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Trigonometry - Pythagoras’ Theorem

Grade 9IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Pythagoras' Theorem applies only to right-angled triangles.

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The hypotenuse is the longest side of a right-angled triangle and is always opposite the 90∘90^\circ angle.

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The theorem states that the square of the hypotenuse is equal to the sum of the squares of the other two sides.

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Pythagorean Triples are sets of three integers that satisfy the theorem (e.g., 3, 4, 5 and 5, 12, 13).

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The converse of the theorem: If a2+b2=c2a^2 + b^2 = c^2, then the triangle must be right-angled.

📐Formulae

a2+b2=c2a^2 + b^2 = c^2 (where cc is the hypotenuse)

c=a2+b2c = \sqrt{a^2 + b^2}

a=c2−b2a = \sqrt{c^2 - b^2}

💡Examples

Problem 1:

A right-angled triangle has two shorter sides of length 5 cm5\text{ cm} and 12 cm12\text{ cm}. Calculate the length of the hypotenuse.

Solution:

13 cm13\text{ cm}

Explanation:

Using the formula c2=a2+b2c^2 = a^2 + b^2, we substitute the values: c2=52+122=25+144=169c^2 = 5^2 + 12^2 = 25 + 144 = 169. Taking the square root, c=169=13c = \sqrt{169} = 13.

Problem 2:

The hypotenuse of a right-angled triangle is 10 cm10\text{ cm} and one of the other sides is 6 cm6\text{ cm}. Find the length of the third side.

Solution:

8 cm8\text{ cm}

Explanation:

Using the rearranged formula a2=c2−b2a^2 = c^2 - b^2, we substitute: a2=102−62=100−36=64a^2 = 10^2 - 6^2 = 100 - 36 = 64. Taking the square root, a=64=8a = \sqrt{64} = 8.

Problem 3:

A ladder of length 5 m5\text{ m} is leaned against a vertical wall. If the base of the ladder is 3 m3\text{ m} away from the wall, how high up the wall does the ladder reach?

Solution:

4 m4\text{ m}

Explanation:

The ladder forms a right-angled triangle where the ladder is the hypotenuse (c=5c=5) and the distance from the wall is the base (b=3b=3). We need to find the height (aa). a2=52−32=25−9=16a^2 = 5^2 - 3^2 = 25 - 9 = 16. Therefore, a=16=4a = \sqrt{16} = 4.