Step-by-Step Proof of Pythagoras Theorem

Additional Problems and Applications of Pythagoras Theorem

1. Problem: Length of a Diagonal in a Rectangle

A rectangle has a length of 8 cm8 \, \text{cm} and a width of 6 cm6 \, \text{cm}. Find the length of its diagonal.

Solution

The diagonal divides the rectangle into two right triangles.
Using Pythagoras Theorem: d2=l2+w2d^2 = l^2 + w^2

Substitute l=8 cml = 8 \, \text{cm} and w=6 cmw = 6 \, \text{cm}: d2=82+62d^2 = 8^2 + 6^2 d2=64+36=100d^2 = 64 + 36 = 100 d=100=10 cmd = \sqrt{100} = 10 \, \text{cm}

The diagonal is 10 cm10 \, \text{cm}.

2. Problem: Distance Between Two Points

Find the distance between the points (2,3)(2, 3) and (6,7)(6, 7) using the Pythagoras Theorem.

Solution

The distance formula is derived from Pythagoras Theorem: d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}

Substitute (x1,y1)=(2,3)(x_1, y_1) = (2, 3) and (x2,y2)=(6,7)(x_2, y_2) = (6, 7): d=(6−2)2+(7−3)2d = \sqrt{(6 – 2)^2 + (7 – 3)^2} d=42+42d = \sqrt{4^2 + 4^2} d=16+16=32=42 unitsd = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \, \text{units}

3. Problem: Ladder Against a Wall

A ladder 10 m10 \, \text{m} long is leaning against a wall. The base of the ladder is 6 m6 \, \text{m} away from the wall. How high does the ladder reach on the wall?

Solution

This forms a right triangle where:

• Hypotenuse = 10 m10 \, \text{m} (ladder).
• Base = 6 m6 \, \text{m} (distance from the wall).
• Height = hh (to be found).

Using Pythagoras Theorem: h2+62=102h^2 + 6^2 = 10^2 h2+36=100h^2 + 36 = 100 h2=64h^2 = 64 h=64=8 mh = \sqrt{64} = 8 \, \text{m}

The ladder reaches 8 m8 \, \text{m} high on the wall.

4. Problem: Is it a Right Triangle?

Check whether the triangle with sides 7 cm,24 cm,and 25 cm7 \, \text{cm}, 24 \, \text{cm}, \text{and } 25 \, \text{cm} is a right triangle.

Solution

For a right triangle, the square of the hypotenuse should equal the sum of the squares of the other two sides.
Here, 25 cm25 \, \text{cm} is the largest side (hypotenuse).

Check: 252=242+7225^2 = 24^2 + 7^2 625=576+49625 = 576 + 49 625=625625 = 625

Since the equation holds, it is a right triangle.

Applications in Real Life

1. Navigation: Used in GPS to calculate the shortest distance between two points.
2. Construction: To ensure stability and symmetry in buildings, bridges, and roof designs.
3. Art and Design: To create geometric patterns or verify proportions.
4. Astronomy: For calculating distances between celestial bodies.

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Statement

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Mathematically: AC2=AB2+BC2AC^2 = AB^2 + BC^2

Step-by-Step Derivation

1. Construction

• Consider a right triangle △ABC\triangle ABC, where ∠A=90∘\angle A = 90^\circ.
• ABAB and BCBC are the two perpendicular sides, and ACAC is the hypotenuse.

2. Algebraic Representation

The coordinates of the points are:

• A(0,0)A(0, 0): Origin.
• B(b,0)B(b, 0): On the x-axis (AB=bAB = b).
• C(0,h)C(0, h): On the y-axis (BC=hBC = h).

Using the distance formula to find the hypotenuse ACAC: AC=(b−0)2+(h−0)2AC = \sqrt{(b – 0)^2 + (h – 0)^2} AC=b2+h2AC = \sqrt{b^2 + h^2}

3. Squaring Both Sides

To eliminate the square root: AC2=b2+h2AC^2 = b^2 + h^2

Here, b2b^2 represents AB2AB^2, and h2h^2 represents BC2BC^2. Thus: AC2=AB2+BC2AC^2 = AB^2 + BC^2

Example Problem

Question

In a right triangle, the lengths of the perpendicular sides are 3 cm3 \, \text{cm} and 4 cm4 \, \text{cm}. Find the length of the hypotenuse.

Solution

Using Pythagoras Theorem: AC2=AB2+BC2AC^2 = AB^2 + BC^2

Substitute AB=3 cmAB = 3 \, \text{cm} and BC=4 cmBC = 4 \, \text{cm}: AC2=32+42AC^2 = 3^2 + 4^2 AC2=9+16=25AC^2 = 9 + 16 = 25 AC=25=5 cmAC = \sqrt{25} = 5 \, \text{cm}

Thus, the hypotenuse is 5 cm5 \, \text{cm}.

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