# Video: AQA GCSE Mathematics Higher Tier Pack 5 β’ Paper 3 β’ Question 13

Consider the quadrilateral in the diagram. Given that π΄π΅ : π΅πΆ = 3 : 2, Show that π·πΆ : π΅πΆ = 1 : 2.

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### Video Transcript

Consider the quadrilateral in the diagram. Given that the ratio of π΄π΅ to π΅πΆ is three to two. Show that the ratio of π·πΆ to π΅πΆ is one to two.

The lengths of three of the sides of this quadrilateral are given as expressions involving the variable π₯. Itβs likely that in order to answer this question, weβre going to need to work out the value of π₯, which means weβre going to need to set up some equations.

Weβre told in the question that the ratio of the side π΄π΅ to the side π΅πΆ is three to two. We can convert this information from a ratio into an equation involving fractions by dividing one side of each ratio by the other. If we divide the left of the top ratio by the right, we have π΄π΅ over π΅πΆ. And this will be the same as if we divide the left-hand side of the second ratio by the right-hand side, which will give three over two. So we have the equation π΄π΅ over π΅πΆ equals three over two.

We can multiply both sides of this equation by two in order to eliminate the denominator on the right-hand side. And it gives two π΄π΅ over π΅πΆ equals three. We can also multiply both sides of this equation by π΅πΆ to eliminate the denominator on the left-hand side. And this will give two π΄π΅ equals three π΅πΆ. We could also have achieved this in one step by cross multiplying.

Now remember weβve been given expressions for both π΄π΅ and π΅πΆ in terms of π₯. π΄π΅ is three π₯ plus three and π΅πΆ is four π₯ minus two. So we can substitute these expressions into our equation. And we have two multiplied by three π₯ plus three is equal to three multiplied by four π₯ minus two. This is now an equation that we can solve in order to find the value of π₯.

First, we expand the bracket on the left-hand side. Two multiplied by three π₯ is six π₯ and two multiplied by three is six. Then, we expand the bracket on the right-hand side. Three multiplied by four π₯ is 12π₯ and three multiplied by negative two is negative six. Notice that we still have terms involving π₯ on both sides of the equation. So to solve it, we need to collect all of these like terms on the same side. Weβll collect them on the right of the equation because we currently have a higher number of π₯s on this side, which means that weβll end up with a positive number of π₯s when we collect.

To do so, we need to eliminate the π₯s on the left-hand side which we do by subtracting six π₯ from each side of this equation. On the left-hand side, weβre just left with six. And on the right-hand side, 12π₯ minus six π₯ is six π₯ and we bring down the negative six. Remember weβre trying to solve for π₯. So we need to leave π₯ on its own on the right-hand side. The next step is to add six to each side of this equation, giving 12 equals six π₯. And finally, we can divide both sides of this equation by six to give two equals π₯.

So we found the value of π₯: π₯ is equal to two. We can now substitute this value of π₯ into the expressions we were given for each side of the quadrilateral. π·πΆ was five π₯ minus seven. So thatβs five multiplied by two which is 10 minus seven which gives three. π΅πΆ is four π₯ minus two. Thatβs four multiplied by two which is eight minus two which gives six. And finally, π΄π΅ is three π₯ plus three. Thatβs three multiplied by two which is six plus three which is equal to nine.

We can perform a quick check of our work so far. Weβve worked out that π΄π΅ was nine units and π΅πΆ was six units. So we have that the ratio of π΄π΅ to π΅πΆ is nine to six. By dividing both sides of this ratio by three, it simplifies to three to two. So our lengths for π΄π΅ and π΅πΆ are in the correct ratio.

We were asked to show though that the lengths π·πΆ and π΅πΆ are in the ratio one to two, which in practical terms means that π΅πΆ is twice as long as π·πΆ. From our diagram, we found that the length of π·πΆ was three and the length of π΅πΆ was six. So our ratio is three to six. By dividing both sides of this ratio by three, we can see that it does indeed simplify to one to two as required.