SOLUTION: 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 cent

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: 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 cent      Log On


   



Question 1203424: 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.

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Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52858) About Me  (Show Source):
You can put this solution on YOUR website!
.

H I N T


As first step, find the radius of the greatest arc.

It is exactly half of the hypotenuse of the given triangle, i.e. R = %286%2Asqrt%282%29%29%2F2 = 3%2Asqrt%282%29 cm.


What follows, is just simple arithmetic to do it step by step - there is nothing interesting or difficult in it.



Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

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.