geometry Q

[RIJU]

In a circumference with center O and radius r, what is the area of triangle KLO? (KLMN is a rectangle)


Note: Figure not drawn to scale.

[TwinnSplitter]

Angle OKN + Angle ONK = 180 - 60 = 120

Angle OKN = Angle ONK (since OK and ON are both radii, and thus the angles opposite them are equal)

Thus Angle OKN = 120/2 = 60

Angle OKN and OKL add up to 90, so angle OKL = 90 - 60 = 30

Thus OKL is a 30-60-90 triangle, and their sides have the ratio 1: root3 : 2.

The hypotenuse is r, and thus side OL = r/2
and side LK = root 3 (r/2)

Thus the area of trianle OLK is (1/2)(r/2)(root 3)(r/2)

= (root3)(r^2) / 8

[scoot]

Another way of doing this.

one side of equ triangle(ok) = r

area of equ. triangle= root(3)*r^2/4.

retangle form 2 equ triangle here.

area of olk = area of equ.triangle/2 = root(3)*r^2/8.

[malkasri]

yet another way:

OK = ON = r. Hence angles opposite them r equal which i.e angle OKN = angle ONK = 60degrees. Therefore it is now equilateral triangle with all sides equal.
Hence KN = r which is equal to LM.

Therefore LO = r/2.

Area of KLO = 1/2*LO*KL

OK is the hypotenuse of right triangle OLK. Hence OK = root(r*r - r*r/4) = root 3(r/2).

Area = 1/2*r/2*root3*r/2 = root 3 *r^2/8.

[lego2401]

the fastest way of doing this:

It is immediately apparent that if we draw a perp bisector of the equilateral triangle, then Angle OKL = 30, which makes triangle KLO a 60-30-90 triangle with hypothenues = r.

So then the sides are in the ratio 1:root(3):2

=> sides are r/2 and root(3)/2*r and r

so area = (r/2 * root(3)/2*r)*0.5 = (root(3)*r^2)/8