[asy]
pair B = (0,0), C = (6,0), A = (0,8), O = (1.91,3.82);
draw(A--B--C--cycle);
draw(B..O--C);
draw(B--extension(B,O,A,C));
draw(A..O);
draw(arc(A,3.82,0,180));
draw(arc(B,5.66,128.66,308.66));
draw(arc(C,5.66,231.34,51.34));
label("$A$",A,N);
label("$B$",B,SW);
label("$C$",C,SE);
label("$D$",B+(0,1.99),W);
label("$O$",O,(1,1));
fill(anglemark(B,C,A)*arc(B,C,A,1.5),lightgray);
[/asy]
Tangent segments are congruent, so $DB = DC = 1.99$ cm. Noting that the radius of $\overbigarc{BC}$ is $\tfrac{6}{2} = 3,$ thus $BA = 3BD,$ so triangle $BAD$ is a $30^\circ$-$60^\circ$-$90^\circ$ triangle with $AD = 3\cdot 1.99 = 5.97$ cm. Therefore, the shaded area is \begin{align*}
&[\triangle ABC] - [\triangle BAD] - [\triangle CAD] - [BCDA] \\
&= \frac{1}{2}(6)(6) - \frac{1}{2}\cdot 5.97\cdot 1.99 - \frac{1}{2}\cdot 5.97\cdot 3 - \frac{1}{2}\cdot 1.99\cdot 3 \\
&= 18 - 5.97 \cdot (1.99 + 3) - 1.99 \cdot \frac32 \\
&= 18 - 33.77 \\
&= -\underbrace{15.77}_{-[\triangle CBA]}.
\end{align*} They want the perimeter, so all we have to do is sum the lengths around the shaded region, i.e. $(-15.77) + 6 + 3 + (5.97 + 1.99 + 3) = \boxed{04.78}$ centimeters.
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.
1 answer
1 answer