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Trigonometry - Pythagoras' theorem

Grade 11A Level

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

🔑Concepts

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Pythagoras' Theorem applies specifically to right-angled triangles. The side opposite the right angle (90∘90^{\circ}) is called the hypotenuse and is always the longest side. In a triangle with sides aa, bb, and hypotenuse cc, the relationship is a2+b2=c2a^2 + b^2 = c^2.

A right-angled triangle with legs a and b, and hypotenuse c.
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The converse of Pythagoras' Theorem states that if the sum of the squares of two sides equals the square of the third side (a2+b2=c2a^2 + b^2 = c^2), then the triangle must be right-angled.

A 3-4-5 triangle used to demonstrate the converse of Pythagoras' theorem.
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To find the distance between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) on a Cartesian plane, we treat the horizontal and vertical differences as the legs of a right-angled triangle.

Distance between points A and B visualized as the hypotenuse of a right-angled triangle.
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In 3D shapes like cuboids, Pythagoras' theorem can be applied twice to find the space diagonal: once to find the diagonal of the base, and then again using that result and the height.

A 3D cuboid showing the internal space diagonal from a bottom-front corner to a top-back corner.

📐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.

Problem 4:

Calculate the length of the diagonal of a rectangle with a width of 8 cm8\text{ cm} and a height of 15 cm15\text{ cm}.

Rectangle showing sides of 15 and 8 with a diagonal d.

Solution:

d2=82+152d^2 = 8^2 + 15^2 d2=64+225d^2 = 64 + 225 d2=289d^2 = 289 d=289d = \sqrt{289} d=17 cmd = 17\text{ cm}

Explanation:

A rectangle can be divided into two right-angled triangles by its diagonal. We use the side lengths 88 and 1515 as aa and bb to solve for the hypotenuse cc (the diagonal).

Problem 5:

An isosceles triangle has a base of 12 m12\text{ m} and two equal sides of 10 m10\text{ m}. Find the perpendicular height of the triangle.

Isosceles triangle split into two right-angled triangles to find height h.

Solution:

h2+62=102h^2 + 6^2 = 10^2 h2+36=100h^2 + 36 = 100 h2=100−36h^2 = 100 - 36 h2=64h^2 = 64 h=64h = \sqrt{64} h=8 mh = 8\text{ m}

Explanation:

In an isosceles triangle, the perpendicular height bisects the base. This creates two right-angled triangles with a base of 6 m6\text{ m} (12÷212 \div 2) and a hypotenuse of 10 m10\text{ m}. We then solve for the vertical leg.