Cut Right Triangle Problem

So, the other day, i was programming something and got a geometry doubt. I apologize for the bad drawing. But anyway, this is a 90-45-45 triangle with legs of 1 unit. It is cut from the 45° point to the exact middle of each of the legs. The red lines are perpendicular to the legs and extends till the meeting point of both the cuts. What is the length of the red lines (only one of them as they form a square) ?

(diagram not to scale)

Note by Akshaj Gopalakrishnan
7 months, 3 weeks ago

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I am getting the lenghts to be 13\displaystyle \frac{1}{3}

Aaghaz Mahajan - 7 months, 3 weeks ago

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You can also use similar triangles. Let the length of the red lines be xx. Consider the right triangle with the vertical red line as it's vertical leg. This right triangle has height xx and base length 1x1 - x, and is similar to the right triangle with height 1/21/2 and base length 11. By similarity we then have that

x1x=1/212x=1x3x=1x=13\dfrac{x}{1 - x} = \dfrac{1/2}{1} \Longrightarrow 2x = 1- x \Longrightarrow 3x = 1 \Longrightarrow x = \dfrac{1}{3}.

Brian Charlesworth - 7 months, 3 weeks ago

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Simply use co-ordinate geometry.....

Aaghaz Mahajan - 7 months, 3 weeks ago

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Could you please explain that? I am really interested.

Akshaj Gopalakrishnan - 7 months, 3 weeks ago

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Here is an outline of my method.....

Consider the origin to be the right angled vertex. The remaining two points are located on (1,0) & (0,1)\left(1,0\right)\ \&\ \left(0,1\right)
Hence, the midpoints will be located at (12,0) & (0,12)\left(\frac{1}{2},0\right)\ \&\ \left(0,\frac{1}{2}\right)
Now, simply write the equations of the lines joining the points and you'll arrive at the answer......

Aaghaz Mahajan - 7 months, 3 weeks ago

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@Aaghaz Mahajan Thanks!

Akshaj Gopalakrishnan - 7 months, 3 weeks ago

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@Akshaj Gopalakrishnan You are welcome!!

Aaghaz Mahajan - 7 months, 3 weeks ago

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