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Old 17th May 2012, 05:16 PM   #1 (permalink)
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Maths quiz help

Quiz set by my college.
I named some variable but still couldn't get the answer.
I am thinking using Pythagoras theorem as well as sin rule.
But everything is too messy. :/


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Old 17th May 2012, 06:16 PM   #2 (permalink)
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Re: Maths quiz help

Originally Posted by .Memo View Post
Quiz set by my college.
I named some variable but still couldn't get the answer.
I am thinking using Pythagoras theorem as well as sin rule.
But everything is too messy. :/

Answer is 8.

The whole area is fixed at 20. However, the base and the height of the triangles are not fixed.

So one of the possible dimension is that if the rectangle has length = 5 and breadth = 4, you will realise that the bottom length BE: EC ratio is 2:3 which is not like what is drawn in the question, yet it can fulfill all the given condition.

I may not have the best solution but here is how i calculate the total area is 20.
Let the length to be l, height to be h, DF to be a and BE to be b.
Hence we have,
al = 10 --------(1)
hb = 8 --------(2)
(h-a)(l-b) = 6 --------(3)
Expand (3),
hl - al - hb + ab = 6
hl - 10 - 8 + ab = 6 (Sub (2) and (3))
hl + ab = 24
Consider (1) x (2), we have
hl x ab = 80
ab = 80 / hl --------(4)
Sub (4) into hl + ab = 24,
hl + 80 / hl = 24
hl^2 - 24hl + 80 = 0
(hl -4)(hl-20) = 0
hl = 20 (reject hl = 4)
Since hl is the area of the rectangle, area of rectangle = 20.

Btw it is a nice question.


Last edited by namelessname; 17th May 2012 at 06:19 PM.
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Old 18th May 2012, 08:32 PM   #3 (permalink)
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Re: Maths quiz help

Thank you! Hope I'm the first 3 tho! (:
I'm hungry! Hahas!

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