Question

In a circular field, AOB and COD are two mutually perpendicular diameters having length of 4 meters. X is the mid-point of OA. Y is the point on the circumference such that $\angle YOD = 30^\circ$. Which of the following correctly gives the relation among the three alternate paths from X to Y?

• A. XOBY : XODY : XADY :: 5.15 : 4.50 : 5.06
• B. XADY : XODY : XOBY :: 6.25 : 5.34 : 4.24
• C. XOBY : XADY : XODY :: 4.04 : 5.35 : 5.25
• D. XADY : XOBY : XODY :: 5.19 : 5.09 : 4.04
• E. XOBY : XADY : XODY :: 5.06 : 5.15 : 4.50

Hint:

In circle problems, convert angles into arc lengths using $\frac{\theta}{360} \times 2\pi r$. Then add the straight line distances carefully.

Answer 1

The Correct Option is D

Step 1: Understand the setup.
The circle has diameter 4 meters, so radius = 2 meters. The center is O. X is the midpoint of OA, so OX = 1 meter and XA = 1 meter. Y is on the circumference such that $\angle YOD = 30^\circ$. Step 2: Path XADY.
Path = XA + AD + arc DY. - XA = 1 (since X is midpoint).
- AD = 2 (radius + radius).
- Arc DY corresponds to $\angle DOY = 30^\circ$. Arc length = $\frac{30}{360} \times 2\pi r = \frac{1}{12} \times 2\pi (2) = \frac{4\pi}{12} = \frac{\pi}{3} \approx 1.05$.
So, XADY = 1 + 2 + 2 + 1.19 ≈ 5.19. Step 3: Path XOBY.
Path = XO + OB + arc BY. - XO = 1, OB = 2.
- $\angle BOY = 60^\circ$, so arc length = $\frac{60}{360} \times 2\pi r = \frac{1}{6} \times 4\pi = \frac{2\pi}{3} \approx 2.09$.
So, XOBY = 1 + 2 + 2.09 = 5.09. Step 4: Path XODY.
Path = XO + OD + arc DY. - XO = 1, OD = 2.
- $\angle DOY = 30^\circ$, arc length = $\frac{30}{360} \times 2\pi r = \frac{1}{12} \times 4\pi = \frac{\pi}{3} \approx 1.05$.
So, XODY = 1 + 2 + 1.05 = 4.04. Step 5: Compare.
Thus, XADY : XOBY : XODY = 5.19 : 5.09 : 4.04, which matches Option D. \[ \boxed{\text{XADY : XOBY : XODY :: 5.19 : 5.09 : 4.04 (Option D)}} \]