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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 only applies to right-angled triangles, which are triangles containing one 90∘90^\circ angle. The longest side, located directly opposite the right angle, is called the hypotenuse (cc). The two shorter sides (aa and bb) meet at the right angle.

A right-angled triangle with sides labeled a, b, and hypotenuse c.
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The theorem states that in any right-angled triangle, the area of the square on the hypotenuse is equal to the sum of the areas of the squares on the other two sides: a2+b2=c2a^2 + b^2 = c^2.

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To find the hypotenuse cc, you must add the squares of the other two sides and then take the square root: c=a2+b2c = \sqrt{a^2 + b^2}.

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To find a shorter side (e.g., aa), you must subtract the square of the known shorter side from the square of the hypotenuse and then take the square root: a=c2−b2a = \sqrt{c^2 - b^2}.

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The Converse of Pythagoras' Theorem: If the sum of the squares of the two shorter sides equals the square of the longest side, 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.

Problem 4:

A rectangle has a width of 8 cm8\text{ cm} and a diagonal of 17 cm17\text{ cm}. Calculate the height of the rectangle.

A rectangle with a diagonal of 17cm and base of 8cm.

Solution:

h2+82=172h^2 + 8^2 = 17^2 h2+64=289h^2 + 64 = 289 h2=289−64h^2 = 289 - 64 h2=225h^2 = 225 h=225h = \sqrt{225} h=15 cmh = 15\text{ cm}

Explanation:

In a rectangle, the diagonal forms a right-angled triangle with the adjacent sides. We use the subtraction form of Pythagoras' Theorem because we are finding a shorter side.

Problem 5:

Calculate the distance between point A(2,2)A(2, 2) and point B(8,10)B(8, 10) on a coordinate plane.

Coordinate grid showing a line segment between (2,2) and (8,10) as the hypotenuse of a right triangle.

Solution:

Horizontal distance (x)=8−2=6\text{Horizontal distance } (x) = 8 - 2 = 6 Vertical distance (y)=10−2=8\text{Vertical distance } (y) = 10 - 2 = 8 Distance AB=62+82\text{Distance } AB = \sqrt{6^2 + 8^2} AB=36+64AB = \sqrt{36 + 64} AB=100AB = \sqrt{100} AB=10 unitsAB = 10\text{ units}

Explanation:

The distance between two points can be found by creating a right-angled triangle where the horizontal change is the base and the vertical change is the height. The distance ABAB is the hypotenuse.