Question

In the figure, two circular arcs subtend $60^\circ$ and $90^\circ$ at their respective centres. If the length of the bottom arc $Y$ is $10\pi$, find the length of the other arc $X$.

• A. $15\pi/\sqrt{2}$
• B. $20\pi\sqrt{2}/3$
• C. $60\pi/\sqrt{2}$
• D. $20\pi/3$
• E. $15\pi$

Hint:

For arcs with different central angles sharing the same endpoints, equate their {chord} lengths using \(c=2R\sin(\theta/2)\) to move from one radius to the other; then compute the needed arc length.

Answer 1

The Correct Option is A

Step 1: Express the given bottom arc \(Y\).
Let the radius of the bottom arc be \(R_Y\). Since \(Y\) subtends \(60^\circ\) and its length is \(10\pi\), \[ \text{Arc length }Y = \frac{60^\circ}{360^\circ}\,2\pi R_Y = \frac{\pi}{3}R_Y = 10\pi \ \Rightarrow\ R_Y=30. \] Step 2: Use the common chord between the two arcs.
Both arcs share the same endpoints, hence the chord length is common. For a central angle \(\theta\) and radius \(R\), chord \(c=2R\sin\!\left(\frac{\theta}{2}\right)\). Bottom arc \(Y\) (\(60^\circ\)): \[ c=2R_Y\sin 30^\circ = 2(30)..... \tfrac{1}{2} = 30. \] Step 3: Find the radius of the other arc \(X\) (subtending \(90^\circ\)).
Let its radius be \(R_X\). With the same chord, \[ c=2R_X\sin 45^\circ = 2R_X..... \frac{\sqrt{2}}{2}=\sqrt{2}\,R_X=30 \ \Rightarrow\ R_X=\frac{30}{\sqrt{2}}=15\sqrt{2}. \] Step 4: Length of arc \(X\).
\[ \text{Arc }X=\frac{90^\circ}{360^\circ}\,2\pi R_X =\frac{\pi}{2}\,R_X =\frac{\pi}{2}\,(15\sqrt{2}) =\boxed{\frac{15\pi}{\sqrt{2}}}. \]