Help in b) iii) only thanks :D

- anonymous

Help in b) iii) only thanks :D

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- chestercat

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- anonymous

http://screencast.com/t/Q0eyWYsQ

- anonymous

@nbouscal @satellite73 @TuringTest

- anonymous

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## More answers

- anonymous

@jim_thompson5910 @dpaInc @KingGeorge @AccessDenied @SmoothMath

- KingGeorge

I'm pretty sure I have the answer, but I'm not sure yet. What lengths did you get for KM and KN?

- anonymous

for KM i got 27 and KN i got 15

- KingGeorge

Give me a minute to see if I was correct.

- anonymous

okay

- KingGeorge

So I have the answer, but I don't think I did it a very good way. Basically, I found the area of KLM using herons formula and the area of LNM using heron's formula.
You also know using some alternating angles of the parallel lines, that triangle LNM is twice the area of triangle NPM, so that let's you find the area of NPM, and then you subtract that off from the area of KLM to find the area of the trapezium KLPN

- anonymous

what iis herons formula? can u show the woking?

- KingGeorge

Heron's formula is a way to find the area of any triangle if you know the side lengths. If you want the general formula, look here: http://en.wikipedia.org/wiki/Heron's_formula
Otherwise, I'll just skip the more tedious work and show you how I got the solution.

- KingGeorge

(I'm skipping the simplification here, but you should check it on your own)
Using that, we get that the area of KLM is \[90\sqrt{2}\]and the area of LNM is \[40\sqrt2\]

- KingGeorge

Thus, the area of NPM is \[\frac{40\sqrt2}{2}=20\sqrt2\]Now, we just put all these numbers in the ratio \[\frac{20\sqrt2}{90\sqrt2-20\sqrt2}=\frac{\sqrt2}{\sqrt2}\cdot\frac{20}{70}=\frac{2}{7}\]And there's the solution.

- anonymous

but the solution actually is 16/65...according to the markin gsolution...hmm

- KingGeorge

It might be a typo since \[\frac{2}{7}=\frac{16}{56}\]So the two possibilities are that there's a typo, or your lengths for KM and KN were incorrect.

- KingGeorge

Could you briefly explain to me how you got the length of KN?

- anonymous

i used similarity triangle method...like il show the working...first i found
KM
LK/LN=KM/LM
15/10=KM/18
KM=27
then i found
NM=
LK/LM=LM/NM
15/10=18/NM
NM=12
then KN= KM-NM
27-12= 15cm :)

- KingGeorge

Definitely no mistakes there. That means that I'm pretty sure it's a typo in the book.

- anonymous

just check if ur answer is wrong, coz it has to be 16/65

- anonymous

okay can u help in another one..its algebra?

- anonymous

\[\frac{x^{2}}{5} + \frac{5}{x} = \frac{1}{2}x +3\]
can u equal them so that one side will have 0

- KingGeorge

Sure, give me a few more seconds to finish checking my first solution. I came up with another, far easier way to find the ration.

- KingGeorge

I might have found a better solution... It's not working out as well as I had anticipated.
As for the algebra question, multiply everything by \(x\). You'll get a cubic where you can easily move everything to one side, and hopefully be able to factor.

- anonymous

can u do it, im stuggling like showing a working

- KingGeorge

It might even be better to multiply the thing by 10x to get rid of the 5 and the 2 in the denominators as well. That would get you \[2x^3+50=5x^2+30x\]Move everything to one side, \[2x^3-5x^2-30x+50=0\]

- KingGeorge

Unfortunately, that doesn't seem to be factoring very easily.
And you were right to call me out on my previous solution to the triangle question. I don't think I'm right.

- anonymous

well how did u get 2x^3 + 50?

- anonymous

any other methods? to do that algebra?

- KingGeorge

\[x(\frac{x^{2}}{5} + \frac{5}{x}) = x(\frac{1}{2}x +3)\]Distribute the x\[(\frac{x^{3}}{5} + 5) = (\frac{1}{2}x^2 +3x)\]Now mulitply everything by 10 to get \[(\frac{10x^{3}}{5} + 50) = (\frac{10x^2}{2} +30x)\]Simplify things, \[2x^3 + 50 = 5x^2 +30x\]and move everything to one side.

- anonymous

any other easy method, because i can remembr once i did it in an easy way?

- anonymous

it is just 1 mark question?

- KingGeorge

I'm not sure I understand. Do you just want to set one side equal to 0 and solve for x on the other side?

- anonymous

yep?

- anonymous

Hallo?

- KingGeorge

I'm still here. So if we set one side equal to 0, we want to find solutions to the equations \[\frac{x^2}{5}+\frac{5}{x}=0\]and \[0=\frac{x}{2}+3\] correct?

- anonymous

no i mean, both the equations should be simplified and be equal = 0

- KingGeorge

Can we come back to that in a minute, I just figured out the easy way to do the triangle problem, and the book's correct. (It really shouldn't have taken me this long).

- anonymous

okay

- KingGeorge

Since NP is parallel to KL, you can show that triangle NPM is similar to triangle KLM. You know that angle K is the same as angle MNP, by corresponding angles. Also, for the same reason, angle KLM is the same as angle NPM, so this means that the triangles are similar.
Now let the area of triangle NPM=A. The ratio between the sidelengths of NPM and KLM is exactly 9/4. Since the formula for area of a triangle is 1/2 bh the area vary as a square, so we square it to get that the area of KLM=81/16 A.

- KingGeorge

The area of the trapezium is the area of KLM minus the area of NPM. So that area is given by \[\frac{81A}{16}-A=\frac{65A}{16}\]Now we take A over that, and we get \[\frac{A}{\frac{65A}{16}}=\frac{16A}{65A}=\frac{16}{65}\]is the ratio.

- anonymous

how did u get 65/16 in the first place

- KingGeorge

\[\frac{81A}{16}-A=\frac{81A}{16}-\frac{16A}{16}=\frac{81A-16A}{16}=\frac{A(81-16)}{16}=\frac{65A}{16}\]

- KingGeorge

And I got the 81/16 because the ratio of the side lengths is 9/4, and the ratio of the areas (of the triangles) is the ration of the side lengths squared. So \[\left(\frac{9}{4}\right)^2=\frac{81}{16}\]

- anonymous

how did u get 9/4?

- KingGeorge

\[\frac{27}{12}=\frac{9}{4}\]I just simplified the ratio between KM and NM

- anonymous

oh okay thats the point..

- KingGeorge

Yup. I should have seen that earlier :(

- anonymous

alright thanks alot man, for ur time...ur a great teacher :)

- KingGeorge

thanks for the compliment :)

- KingGeorge

I've got to go, so if you still need help with the algebra question, just post it in another question, and hopefully some one will answer it. Good luck!

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