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2002 Paper 2 Question 10


Posted 17 May 2006 - 07:17 PM

How do you do part A in this question?

Mr H

Posted 17 May 2006 - 08:06 PM

I don't have access to that paper at the moment to give you my solution so here's someone else's:



Posted 17 May 2006 - 08:12 PM

i really didnt get that question at all...im surprised it hasnt been asked before :S


Posted 17 May 2006 - 08:19 PM

I'm still puzzled by it.

Why l/a?

when you can could you talk me through solving it?

ad absurdum

Posted 17 May 2006 - 11:21 PM

Hmm there's apparently another way to do it with similar shapes and the like, but I prefer to use stuff that's fresh in my memory.

First, find the gradient of the longest side (from (0,6) to (8,0)):

\begin{align*}m &= \frac{y_2-y_1}{x_2-x_1} \\
&= \frac{0-6}{8-0} \\
&= \frac{-3}{4} \\\end{align*}

Now the other side l is parallel to this, so it has the same gradient, and it also touches the point (a,0), so finding an equation for l:

\begin{align*}y-b &= m(x-a) \\
y-0 &= \frac{-3}{4}(x-a) \\
4y &= -3x + 3a \\\end{align*}

Now find the y intercept, i.e. when x= 0:

\begin{align*}4y = 3a \\
y = \frac{3a}{4} \\ \end{align*}

So the y intercept is the point (0, 3a/4)

Now, we want the length of l, so distance formula between the point (0, 3a/4) and (a,0):

\begin{align*}d &= \sqrt{(x_2-x_1)^2 - (y_2-y_1)^2} \\
&= \sqrt{(a-0)^2 - (0-\frac{3a}{4})^2} \\
&= \sqrt{a^2 + \frac{9a^2}{16}} \\
&= \sqrt{\frac{16a^2}{16} + \frac{9a^2}{16}} \\
&= \sqrt{\frac{25a^2}{16}} \\
&= \frac{5a}{4} \\\end{align*}

tricky question.


Posted 18 May 2006 - 12:17 AM

i like the similar triangles option and its one i would of went for when i did that question as a past paper since its only for 2 marks at most this question


Posted 18 May 2006 - 12:41 AM

The way ad_absurdum did it was the way i was going for, that way makes alot more sense to me than the similar triangles one.

Mr H

Posted 18 May 2006 - 08:00 AM

I would also solve part (a) using similar triangles.

Here is my solution if it helps:


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