Geometry
This page is part of the section “USE basic level. Geometry”. Here are solutions to typical geometry problems: area of a triangle, Pythagorean theorem, area of a circle, volume of a cylinder. All problems are written independently, with neat answers.
Problems with solutions
Problem 1. Area of a triangle
Find the area of a triangle if two of its sides are 3 and 4, and the angle between them is 30°.
The area of a triangle in terms of two sides and the angle between them:
Substitute sin 30° = 1/2:
Answer: 3.
Problem 2. Pythagorean theorem
In a right triangle the legs are 6 and 8. Find the hypotenuse.
By the Pythagorean theorem, the square of the hypotenuse equals the sum of the squares of the legs:
Answer: 10.
Problem 3. Area of a circle
Find the area of a circle of radius 5.
The area of a circle is given by S = πR²:
Answer: 25π.
Problem 4. Volume of a cylinder
Find the volume of a cylinder if the radius of its base is 2 and the height is 5.
The volume of a cylinder is given by V = πR²h:
Answer: 20π.
Frequently asked questions
How to find the area of a triangle from two sides and the included angle?
Use the formula S = ½·a·b·sin γ, where a and b are the sides and γ is the angle between them. For example, S = ½·3·4·sin 30° = ½·3·4·½ = 3.
How to apply the Pythagorean theorem?
In a right triangle, the square of the hypotenuse equals the sum of the squares of the legs: c² = a² + b². Knowing two elements, you can find the third. For legs 6 and 8: c² = 36 + 64 = 100, c = 10.
How to find the area of a circle?
Use the formula S = πR², where R is the radius. For R = 5: S = π·25 = 25π. The answer can be left with π or approximated with π ≈ 3.14 — but in the USE basic level it is usually left with π.
How to find the volume of a cylinder?
Use the formula V = πR²h, where R is the base radius and h is the height. For R = 2 and h = 5: V = π·4·5 = 20π.
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