Angles, areas, perimeters, circles, and solids. Geometry is the art of seeing shape and finding the numbers that describe it.
1
Angles in a triangle
Key fact
The angles of any triangle add up to \(180°\)
Isosceles triangle
Two equal sides → the two base angles are equal.
If the angle between the equal sides is \(80°\), then each base angle is \(\dfrac{180° - 80°}{2} = 50°\).
Equilateral triangle
All three sides are equal → all three angles are \(60°\).
Try it
In an isosceles triangle, the angle between the equal sides is \(40°\). Each base angle is:
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2
Area and perimeter
Perimeter — the sum of all sides. Area — how many square units fit inside.
A square has area \(64\;\text{cm}^2\). Its perimeter is:
Side: \(a = \sqrt{64} = 8\) cm.
Perimeter: \(P = 4 \times 8 = 32\) cm.
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A rectangle has side lengths 7 cm and 3 cm. Its area and perimeter are:
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A square has perimeter 48 cm. Its area is:
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3
Circle: circumference and area
Circle formulas
Circumference: \(C = 2\pi r = \pi d\) Area of the circle: \(S = \pi r^2\) The number \(\pi \approx 3.14\) (but in CMS problems answers are usually left in terms of \(\pi\))
In CMS Olympiads
Answers are almost always written using \(\pi\), for example \(25\pi\) or \(5\pi + 10\). You do not need decimals — treat \(\pi\) like an ordinary symbol.
Try it
The area of a circle with radius 6 cm is:
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4
Semicircle
A semicircle is half of a circle. Be careful with the perimeter: to half of the circumference you must add the diameter (the straight part)!
Perimeter of a semicircle
\(P = \pi r + d = \pi r + 2r\)
Curved part (\(\pi r\)) + straight part (diameter \(d\))
Test problem — Q11
Perimeter of a semicircle with diameter 10 cm:
Radius \(r = 5\) cm.
Curved part: \(\pi \times 5 = 5\pi\).
Straight part: diameter = 10.
Perimeter: \(5\pi + 10\) cm.
Typical mistake
Do not forget to add the diameter. If you write only \(5\pi\), that is only the curved part, not the whole perimeter!
Try it
The perimeter of a semicircle with diameter 14 cm is:
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5
Composite figures
A very common CMS type: “a path around a pool” or “a frame around a picture.” The key is area of the large figure minus area of the small one.
Faces: 6 (each is a square) Edges: 12 Vertices: 8 Volume: \(V = a^3\) Surface area: \(S = 6a^2\)
Test problem — Q13
A cube is painted and cut into 27 small cubes. How many have exactly 2 painted faces?
Cubes with 2 painted faces are those on the edges (but not at the corners).
A cube has 12 edges. On each edge there is 1 middle cube → \(12 \times 1 = 12\).
Try it
A cube is painted and cut into 27 small cubes. How many small cubes have exactly 3 painted faces?
Try it
How many small cubes have exactly 1 painted face?
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7
Olympiad practice
Problem 1 · Easy
The angles of a triangle are in the ratio \(1 : 2 : 3\). The largest angle is:
Let the angles be \(x, 2x, 3x\). Sum: \(x + 2x + 3x = 6x = 180°\).
\(x = 30°\). Largest angle: \(3 \times 30° = 90°\).
Problem 2 · Easy
The circumference of a circle with radius 7 cm is:
Problem 3 · Medium
A 15×10 m rectangular pool is surrounded by a path 3 m wide. The area of the path is:
Outer rectangle: \((15+6) \times (10+6) = 21 \times 16 = 336\) m².
Pool: \(15 \times 10 = 150\) m².
Path: \(336 - 150 = 186\) m².