IMO Practice Test — Areas Related to Circles
6 Questions • 15 min • Olympiad level
15:00
Question 1 of 6
medium
A square is inscribed perfectly inside a circle of radius $r$. Find the total area of the region left inside the circle outside the square.
$\pi r^2 - r^2$
$\pi r^2 - 2r^2$
$2r^2 - \pi r^2$
$\pi r^2 - 4r^2$
Explanation: Diagonal of square is $2r$. Area of square = $\frac{1}{2} \cdot d^2 = \frac{1}{2} \cdot 4r^2 = 2r^2$. Shaded area = $\pi r^2 - 2r^2$.
Question 2 of 6
medium
Three identical coins of radius 7 cm each are placed flat on a table so that they touch each other. Find the area of the tiny trapped gap space between them. ($\sqrt{3}=1.73$, $\pi=22/7$).
7.77 sq. cm
1.45 sq. cm
3.43 sq. cm
5.21 sq. cm
Explanation: Joining the three centres forms an equilateral triangle of side $14$ cm, with area $\frac{\sqrt{3}}{4} \cdot 14^{2} = \frac{1.73}{4} \cdot 196 = 84.77$. The three $60^\circ$ sectors (radius $7$) together make a semicircle of area $\frac{1}{2} \cdot \frac{22}{7} \cdot 7^{2} = 77$. Gap $= 84.77 - 77 = 7.77$ sq. cm.
Question 3 of 6
medium
The short hour hand of a massive tower clock is 6 cm long and the minute hand is 10 cm long. Find the sum of distances covered by their tips in 24 hours.
$416\pi$ cm
$496\pi$ cm
$504\pi$ cm
$240\pi$ cm
Explanation: The hour hand completes $2$ turns: $2 \cdot (2\pi \cdot 6) = 24\pi$. The minute hand completes $24$ turns: $24 \cdot (2\pi \cdot 10) = 480\pi$. Sum $= 504\pi$ cm.
Question 4 of 6
medium
A wire is bent into the shape of an equilateral triangle enclosing an area of $121\sqrt{3}$ sq. cm. If the same wire is unbent and re-bent into a perfect circle, find the area of the circle.
154 sq. cm
346.5 sq. cm
616 sq. cm
246.4 sq. cm
Explanation: Triangle area = $\frac{\sqrt{3}}{4}s^2 = 121\sqrt{3} \implies s=22$. Perimeter = 66. Circle perimeter $2\pi r = 66 \implies r = 10.5$. Area = $\pi r^2 = 346.5$.
Question 5 of 6
medium
Four equal-sized circular running tracks are designed inside a square park of side 28 meters such that each circle touches two other circles. Find the area of the leftover central space.
168 sq. cm
42 sq. cm
84 sq. cm
126 sq. cm
Explanation: Park area = $28^2 = 784$. The 4 circles each have radius 7 m. Total area of 4 circles = $4 \cdot (22/7) \cdot 7^2 = 616$. Leftover area = $784 - 616 = 168$.
Question 6 of 6
medium
A chord of a circle of radius 12 cm subtends an angle of 120 degrees at the center. Find the area of the corresponding minor segment. ($\pi=3.14$, $\sqrt{3}=1.73$).
88.44 sq. cm
62.28 sq. cm
150.72 sq. cm
44.22 sq. cm
Explanation: Sector area $= \frac{120}{360} \cdot 3.14 \cdot 144 = 150.72$. Triangle area $= \frac{1}{2}r^{2}\sin 120^\circ = \frac{1}{2} \cdot 144 \cdot \frac{\sqrt{3}}{2} = 36 \cdot 1.73 = 62.28$. Minor segment $=$ sector $-$ triangle $= 150.72 - 62.28 = 88.44$ sq. cm.