Bunuel wrote:

A semicircle with area of \(x \pi\) is marked by seven points equally spaced along the half arc of the semicircle, such that two of the seven points form the endpoints of the diameter. What is the probability of forming a triangle with an area less than x from the total number of triangles formed by combining two of the seven points and the center of the diameter?

A. 4/5

B. 6/7

C. 17/21

D. 19/21

E. 31/35

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 = \(\frac{\sqrt{3[}{square_root]/4}*a^2=\frac{[square_root]3[}{square_root]/4}*[square_root]2x}^2=\frac{\sqrt{3[}{square_root]/2}*x\), 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 \([square_root]2x}\)..

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.