Math Help Forum

Math Help Forum Feed Site Feed

Go Back   Math Help Forum > Pre-University Math Help > Geometry
Reply
 
Thread Tools Display Modes
  #1  
Old December 2nd, 2008, 10:13 PM
realintegerz's Avatar
Member
 
Join Date: Aug 2008
Posts: 109
Country:
Thanks: 42
Thanked 0 Times in 0 Posts
realintegerz is on a distinguished road
Default Square inscribed in triangle

A square is inscribed in a right triangle whose short sides are in the ratio of 1:2. What is the length of the side of the square in terms of the length of the shortest side of the circumscribed triangle?

I tried assuming that the shortest sides were 1 and 2, which makes the hypotenuse sq. root 5

Then I solved for the area of the whole triangle and I got 1

So then I made 1 = (1/2)(1-w)(2-w)

But for some reason I got w = 3 or 0 which doesn't work out...
Reply With Quote
Advertisement
 
  #2  
Old December 2nd, 2008, 10:35 PM
Newbie
 
Join Date: Dec 2008
Posts: 1
Country:
Thanks: 0
Thanked 0 Times in 0 Posts
dtmcnamara is on a distinguished road
Default

i may be making up some BS but does this look correct/true?

Reply With Quote
  #3  
Old December 3rd, 2008, 03:46 AM
Newbie
 
Join Date: Nov 2008
Posts: 4
Country:
Thanks: 0
Thanked 2 Times in 2 Posts
Shoreline is on a distinguished road
Default

Quote:
Originally Posted by realintegerz View Post
A square is inscribed in a right triangle whose short sides are in the ratio of 1:2. What is the length of the side of the square in terms of the length of the shortest side of the circumscribed triangle?

I tried assuming that the shortest sides were 1 and 2, which makes the hypotenuse sq. root 5

Then I solved for the area of the whole triangle and I got 1

So then I made 1 = (1/2)(1-w)(2-w)

But for some reason I got w = 3 or 0 which doesn't work out...

I will try to explain my solution without a picture... I somehow cannot upload a picture from my school...

Draw a right triangle, B = 90 degrees AB = 1 BC = 2
Draw inside the triangle a square label the square EBFG with E on AB en F on BC and G on AC

EB = BF = x
Then AE = 1 - x and FC = 2 - x

You've got two similar triangles AEG and GFC there is a ratio between the two triangles and because of that

AE : GF = EG : FC

but also AE x FC = GF x EG and
(1 - x)(2 - x ) = x*x
2 - 3x + x^2 = x^2
2 - 3x = 0
and x = 2/3

so we get the ratio 2/3 : 1 : 2
[in my school we have to multiply by 3 ..... 2 : 3 : 6]

Last edited by Shoreline; December 3rd, 2008 at 04:38 AM.
Reply With Quote
Reply

Thread Tools
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are Off
Pingbacks are Off
Refbacks are Off
Forum Jump


All times are GMT -7. The time now is 10:25 PM.


Powered by vBulletin® Version 3.7.3
Copyright ©2000 - 2009, Jelsoft Enterprises Ltd.
SEO by vBSEO 3.2.0 ©2008, Crawlability, Inc.
©2005 - 2009 Math Help Forum


Math Help Forum is a community of maths forums with an emphasis on maths help in all levels of mathematics.
Register to post your math questions or just hang out and try some of our math games or visit the arcade.