be found.
Let’s start by multiplying the numerators and the denominators.
\(\frac{x + 3}{x + 7} \times \frac{x – 3}{2x + 1} = \frac{(x + 3)(x – 3)}{(x + 7)(2x + 1)} \)
The question asks us to write the fraction in a specific form. To do this, we simply expand the brackets. It can help to consider the numerator and denominator separately:
\(\begin{aligned} (x + 3)(x – 3) &= x^2 + 3x – 3x – 9 \\ &= x^2 – 9\end{aligned} \)
\(\begin{aligned} (x + 7)(2x + 1) &= 2x^2 + 14x + x + 7 \\ &= 2x^2 + 15x + 7\end{aligned} \)
Putting this back in the fraction:
\(\frac{x + 3}{x + 7} \times \frac{x – 3}{2x + 1} = \frac{x^2 – 9}{2x^2 + 15x + 7} \)
Question 3: Translation
Translation keeps a shape the same size and orientation but simply changes its location. Usually, translations are given using vectors.
Translate the shape below by the vector \(\begin{pmatrix} -2 \\ 5 \end{pmatrix} \).
The top number on the vector tells us how far to move the shape horizontally. If the number’s positive, we move the shape to the right and if it’s negative, we move the shape to the left. The bottom number tells us how far to move the shape vertically. If the number’s positive, we move the shape up and if it’s negative, we move the shape down. In this case, we move the shape 2 units left and 5 units up.
Pick a vertex (corner) and count the number of squares we need. This will give us the new location of the vertex.
Now, we can draw in the rest of the shape. Make sure it’s exactly the same size.
If asked to describe a transformation, we use a very similar method.
Describe the transformation that maps shape A onto shape B.
As before, pick a vertex on shape A and count the squares to get to the corresponding vertex on shape B.
So, the shape has been translated 3 units to the right and 4 units down. We can also write this as a vector: \(\begin{pmatrix} 3 \\ -4 \end{pmatrix} \).
Question 4: Plotting Quadratics
When asked to draw a quadratic graph, you will usually be given a table to complete. Unlike linear graphs, these graphs won’t be straight lines. Instead, they will be a smooth curve called a parabola.
Let’s use the table below to draw the graph with equation \(y = x^2 + 2x – 3\).
We simply need to substitute each value of \(x\) into the equation \(y = x^2 + 2x – 3\). Be particularly careful with negative signs. If you are putting the values into a calculator, put brackets around any negative values.
\(\begin{aligned} x &= -3 & y&=(-3)^2+2\times(-3)-3 \\&&&=9-6-3\\&&&=0\end{aligned}\)
\(\begin{aligned} x &= 0 & y&=(0)^2+2\times(0)-3 \\&&&=0+0-3\\&&&=-3\end{aligned}\)
\(\begin{aligned} x &= 3 & y&=(3)^2+2\times(3)-3 \\&&&=9+6-3\\&&&=12\end{aligned}\)
Now, we can plot these points on a set of axes.
Finally, we can join the coordinates. Make sure you draw a smooth curve rather than lots of short, straight lines. This includes at the bottom of the graph – it continues to curve as it changes direction.
Question 5: Completing the Square
Completing the square can be quite tricky – especially if the quadratic is in the form \(ax^2 + bx + c\). Let’s start with a slightly easier expression.
Write \(x^2 + 5x – 2\) in the form \((x + a)^2 + b\).
First, we halve the coefficient of \(x\). In this case, that’s 5. This gives us the value of \(a\): \((x + \frac{5}{2})^2\). Let’s expand this to see the expression:
\(\begin{aligned} \left(x + \frac{5}{2}\right)^2 &= \left(x + \frac{5}{2}\right)\left(x + \frac{5}{2}\right)\\ &= x^2 +\frac{5}{2}x+\frac{5}{2}x+ \frac{25}{4}\\ &= x^2 + 5x + \frac{25}{4}\end{aligned}\)
This is similar to the expression we started with. In fact, the first two terms are identical. The value of \(b\) will adjust this expression to get the expression we started with. To find \(b\), we will subtract \(\frac{25}{4}\) from \(\left(x + \frac{5}{2}\right)^2\). We should also include \(– 2\) at this point.
\(\begin{aligned} x^2 + 5x – 2 &= \left(x + \frac{5}{2}\right)^2-\frac{25}{4} – 2 \\ &= \left(x + \frac{5}{2}\right)^2-\frac{33}{4}\end{aligned}\)
Write \(2x^2 – 20x + 7\) in the form \(p(x + q)^2 + r\) where \(p\), \(q\) and \(r\) are integers to be found.
We will start by factorising out 2 from the expression before continuing as before:
\(\begin{aligned} 2x^2 – 20x + 7 &= 2\left[x^2-10x+\frac{7}{2}\right] \\ &= 2\left[\left(x-5\right)^2 -25 +\frac{7}{2} \right] \\&= 2\left[\left(x-5\right)^2 -\frac{43}{2} \right] \\ &= 2\left(x-5\right)^2 -43\end{aligned}\)
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