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Trigonometry - Pythagoras' theorem

Grade 11IGCSE

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

🔑Concepts

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Pythagoras' Theorem applies specifically to right-angled triangles.

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The hypotenuse is the longest side and is always opposite the right angle (90∘90^{\circ}).

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

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The converse of Pythagoras' Theorem can be used to check if a triangle is right-angled: if a2+b2=c2a^2 + b^2 = c^2, then the triangle must have a 90∘90^{\circ} angle.

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3D Pythagoras is used to find lengths in three-dimensional shapes like cuboids and pyramids by applying the theorem twice or using the extended formula.

📐Formulae

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

c=a2+b2c = \sqrt{a^2 + b^2} (finding the hypotenuse)

a=c2−b2a = \sqrt{c^2 - b^2} (finding a shorter side)

d2=x2+y2+z2d^2 = x^2 + y^2 + z^2 (3D Pythagoras for a space diagonal dd in a cuboid with dimensions x,y,zx, y, z)

Distance=(x2−x1)2+(y2−y1)2Distance = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} (Distance between two points on a coordinate plane)

💡Examples

Problem 1:

A ladder of length 5m is leaned against a vertical wall. The base of the ladder is 3m away from the wall on horizontal ground. How high up the wall does the ladder reach?

Solution:

Let the height be hh. Using a2+b2=c2a^2 + b^2 = c^2, we have 32+h2=523^2 + h^2 = 5^2. 9+h2=259 + h^2 = 25. h2=25−9=16h^2 = 25 - 9 = 16. h=16=4h = \sqrt{16} = 4.

Explanation:

In this scenario, the ladder acts as the hypotenuse (c=5c=5) and the distance from the wall is one of the shorter sides (b=3b=3). We rearrange the formula to solve for the missing vertical side aa.

Problem 2:

A triangle has side lengths of 7cm, 24cm, and 25cm. Determine if this triangle is right-angled.

Solution:

72+242=49+576=6257^2 + 24^2 = 49 + 576 = 625. The square of the longest side is 252=62525^2 = 625. Since 49+576=62549 + 576 = 625, the condition a2+b2=c2a^2 + b^2 = c^2 is satisfied.

Explanation:

To check for a right angle, square the two shorter sides and sum them. If the result equals the square of the longest side (the converse of Pythagoras' Theorem), the triangle is right-angled.

Problem 3:

Find the length of the internal diagonal of a cuboid with dimensions 3cm, 4cm, and 12cm.

Solution:

d2=32+42+122=9+16+144=169d^2 = 3^2 + 4^2 + 12^2 = 9 + 16 + 144 = 169. d=169=13d = \sqrt{169} = 13cm.

Explanation:

In 3D Pythagoras, the squared length of the space diagonal is the sum of the squares of the length, width, and height. This is equivalent to applying Pythagoras twice: once to find the diagonal of the base, and then again to find the diagonal of the cuboid.