+0  
 
0
1043
2
avatar

A square is inscribed in a right triangle, as shown below. The legs of the triangle are 2 and 3. Find the side length of the square.

 Mar 30, 2020
 #1
avatar+663 
+1

Hello Guest! Make an account! cheeky

Hypotenuse of big triangle: \(\sqrt{13}\)

 

This is just a compilation of a bunch of equations!

By Pythagorean Theorem:

(1) \((2-b)^2+(3-d)^2=b^2-a^2\)

(2) \(b^2-a^2=d^2-c^2\)

(3) \(d^2-c^2=\sqrt{13}-a-c\)

 

By similarity where \(s\) is the side length of the square:

(4)  \(\frac{a}{b}=\frac{d}{s}\)

(5) \(\frac{2}{3}=\frac{2-b}{3-d}\)

(6) \(\frac{2}{3}=\frac{s}{c}\)

(7) \(\frac{2}{3}=\frac{a}{s}\)

 

General:

(8) \(a+c+s=\sqrt{13}\)

 Mar 30, 2020
 #2
avatar+25595 
+2

A square is inscribed in a right triangle, as shown below.

The legs of the triangle are 2 and 3. 

Find the side length of the square.

 

 

\(\text{Let $A=(0,\ 0)$ } \\ \text{Let $B=(x_b,\ y_b)=(\sqrt{13},\ 0)$ } \\ \text{Let $C=(x_c,\ y_c)=\Big(2\cos(A),\ 2\sin(A)\Big)$ } \\ \text{Let $ \cos(A)=\dfrac{2}{\sqrt{13}},\ \sin(A)=\dfrac{3}{\sqrt{13}}$ } \)

 

\(\begin{array}{|lrcll|} \hline & \tan{\varphi} &=& \dfrac{2}{2+3} \\\\ & \tan{\varphi} &=& \dfrac{2}{5} \\\\ \text{line }1: & y &=& \tan{\varphi}*x \\\\ & \mathbf{y} &=& \mathbf{\dfrac{2}{5}*x} \\ \\ \hline \text{line }2: & \dfrac{y-y_b}{x-x_b} &=& \dfrac{y_c-y_b}{x_c-x_b} \\\\ &&& \boxed{x_b=\sqrt{13},\ y_b = 0,\quad x_c=\dfrac{4}{\sqrt{13}},\ y_c = \dfrac{6}{\sqrt{13}} } \\\\ & \dfrac{y}{x-\sqrt{13}} &=& \dfrac{\dfrac{6}{\sqrt{13}}}{\dfrac{4}{\sqrt{13}}-\sqrt{13}} \\\\ & \dfrac{y}{x-\sqrt{13}} &=& \dfrac{6}{4-13} \\\\ & \dfrac{y}{x-\sqrt{13}} &=& -\dfrac{2}{3} \\\\ & \mathbf{y} &=& \mathbf{-\dfrac{2}{3}(x-\sqrt{13})} \\ \hline \end{array}\)

 

The intersection point of both lines:

\(\begin{array}{|rcll|} \hline y= \dfrac{2}{5}*x &=& -\dfrac{2}{3}(x-\sqrt{13}) \\\\ \dfrac{2}{5}*x &=& -\dfrac{2}{3}(x-\sqrt{13}) \\\\ \dfrac{9}{10}*x &=& -(x-\sqrt{13}) \\\\ \dfrac{9}{10}*x &=& -x+\sqrt{13} \\\\ x+ \dfrac{9}{10}*x &=& \sqrt{13} \\\\ \dfrac{19}{10}*x &=& \sqrt{13} \\\\ x &=& \dfrac{10\sqrt{13}}{19} \\\\ y = s &=& \dfrac{3}{5}x \\\\ s &=& \dfrac{3}{5}*\dfrac{10\sqrt{13}}{19} \\\\ s &=& \dfrac{6}{19} \sqrt{13} \\ \mathbf{s} &=& \mathbf{1.13859513962} \\ \hline \end{array}\)

 

The side length of the square is \(\mathbf{\dfrac{6}{19} \sqrt{13} = 1.13859513962}\)

 

laugh

 Mar 30, 2020

10 Online Users