Question

Prove that the purple shaded area is equal to the area of incircle of the equilateral triangle (yellow striped circle)?

Answer 1

Explanation:

The area of the incircle is `pir^2`.

Noting the right triangle with hypotenuse `R` and leg `r` at the base of the equilateral triangle, through trigonometry or the properties of `30˚-60˚-90˚` right triangles we can establish the relationship that `R=2r`.

Note that the angle opposite `r` is `30˚` since the equilateral triangle's `60˚` angle was bisected.

This same triangle can be solved through the Pythagorean theorem to show that half the side length of the equilateral triangle is `sqrt(R^2-r^2)=sqrt(4r^2-r^2)=rsqrt3`.

Now examining half of the equilateral triangle as a right triangle, we see that the height `h` of the equilateral triangle can be solved for in terms of `r` using the relationship `tan(60˚)=h/(rsqrt3)`. Since `tan(60˚)=sqrt3`, this becomes `h/(rsqrt3)=sqrt3` so `h=3r`.

The area of the equilateral triangle is then `1/2bh`, and its base is `2rsqrt3` and its height `3r`. Thus, its area is `1/2(2rsqrt3)(3r)=3r^2sqrt3`.

The area of the smaller shaded region is equal to one-third the area of the equilateral triangle minus the incircle, or `1/3(3r^2sqrt3-pir^2)` which is equivalent to `r^2((3sqrt3-pi)/3)`.

The area of the larger circle is `piR^2=pi(2r)^2=4pir^2`.

The area of the larger shaded region is one-third the larger circle's area minus the area of the equilateral triangle, or `1/3(4pir^2-3r^2sqrt3)` which simplifies to be `r^2((4pi-3sqrt3)/3)`.

The total area of the shaded area is then `r^2((3sqrt3-pi)/3)+r^2((4pi-3sqrt3)/3)=r^2((3sqrt3-3sqrt3-pi+4pi)/3)=r^2((3pi)/3)=pir^2`, which is equivalent to the incircle's area.

Answer 2

Explanation:

For an equilateral triangle center of gravity , center of circumcircle and orthocenter coincide.
So Radius of cicumcircle (R) and radius of incircle (r) will have following relation

`R:r=2:1=>R=2r`

Now from the figure it is obvious that area of the BIG purple shaded region`=1/3(piR^2-Delta)`

And area of the SMALL purple shaded region`=1/3(Delta-pir^2)`

where `Delta` represents the area of the equilateral triangle.

So

`color(purple)("TOTAL area of the BIG and SMALL purple shaded region"`

`=1/3(piR^2-Delta)+1/3(Delta-pir^2)`

`=1/3(piR^2-cancelDelta+cancelDelta-pir^2)`

Inserting R =2r

`=1/3(pi(2r)^2-pir^2)`

`=1/3(4pir^2-pir^2)`

`=1/cancel3xxcancel3pir^2`

`=pir^2->color(orange)"Area of yellow striped circle"`