Pythagoras theorem — proof and numerical problems with right triangles

Question

State and prove the Pythagoras theorem using the concept of similar triangles. Then solve: In a right triangle, the two legs are 6 cm and 8 cm. Find the hypotenuse. Also, if the hypotenuse is 13 cm and one leg is 5 cm, find the other leg.

(NCERT Class 10 — essential for CBSE boards)

Solution — Step by Step

In a right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides.

c^2 = a^2 + b^2

where $c$ is the hypotenuse (the side opposite the right angle), and $a$ , $b$ are the other two sides (legs).

Let triangle $ABC$ be right-angled at $B$ . Draw the altitude $BD$ from $B$ to hypotenuse $AC$ .

This altitude creates two smaller triangles: $\triangle ADB$ and $\triangle BDC$ .

Key result: $\triangle ADB \sim \triangle ABC \sim \triangle BDC$ (all three are similar, by AA similarity).

From $\triangle ADB \sim \triangle ABC$ :

\frac{AD}{AB} = \frac{AB}{AC} \implies AB^2 = AD \cdot AC \quad \text{...(i)}

From $\triangle BDC \sim \triangle ABC$ :

\frac{DC}{BC} = \frac{BC}{AC} \implies BC^2 = DC \cdot AC \quad \text{...(ii)}

Adding (i) and (ii):

AB^2 + BC^2 = AD \cdot AC + DC \cdot AC = AC(AD + DC) = AC \cdot AC = AC^2

\boxed{AB^2 + BC^2 = AC^2}

Legs: $a = 6$ cm, $b = 8$ cm.

c^2 = 6^2 + 8^2 = 36 + 64 = 100

c = \sqrt{100} = \mathbf{10 \text{ cm}}

This is the classic 3-4-5 Pythagorean triple scaled by 2: $(6, 8, 10)$ .

Hypotenuse: $c = 13$ cm, one leg: $a = 5$ cm.

b^2 = c^2 - a^2 = 169 - 25 = 144

b = \sqrt{144} = \mathbf{12 \text{ cm}}

This is the $(5, 12, 13)$ Pythagorean triple.

Why This Works

The Pythagoras theorem connects geometry to algebra. The altitude from the right angle vertex to the hypotenuse creates similar triangles (same angles, different sizes). Similar triangles give us proportional sides, and those proportions lead directly to the algebraic relationship $a^2 + b^2 = c^2$ .

Geometrically, $a^2$ represents the area of a square built on side $a$ , and similarly for $b^2$ and $c^2$ . The theorem says: the area of the square on the hypotenuse equals the total area of the squares on the two legs.

Alternative Method — Common Pythagorean Triples

Memorise these triples to speed up calculations:

Triple	Multiples
$(3, 4, 5)$	$(6, 8, 10)$ , $(9, 12, 15)$ , $(15, 20, 25)$
$(5, 12, 13)$	$(10, 24, 26)$
$(8, 15, 17)$	$(16, 30, 34)$
$(7, 24, 25)$	—

For CBSE Class 10 boards: the proof using similar triangles is the NCERT-prescribed proof. Write it with a clear figure and mention “AA similarity” as the reason. The two-column proof format (statement-reason) is preferred by examiners. Also, the converse is important: if $a^2 + b^2 = c^2$ , then the triangle is right-angled. CBSE sometimes asks to verify whether a triangle with given sides is right-angled.

Common Mistake

The most common error: applying the theorem to non-right triangles. The Pythagoras theorem works ONLY for right-angled triangles. If the triangle doesn’t have a $90°$ angle, you need the cosine rule instead: $c^2 = a^2 + b^2 - 2ab\cos C$ . Also, always identify the hypotenuse correctly — it’s the longest side and is opposite the right angle. Students sometimes square the wrong side, leading to incorrect results.

Question

Solution — Step by Step

Why This Works

Alternative Method — Common Pythagorean Triples

Common Mistake

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