Right triangle $ABC$ has $AB = AC = 6$ cm. Circular arcs are drawn with centers at $A, B$ and $C,$ so that the arc centered at $A$ is tangent to side $BC$ and so that the arcs centered at $B$ and $C$ are tangent to the arc centered at $A,$ as shown. What is the perimeter of the shaded region? Express your answer as a decimal to the nearest hundredth.
The triangle is isosceles so angles B and C = 45°
The large quarter circle will be tangent to BC at its midpoint = (3,3)
The radius of this circle = sqrt (3^2 + 3^2) = sqrt (18) = 3sqrt (2)
The arc length of this circle is (1/4) *2 * pi * ( 3sqrt (2)) = (3/2)sqrt (2) pi (1)
The radius of each of the smaller arcs at B and C = 6 - (3)sqrt (2)
And the combined length of the arcs at B and C = (1/4) * 2pi * ( 6 - 3sqrt 2) =
3pi - (3/2)sqrt (2)pi (2)
And the length of the shaded area bordering the hypotenuse = sqrt ( 6^2 + 6^2) - 2 (6 - 3sqrt 2) =
6sqrt2 - 12 + 6sqrt 12 = 12 sqrt 2 - 12 (3)
The total perimeter of the shaded area = (1) + (2) + (3) =
(3/2)sqrt( 2) pi + 3pi - (3/2)sqrt (2) pi + 12sqrt (2) - 12 =
3pi + 12(sqrt 2 - 1) = 14.395 cm ≈ 14.4 cm
Use the pythagorean theorem to find that hypotenuse BC = 6*sqrt(2)
Or you can use the 45-45-90 triangle template.
Let D = midpoint of B and C
It turns out that any right triangle will have its circumcenter at the midpoint of the hypotenuse.
This means we can draw a circle centered at D, and radius 3*sqrt(2).
This circumcircle passes through A, B, and C.
The distance from A to D is 3*sqrt(2), aka the radius of this new circle.
Therefore, segment AD = 3*sqrt(2)
Use this info to determine the radius of each small arc.
To get the distance along a circle's edge, you'll need this formula
arc length = (angle in radians)*r
45 degrees = pi/4 radians
90 degrees = pi/2 radians
I'll let the student finish up from here.