# Chapter 36 Assorted problems

Solutions to problems which have yet to be categorised.

## 36.0.1 Cones problem

Problem: Two similar cones, $X$ and $Y$, have surface areas $270\text{cm}^{2}$ and $120\text{cm}^{2}$ respectively.

The volume of cone $X$ is $1215\text{cm}^{3}$.

Show that the volume of cone Y is $360\text{cm}^{3}$.

Solution:

We have two cones, let’s let $X$ be this one, with radius $r$ and height $h$

Then we can draw $Y$ (which is smaller than $X$, but not drawn to scale here). As they are similar cones, the radius of $Y$ is equal to $ar$ (where $a$ is the scale factor between the two cones) and the height of $Y$ is equal to $ah$

First, the volume of a cone is $\frac{\pi r^{2}h}{3}$ and the surface area of a cone is $\pi r(r+\sqrt{h^{2}+r^{2}})$. These can be derived using integration (but not in Single Maths A Level at least).

Using this, we can write this equation relating the radius and height of $X$ to the value given in the question

$\pi r(r+\sqrt{h^{2}+r^{2}})=270$ (36.1)

We can also do the same for $Y$

$\pi ar(ar+\sqrt[]{(ah)^{2}+(ar)^{2}})=120$ (36.2)

This can be simplified a bit by factoring the $a^{2}$ from the square rooted part of the equation, giving

$\pi ar(ar+\sqrt{a^{2}}\sqrt[]{(h)^{2}+(r)^{2}})=120$ (36.3)

and thus that

$\pi a^{2}r(r+\sqrt{(h)^{2}+(r)^{2}})=120$ (36.4)

We can solve this for $a$ by dividing the two equations, giving that

$\frac{\pi a^{2}r(r+\sqrt{(h)^{2}+(r)^{2}})}{\pi r(r+\sqrt{h^{2}+r^{2}})}=\frac% {120}{270}$ (36.5)

and thus, after cancelling, that

$a^{2}=\frac{9}{4}$ (36.6)

which means that

$a=\frac{3}{2}$ (36.7)

Given this, we can then turn to the volume formula.

We know that the volume of $X$ is

$\pi r^{2}h=1215$ (36.8)

and the formula for the volume of $Y$ is equal to

$\pi(ar)^{2}(ah)=a^{3}\pi r^{2}h$ (36.9)

Thus, by multiplying Equation 36.8 through by $a^{3}$, we have

$a^{3}\pi r^{2}h=a^{3}1215$ (36.10)

The left-hand side is just the volume of $Y$, and the right-hand side is just

 $\displaystyle a^{3}1215$ $\displaystyle=\left(\frac{2}{3}\right)^{3}\cdot 1215$ (36.11) $\displaystyle=360$ (36.12)

Which is what we needed to show.

## 36.0.2 Divisibility of $abc_{10}$

###### Example 36.0.1
\say

When we represent 789 in decimal, we mean 7 hundreds plus 8 tens plus 9 units. If $abc_{10}$ is taken to represent a decimal number (not $a$ times $b$ times $c$ as in algebra), we mean the value $100a+10b+c$. Show that if $a+b+c$ is divisible by 9, then $abc_{10}$ is divisible by 9.

Solution: Anything that is divisible by 9 can be written as $9\cdot\text{something}$. As this is algebra, we can use a letter (e.g. $m$) to represent \saysomething. As we are assuming that $a+b+c$ is divisible by 9, we can write that

$a+b+c=9m$ (36.13)

Where $m$ is some number (which depends on the value of $a$, $b$ and $c$). Then we can consider $abc_{10}$. First, we can write this (as given in the question) as

$abc_{10}=100a+10b+c$ (36.14)

To show that this is divisible by $9$ we need to write it as $9\cdot\text{something}$. We need to apply the fact that $a+b+c=9m$ here (because we are trying to show that, given this assumption, $abc_{10}$ is divisible by $9$). We want to somehow extract an $a+b+c$, here. Currently, we have a $c$, which is great (because it’s what we’re after), and all we need is a $b$ and a $c$, which we can get by breaking $100a$ into $99a+a$ and $10b$ into $9b+b$. When we apply this to the expression as a whole, our desired result pops out fairly quickly.

 $\displaystyle abc_{10}$ $\displaystyle=100a+10b+c$ (36.15) $\displaystyle=99a+a+9b+b+c$ (36.16) $\displaystyle=9(11a+b)+a+b+c$ (36.17) $\displaystyle=9(11a+b)+9m$ (36.18) $\displaystyle=9(11a+b+m)$ (36.19)

Therefore, if $a+b+c=9m$ (i.e. $a+b+c$ is divisible by $9$) $abc_{10}$ is also divisible by $9$.

## 36.0.3 Rationalising

###### Example 36.0.2

Show that $a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$, and hence rationalise the denominator of

$\frac{1}{\sqrt{3}-\sqrt{2}}$ (36.20)

Solution

The first part is just algebraic manipulation, i.e.

 $\displaystyle(a-b)(a^{2}+ab+b^{2})$ $\displaystyle=a^{2}+a^{2}b+ab^{2}-ba^{2}-ab^{2}-b^{3}$ (36.21) $\displaystyle=a^{2}+(a^{2}b-ab^{2})+(ab^{2}-ab^{2})-b^{3}$ (36.22) $\displaystyle=a^{3}-b^{3}$ (36.23)

For the second part, we want to apply this identity to the expression to rationalise. We have an expression in the form

$\frac{1}{a-b}$ (36.24)

And using the identity (Equation 36.20), we know that

$a-b=\frac{a^{3}-b^{3}}{a^{2}+ab+b^{2}}$ (36.25)

Which implies (using the properties of reciprocals and fractions11 1 See Section 4.1.1 if you’re unsure about this) that

 $\displaystyle\frac{1}{a-b}$ $\displaystyle=\frac{a^{2}+ab+b^{2}}{a^{3}-b^{3}}$ (36.26)

Given that our $a=\sqrt{3}$ and $b=\sqrt{2}$, this is exactly what we want! As $a^{3}=3$ (which is rational) and $b^{3}=2$ which is rational, $a^{3}-b^{3}$ is also rational (specifically $a-b=1$). Therefore, we have that

$\frac{1}{\sqrt{3}-\sqrt{2}}=\frac{(\sqrt{3})^{2}+\sqrt{3}\sqrt{% 2}+(\sqrt{2})^{2}}{1}$ (36.27)