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# 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.

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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

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Hello Guest! Make an account!  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
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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 $$\mathbf{\dfrac{6}{19} \sqrt{13} = 1.13859513962}$$ Mar 30, 2020