IMO Practice Test — Circles
6 Questions • 15 min • Olympiad level
15:00
Question 1 of 6
If the circumference of a circle increases from \(4\pi\) to \(8\pi\), by what factor does its interior area increase?
\(2\) times
\(4\) times
\(8\) times
\(16\) times
Explanation: \(C\) doubles \(\implies r\) doubles (\(2\) to \(4\)). \(\text{Area}_1 = 4\pi\), \(\text{Area}_2 = 16\pi\). \(16\pi \div 4\pi = 4\).
Question 2 of 6
A circular grass lawn of radius \(14\text{ m}\) is surrounded by a outer concentric gravel path that is exactly \(7\text{ m}\) wide. Find the area of the gravel path. (Use \(\pi = \frac{22}{7}\))
\(154\text{ m}^2\)
\(462\text{ m}^2\)
\(616\text{ m}^2\)
\(1155\text{ m}^2\)
Explanation: Inner \(r=14\), Outer \(R=21\). \(\text{Path} = \pi(21^2 - 14^2) = \frac{22}{7}(441-196) = \frac{22}{7}(245) = 462\text{ m}^2\).
Question 3 of 6
A wire in the shape of a circle encloses an area of \(1386\text{ cm}^2\). If the exact same wire is unbent and re-formed into a square, what is the area enclosed by the square? (Use \(\pi = \frac{22}{7}\))
\(961\text{ cm}^2\)
\(1089\text{ cm}^2\)
\(1225\text{ cm}^2\)
\(1386\text{ cm}^2\)
Explanation: \(\frac{22}{7}r^2=1386 \implies r=21\). Wire length \(= C = 2\times\frac{22}{7}\times21 = 132\text{ cm}\). Square side \(= 132\div 4 = 33\text{ cm}\). \(\text{Square Area} = 33^2 = 1089\text{ cm}^2\).
Question 4 of 6
A toy car wheel makes \(1000\) rotations to cover a distance of \(880\text{ meters}\). Find the diameter of the wheel in centimeters. (Use \(\pi = \frac{22}{7}\))
\(14\text{ cm}\)
\(28\text{ cm}\)
\(44\text{ cm}\)
\(56\text{ cm}\)
Explanation: Distance \(= 88000\text{ cm}\). \(C = 88000 \div 1000 = 88\text{ cm}\). \(\pi d = 88 \implies \frac{22}{7}d = 88 \implies d = 28\text{ cm}\).
Question 5 of 6
The perimeter of a semi-circular protractor (including its straight diameter base) is \(36\text{ cm}\). Find the radius of the protractor. (Use \(\pi = \frac{22}{7}\))
\(7\text{ cm}\)
\(14\text{ cm}\)
\(18\text{ cm}\)
\(21\text{ cm}\)
Explanation: \(\text{Perimeter} = \pi r + 2r = r(\pi + 2)\). \(r(\frac{22}{7} + 2) = 36 \implies r(\frac{36}{7}) = 36 \implies r = 7\text{ cm}\).
Question 6 of 6
Four identical circular metal washers, each of radius \(7\text{ cm}\), are arranged tightly in a \(2 \times 2\) square block layout so they touch. Find the area of the empty gap left enclosed between the four wheels.
\(42\text{ cm}^2\)
\(112\text{ cm}^2\)
\(154\text{ cm}^2\)
\(196\text{ cm}^2\)
Explanation: Outer bounding square side \(= 2r + 2r = 14 + 14 = 28\text{ cm}\). \(\text{Square Area} = 28^2 = 784\)? No, corner center square side \(= 2r = 14\). \(\text{Area} = 14^2 = 196\). 4 quarter sectors \(= 1\) full circle area \(= 154\). \(\text{Gap} = 196 - 154 = 42\text{ cm}^2\).