Converse of Pythagoras Theorem and Geometry Applications for CBSE Class 10

While the Pythagoras Theorem states that every right-angled triangle satisfies $a^2 + b^2 = c^2$, geometry often presents the inverse challenge: Given three side lengths of a triangle, how can we prove that one of its angles is strictly a right angle ($90^\circ$)?

The Converse of Pythagoras Theorem provides the definitive criterion to verify right angles. Together, the Pythagoras Theorem and its converse form the foundation for solving practical distance problems—from surveying land boundaries and navigating aircraft flights to ladder positioning and proving structural riders in CBSE Class 10 Mathematics.

What You Will Learn

• Statement and rigorous geometric proof of the Converse of Pythagoras Theorem
• Testing whether given numerical side lengths form a right triangle (Pythagorean Triples)
• Application 1: Ladder leaning against a vertical wall
• Application 2: Aircraft flight distance vectors (directional bearings)
• Application 3: The Equilateral Triangle Altitude Theorem ($3AB^2 = 4AD^2$)
• Application 4: The British Flag Theorem rider for a point inside a rectangle
• Board exam presentation templates and common algebraic traps

1. Statement of the Converse of Pythagoras Theorem

Theorem Statement (CBSE Theorem 6.9)

In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

$\text{In } \Delta ABC, \quad \text{if } \mathbf{AC^2 = AB^2 + BC^2}, \quad \text{then } \mathbf{\angle B = 90^\circ}.$

2. Geometric Proof of the Converse

              A                                           P
             /|                                          /|
            / |                                         / |
           /  |                                        /  |
          /   |                                       /   |
         C----B                                      R----Q (90°)
       (Given: AC² = AB² + BC²)                    (Constructed: ∠Q = 90°)

Given:

A triangle $ABC$ in which $AC^2 = AB^2 + BC^2$.

To Prove:

$\angle B = 90^\circ$

Construction:

Construct a right-angled triangle $PQR$, right-angled at $Q$ ($\angle Q = 90^\circ$), such that: $PQ = AB \quad \text{and} \quad QR = BC$

Proof:

1. In the constructed right triangle $\Delta PQR$: Since $\angle Q = 90^\circ$, apply the Pythagoras Theorem: $PR^2 = PQ^2 + QR^2$
2. Substitute $PQ = AB$ and $QR = BC$ from our construction: $PR^2 = AB^2 + BC^2 \quad \text{--- (1)}$
3. Compare with the given condition: We are given that: $AC^2 = AB^2 + BC^2 \quad \text{--- (2)}$
4. Equate Equations (1) and (2): $PR^2 = AC^2 \implies PR = AC \quad \text{--- (3)}$
5. Prove Congruence of $\Delta ABC$ and $\Delta PQR$: In $\Delta ABC$ and $\Delta PQR$:
   • $AB = PQ$ (By construction)
   • $BC = QR$ (By construction)
   • $AC = PR$ (Proved above in (3)) Therefore, by the SSS Congruence Criterion: $\Delta ABC \cong \Delta PQR$
6. Apply Corresponding Parts of Congruent Triangles (CPCT): $\angle B = \angle Q$ Since $\angle Q = 90^\circ$ by construction: $\mathbf{\angle B = 90^\circ}$ Hence, proved.

3. Testing Side Lengths for Right Triangles

To check if sides $a, b, c$ (with $c$ being the longest side) form a right-angled triangle:

1. Calculate $a^2 + b^2$.
2. Calculate $c^2$.
3. If $a^2 + b^2 = c^2$, it is a right triangle (hypotenuse $= c$).

Common Pythagorean Triples:

• $3, 4, 5$ ($3^2 + 4^2 = 9 + 16 = 25 = 5^2$)
• $5, 12, 13$ ($5^2 + 12^2 = 25 + 144 = 169 = 13^2$)
• $7, 24, 25$ ($7^2 + 24^2 = 49 + 576 = 625 = 25^2$)
• $8, 15, 17$ ($8^2 + 15^2 = 64 + 225 = 289 = 17^2$)

4. High-Yield Practical Applications (NCERT Classics)

Application 1: The Leaning Ladder Problem

Problem: A ladder $10\text{ m}$ long reaches a window $8\text{ m}$ above the ground. Find the distance of the foot of the ladder from the base of the wall.

Solution:

1. Let the wall be $AB = 8\text{ m}$, the ladder be $AC = 10\text{ m}$, and distance from wall be $BC$.
2. The wall stands vertically upright on level ground, so $\angle B = 90^\circ$.
3. By Pythagoras Theorem: $AC^2 = AB^2 + BC^2$ $10^2 = 8^2 + BC^2 \implies 100 = 64 + BC^2$ $BC^2 = 100 - 64 = 36 \implies BC = \sqrt{36} = 6\text{ m}$
4. Therefore, <u>the foot of the ladder is $6\text{ metres}$ from the wall</u>.

Application 2: Aeroplane Flight Vectors (NCERT Classic)

Problem: An aeroplane leaves an airport and flies due north at a speed of $1000\text{ km/h}$. At the same time, another aeroplane leaves the same airport and flies due west at a speed of $1200\text{ km/h}$. How far apart will the two planes be after $1\frac{1}{2}\text{ hours}$?

Solution:

1. Time $t = 1\frac{1}{2}\text{ hours} = \frac{3}{2}\text{ hours}$.
2. Distance travelled North ($OA$): $OA = 1000\text{ km/h} \times \frac{3}{2}\text{ h} = 1500\text{ km}$
3. Distance travelled West ($OB$): $OB = 1200\text{ km/h} \times \frac{3}{2}\text{ h} = 1800\text{ km}$
4. North and West directions are perpendicular to each other, so $\angle AOB = 90^\circ$.
5. By Pythagoras Theorem in right triangle $\Delta AOB$: $AB^2 = OA^2 + OB^2$ $AB^2 = (1500)^2 + (1800)^2 = 2250000 + 3240000 = 5490000$ $AB = \sqrt{5490000} = \sqrt{90000 \times 61} = 300\sqrt{61}\text{ km}$
6. Therefore, <u>the two aeroplanes will be $300\sqrt{61}\text{ km}$ apart</u>.

Application 3: Equilateral Triangle Altitude Theorem (CBSE Board Classic)

Problem: In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes ($3AB^2 = 4AD^2$).

Solution:

1. Let $\Delta ABC$ be an equilateral triangle with side $a$ ($AB = BC = CA = a$).
2. Draw altitude $AD \perp BC$.
3. In an equilateral triangle, the altitude bisects the opposite base: $BD = DC = \frac{a}{2}$
4. In right-angled triangle $\Delta ADB$ ($\angle ADB = 90^\circ$): $AB^2 = AD^2 + BD^2$
5. Substitute $AB = a$ and $BD = \frac{a}{2}$: $a^2 = AD^2 + \left(\frac{a}{2}\right)^2 = AD^2 + \frac{a^2}{4}$
6. Transpose: $a^2 - \frac{a^2}{4} = AD^2 \implies \frac{3a^2}{4} = AD^2$ $3a^2 = 4AD^2$
7. Since $a = AB$, substitute $a = AB$: $\mathbf{3AB^2 = 4AD^2}$ Hence, proved.

5. Summary and Examination Tips

Exam Tip: In the proof of the Converse of Pythagoras Theorem, remember that you MUST construct a separate triangle $\Delta PQR$ with a right angle at $Q$. You cannot assume $\Delta ABC$ is a right triangle until the end of the proof!

Common Mistake: When checking triples, students sometimes forget to identify the longest side. In $a^2 + b^2 = c^2$, $c$ must ALWAYS be the largest number among the three sides.