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 April 6th, 2011, 03:04 PM #1 Member   Joined: Feb 2011 Posts: 61 Thanks: 0 Help with Geometry Problem! PLease help.
 April 6th, 2011, 03:51 PM #2 Senior Member   Joined: Nov 2010 Posts: 502 Thanks: 0 Re: Help with Geometry Problem! Well, we know the top triangle is similar to the equilateral triangle, so we know all the angles in the figure. We also have that all the sides of a square are the same... Post your progress and we can recommend paths to take.
 April 6th, 2011, 03:53 PM #3 Member   Joined: Feb 2011 Posts: 61 Thanks: 0 Re: Help with Geometry Problem! Well, the thing is, I have no clue how to figure out one side of the square! If I could have that then the problem would be solved. I tried cutting the triangle in half and finding the height, but that doesn't help me, since I need to limit myself to finding something about the square.
 April 6th, 2011, 03:56 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 522 Math Focus: Calculus/ODEs Re: Help with Geometry Problem! Let t be the side length of the triangle and s be the side length of the square. Drop a vertical from vertex B to the top of the square and label the length of this line segment $h_1$. Thus the height s of the square is:  Let x be the distance from vertex A to the point of intersection of the square on side AB. By similarity, we have: $\frac{x}{s}=\frac{s}{h_1}\:\therefore\:h_1=\frac{s ^2}{x}=\frac{s^2}{t-s}$ This gives:  Multiply through by t - s: $st-s^2=\frac{\sqrt{3}}{2}t(t-s)-s^2$ $s$$\frac{2+\sqrt{3}}{2}$$=\frac{\sqrt{3}}{2}t$ $s$$2+\sqrt{3}$$=\sqrt{3}t$  Then the area A of the square is given by: $A=s^2=\frac{3}{7+4\sqrt{3}}t^2$ With t = 10 in., this gives, to the nearest tenth: $A\approx21.5\text{ in^2}$
 April 6th, 2011, 05:52 PM #5 Member   Joined: Feb 2011 Posts: 61 Thanks: 0 Re: Help with Geometry Problem! Thanks MarkFL.
April 6th, 2011, 09:54 PM   #6
Math Team

Joined: Dec 2006
From: Lexington, MA

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Re: Help with Geometry Problem!

Hello, rokr32!

Another approach . . .

Quote:
 Equilateral triangle with sides of 10 inches, Find the area of the inscribed square to nearest tenth. Code:  A * / \ x / \ x / \ D *-------* E /| x |\ / |x x| \ 10-x / | | \ B *---*-------*---* C y F x G y

$\text{We have: }\:AB \,=\,BC\,=\,CA \,=\,10.$

$\text{Let the side of the square be }x.$

$\text{Since }\Delta ADE\text{ is equilateral: }\:AD\,=\,AE\,=\,x.$
[color=beige]. . [/color]$\text{Then: }\:EC \:=\:10-x$

$\text{Let }BF \,=\,GC\,=\,y.$
$\text{Then: }\:x\,+\,2y \:=\:10\;\;\;\Rightarrow\;\;\;y \:=\:\frac{10-x}{2}$

$\text{In right triangle }EGC:\;\;x^2\,+\,y^2\:=\:EC^2 \;\;\;\Rightarrow\;\;\;x^2\,+\,\left(\frac{10-x}{2}\right)^2 \:=\:(10-x)^2$

$\text{Solve for }x\text{ and we get: }\:x \:=\:\frac{10\sqrt{3}}{2\,+\,\sqrt{3}} \:=\:10(2\sqrt{3}\,-\,3)$

$\text{The area of the square is: }\:x^2 \:=\:\left[10(2\sqrt{3}\,-\,3)\right]^2 \;\approx\; 21.5\text{ in}^2$

 April 7th, 2011, 04:32 AM #7 Global Moderator   Joined: Dec 2006 Posts: 21,105 Thanks: 2324 In [color=#00AA00]MarkFL[/color]'s notation, $h_{\small1}\,=\,\frac{\sqrt{\small3}}{\small2}s,$ so $s\,=\,\frac{\sqrt{\small3}}{\small2}t\,-\,\frac{\sqrt{\small3}}{\small2}s.$ Rearranging gives $s\,=\,\frac{\sqrt{\small3}}{\small2\,+\,\sqrt{\sma ll3}}t\,=\,\small(2\sqrt{\small3}\,-\,3)\large t,$ etc.
 April 7th, 2011, 03:54 PM #8 Member   Joined: Feb 2011 Posts: 61 Thanks: 0 Re: Help with Geometry Problem! Wow thanks for the different approach soroban, it is easier to understand. I just had the test today and got a 93 on this stuff. Not too happy about it, but also not bummed. Thanks to you guys

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