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Trigonometry - Pythagoras’ Theorem

Grade 12A Level

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

🔑Concepts

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Pythagoras' Theorem is only applicable to right-angled triangles. The side opposite the right angle (90∘90^{\circ}) is called the hypotenuse and is always the longest side.

A right-angled triangle with sides labeled a, b and hypotenuse c.
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To find the hypotenuse cc, you square both shorter sides, add them, and take the square root: c=a2+b2c = \sqrt{a^2 + b^2}. To find a shorter side, subtract the square of the known side from the square of the hypotenuse: a=c2−b2a = \sqrt{c^2 - b^2}.

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Pythagoras' Theorem can be extended to 3D shapes. For a rectangular cuboid with length ll, width ww, and height hh, the space diagonal dd is the distance from one corner to the opposite corner through the center.

A 3D cuboid showing the internal space diagonal d.
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The Converse of Pythagoras' Theorem states that if a2+b2=c2a^2 + b^2 = c^2 holds true for a triangle with sides a,b,ca, b, c, 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}

d2=x2+y2+z2d^2 = x^2 + y^2 + z^2 (Diagonal of a cuboid with dimensions x,y,zx, y, z)

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

💡Examples

Problem 1:

A ladder of length 13m leans against a vertical wall. The foot of the ladder is 5m away from the base of the wall. How high up the wall does the ladder reach?

Solution:

12m

Explanation:

Identify the hypotenuse (c=13c = 13) and one side (b=5b = 5). Using a2=c2−b2a^2 = c^2 - b^2, we get a2=132−52=169−25=144a^2 = 13^2 - 5^2 = 169 - 25 = 144. Taking the square root, a=144=12a = \sqrt{144} = 12.

Problem 2:

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

Solution:

13cm

Explanation:

Using the 3D Pythagoras formula d2=l2+w2+h2d^2 = l^2 + w^2 + h^2. Here, d2=32+42+122=9+16+144=169d^2 = 3^2 + 4^2 + 12^2 = 9 + 16 + 144 = 169. Thus, d=169=13d = \sqrt{169} = 13.

Problem 3:

Determine if a triangle with side lengths 7cm, 24cm, and 25cm is a right-angled triangle.

Solution:

Yes, it is right-angled.

Explanation:

Check if a2+b2=c2a^2 + b^2 = c^2. Calculate 72+242=49+576=6257^2 + 24^2 = 49 + 576 = 625. Calculate 252=62525^2 = 625. Since 625=625625 = 625, the converse of Pythagoras' theorem confirms it is a right-angled triangle.

Problem 4:

A rectangular field measures 60 m60\text{ m} by 80 m80\text{ m}. A person walks diagonally across the field from one corner to the opposite corner. Calculate the distance they walk.

A rectangle with a diagonal line labeled x, and sides labeled 80m and 60m.

Solution:

  1. Identify the triangle: The diagonal forms the hypotenuse (cc) of a right-angled triangle with sides a=60a = 60 and b=80b = 80.
  2. Apply formula: c2=a2+b2c^2 = a^2 + b^2
  3. Substitute: c2=602+802c^2 = 60^2 + 80^2
  4. Calculate: c2=3600+6400=10000c^2 = 3600 + 6400 = 10000
  5. Solve for cc: c=10000=100 mc = \sqrt{10000} = 100\text{ m}

The distance walked is 100 m100\text{ m}.

Explanation:

Since the field is rectangular, the corner angle is 90∘90^{\circ}, allowing the use of Pythagoras' Theorem to find the diagonal length.

Problem 5:

An isosceles triangle has a base of 10 cm10\text{ cm} and two equal sides of 13 cm13\text{ cm} each. Calculate the vertical height of the triangle.

An isosceles triangle split into two right-angled triangles to find the height h.

Solution:

  1. Split the isosceles triangle into two identical right-angled triangles by drawing the altitude.
  2. The base of each right-angled triangle is half the total base: 10÷2=5 cm10 \div 2 = 5\text{ cm}.
  3. The hypotenuse cc is 13 cm13\text{ cm} and one side bb is 5 cm5\text{ cm}. Let height be hh.
  4. Apply formula: h2=c2−b2h^2 = c^2 - b^2
  5. Substitute: h2=132−52h^2 = 13^2 - 5^2
  6. Calculate: h2=169−25=144h^2 = 169 - 25 = 144
  7. Solve for hh: h=144=12 cmh = \sqrt{144} = 12\text{ cm}

The vertical height is 12 cm12\text{ cm}.

Explanation:

In an isosceles triangle, the perpendicular height bisects the base, creating two right-angled triangles where the vertical height is one of the legs.