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Geometry

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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
-1

Feb 26, 2024
#2
+14944
+1

Find the area of the square.

Hello magenta!

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

!

Feb 26, 2024
edited by asinus  Feb 27, 2024
#3
+1622
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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
+394
+1

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

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 .

hairyberry  Feb 27, 2024