A semicircle with area of xπ is marked by seven points equally spaced

HI,

Bunuel, pl relook into the choices given..


firstly semi circle has a area of \(x \pi\)..
so \(\frac{\pi*r^2}{2}=x\pi.......r=\sqrt{2x}\)
so when we make 7 points in the way it has been described, there are 6 equal segments which will have area of \(\frac{x \pi}{6}\)..LESS than \(x\pi\)
here central angle at centre is 180/6=30..
How many triangles ? 6

Now when we take two such pieces, the centre angle becomes 60 and other two sides are radius so it becomes EQUILATERAL triangle with each side \(\sqrt{2x}\)..
Area = \(\frac{\sqrt{3}}{4}*a^2=\frac{\sqrt{3}}{4}*\sqrt{2x}^2=\frac{\sqrt{3}}{2}*x\), which is less than x.
How many such triangles? when 1 and 3 is choosen OR 2 and 4 OR 3 and 5 OR 4 and 6 OR 5 and 7------- 5 triangles..

Next when you choose three segments that is 30*3=90, it becomes right angled triangle with sides \(\sqrt{2x}\)..
area = \(\frac{1}{2}*\sqrt{2x}^2=x\) which is NOT less than x.
From here on any triangle with three segments or MORE will have AREA equal to or greater than x..

so total triangles with area less than x is 6+5=11
total triangles possible = 7C2=\(\frac{7!}{5!2!}\)=21..
but it includes the two points 1 and 7 where these three points form a straight line DIAMETER.. so 21-1=20

probability = \(\frac{11}{20}\)

Checked. The options are copied as they show up in the source.