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A square is inscribed in a right triangle, as shown below. The legs of the triangle are 1 and 3. Find the area of the square.

 

 Feb 26, 2024
 #1
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The answer is 90/169

 Feb 26, 2024
 #2
avatar+14944 
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Find the area of the square.

Hello magenta!

 

\(1:3=x:(3-x)\\ 3x=3-x\\ 4x=3\\ x=\frac{3}{4}\\ \color{blue}A=\frac{3}{4}\cdot \frac{3}{4}=\frac{9}{16}\)

 

laugh  !

 Feb 26, 2024
edited by asinus  Feb 27, 2024
 #3
avatar+1622 
+2

Through construction, AB = AF + FB = 3.

AF = EF = x. Since ADEF is a square, DE is parallel to AB, and thus angle CED is congruent to angle CBA by corresponding angles, and ADE = angle EFB because they are both 90 degrees. Thus, triangle CDE is similar to EFB is similar to CAB, and CA : AB = 1 : 3, likewise, FE : FB = 1 : 3. Substituting, we have x : FB = 1 : 3, and therefore FB = 3x. Substituting this into the original equation, AB = x + 3x = 4x = 3. And thus, x = 3/4. The area of the square would be x^2 = 9/16.

 Feb 26, 2024
 #4
avatar+394 
+1

Thanks, I can confirm this is the correct answer, admist this confusion. laugh

We can verify proyaop's answer through similar triangles \(\triangle BEF \sim\triangle ECD\), so to check we make sure \(\frac{BF}{EF}=\frac{ED}{CD}\), by proyaop's answer, \(BF=3-\frac{3}{4}=\frac{9}{4}\)\(EF=\frac{3}{4}\) , \(ED=\frac{3}{4}\)\(CD=1-\frac{3}{4}=\frac{1}{4}\), so \(\frac{\frac{9}{4}}{\frac{3}{4}}=\frac{\frac{3}{4}}{\frac{1}{4}}=3\).

Asinus did nothing majorly wrong, i think they misread the side lengths as 1 and 2 smiley.  

hairyberry  Feb 27, 2024

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