2022-01-20

Three right angled triangles in a semicircle within two parallel lines

This question was uploaded on 20/01/22 on social media accounts.

Solution:

Let the hypotenuse of the blue triangle be 'x' and the hypotenuse of two red triangles be 'a' and 'b'.
All three triangles are similar due to similar angles so their areas can be compared by the length of their hypotenuse.
We know that area of the right angle triangle is proportional to the square of the hypotenuse.
Red/Blue = x2/(a2+b2)
Comparing 'h' in both right angles triangles:
\[\color{Blue} {h^2=\color{#DF00FE}{\left(\frac{a+b}2\right)^2-\left(\frac X2\right)^2}=\color{#1BD62E}{\left(\frac X2\right)^2-\left(\frac{a-b}2\right)^2}}\]
\[\Rightarrow\color{Blue}{X^2}=\color{Red}{a^2+b^2}\]
Now,
\[\frac{\color{Red}{Red}}{\color{Blue}{Blue}}=\frac{\color{Red}{a^2+b^2}}{\color{Blue}{X^2}}=1\]
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