krit.club logo

Geometry and Trigonometry - Pythagorean theorem and its applications

Grade 10IB

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

🔑Concepts

•

The Pythagorean theorem states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. This is expressed as a2+b2=c2a^2 + b^2 = c^2, where cc is the hypotenuse.

A right-angled triangle with sides labeled a, b and hypotenuse c.
•

The Converse of the Pythagorean Theorem allows us to determine if a triangle is right-angled. If the side lengths satisfy the condition a2+b2=c2a^2 + b^2 = c^2, the triangle must contain a right angle (90∘90^\circ) opposite side cc.

•

Pythagorean triples are sets of three positive integers (a,b,c)(a, b, c) that satisfy the theorem. Common examples include (3,4,5)(3, 4, 5), (5,12,13)(5, 12, 13), and (8,15,17)(8, 15, 17).

•

In 3D geometry, the theorem can be extended to find the space diagonal of a rectangular prism (cuboid). The distance from one corner to the opposite corner is given by D=l2+w2+h2D = \sqrt{l^2 + w^2 + h^2}.

A cuboid showing the space diagonal D from the bottom-front-left corner to the top-back-right corner.

📐Formulae

a2+b2=c2a^2 + b^2 = c^2

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

a=c2−b2a = \sqrt{c^2 - b^2}

d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

D=l2+w2+h2D = \sqrt{l^2 + w^2 + h^2}

💡Examples

Problem 1:

A 13m long ladder is leaning against a vertical wall. The base of the ladder is 5m away from the wall on horizontal ground. How high up the wall does the ladder reach?

Solution:

Step 1: Identify the parts of the triangle. The ladder is the hypotenuse (c=13c = 13), and the distance from the wall is one leg (b=5b = 5). We need to find the height (aa). Step 2: Use the rearranged formula a=c2−b2a = \sqrt{c^2 - b^2}. Step 3: Substitute the values: a=132−52a = \sqrt{13^2 - 5^2}. Step 4: Calculate the squares: a=169−25a = \sqrt{169 - 25}. Step 5: Subtract: a=144a = \sqrt{144}. Step 6: Solve the square root: a=12a = 12.

Explanation:

In this real-world application, the wall and the ground form a 90∘90^{\circ} angle. Since we are looking for one of the shorter sides (the height), we subtract the square of the known side from the square of the hypotenuse.

Problem 2:

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

Solution:

Step 1: Identify the longest side as the potential hypotenuse (c=25c = 25) and the other two as legs (a=7,b=24a = 7, b = 24). Step 2: Calculate a2+b2a^2 + b^2: 72+242=49+576=6257^2 + 24^2 = 49 + 576 = 625. Step 3: Calculate c2c^2: 252=62525^2 = 625. Step 4: Compare the results: Since a2+b2=c2a^2 + b^2 = c^2 (625=625625 = 625), the condition is met.

Explanation:

This uses the Converse of the Pythagorean theorem. Because the square of the longest side equals the sum of the squares of the other two sides, the triangle must be right-angled.

Problem 3:

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

A rectangle with a diagonal line labeled d, length 15 cm, and width 8 cm.

Solution:

  1. Let the diagonal be dd, length l=15l = 15, and width w=8w = 8.
  2. Apply the Pythagorean theorem: d2=l2+w2d^2 = l^2 + w^2
  3. Substitute the values: d2=152+82=225+64=289d^2 = 15^2 + 8^2 = 225 + 64 = 289
  4. Solve for dd: d=289=17 cmd = \sqrt{289} = 17 \text{ cm}

Explanation:

A rectangle's diagonal splits it into two identical right-angled triangles. We use the length and width as the two shorter sides to find the hypotenuse.

Problem 4:

A ship travels 24 km24 \text{ km} due North and then 10 km10 \text{ km} due East. What is the direct distance from the starting point to the final position?

A vector diagram showing a movement 24 units up and 10 units right, with the resultant hypotenuse x.

Solution:

  1. The path forms a right-angled triangle where the legs are 24 km24 \text{ km} and 10 km10 \text{ km}.
  2. Let the direct distance be xx: x2=242+102x^2 = 24^2 + 10^2
  3. x2=576+100=676x^2 = 576 + 100 = 676
  4. x=676=26 kmx = \sqrt{676} = 26 \text{ km}

Explanation:

The northward and eastward paths are perpendicular to each other, creating a right angle. The direct distance is the hypotenuse of the triangle formed.