finding lengths of this shape :/

**miranlol** · June 2nd 2009, 06:33 AM

heya everyone

ImageShack - Image Hosting :: problemfbp.jpg

have been puzzling at this for over an hour now :/ tried using simutaneous equations

$20^2 = x^2 + (a+b)^2$ - pithag..
$30^2 = y^2 + (a+b)^2$ - pithag
$x / 8 = (a + b) / b$ - height/length ratios
$y / 8 = (a + b) / a$ - height/length ratios...

where a and b is D split up into the smaller triangles...

but solving these seems very tedious..anyone that can help?

thanks in advance.

oh btw the problem is solve X Y and D

ps. the 8 is the length of the blue line. sorry its a bit sketchy
edit2: math wrapped it lol.

**Amer** · June 3rd 2009, 12:56 AM

this is your picture

finding lengths of this shape :/-good.jpg

- pithag..

- pithag

- height/length ratios

- height/length ratios...

where a and b is D split up into the smaller triangles...

OK let d=a+b ..so b=d-a ok

now you have

$400=x^2+d^2$ .......(1)

$900=y^2+d^2$ ....(2)

$\frac{x}{8}=\frac{d}{d-a}$ ...(3)

$\frac{y}{8}=\frac{d}{a}$ ....(4)

and from (3)

$a=\frac{d(x-8)}{x}$

sub this in equation (4)

$\frac{y}{8}=\frac{d}{\dfrac{d(x-8)}{x}}$

$y(d(x-8))=8dx$

$y=\frac{8x}{x-8}$

you continue it is now obvious right

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