CBSE Class 10 Mathematics Triangles Notes Set 06

Pythagoras Theorem

Theorem 7. In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Given: A right-angled triangle ABC in which ∠B = 90°.
To prove: (Hypotenuse)² = (Base)² + (Perpendicular)², that is AC² = AB² + BC².
Construction: From B draw BD ⊥ AC.

Proof: In ΔADB and ΔABC: ∠ADB = ∠ABC (each 90°) and ∠A is common, so ΔADB ~ ΔABC (AA). Therefore\[ \frac{AD}{AB}=\frac{AB}{AC}\Rightarrow\ AB^2=AD\times AC\qquad\dots(i) \]In ΔBDC and ΔABC: ∠CDB = ∠ABC (each 90°) and ∠C is common, so ΔBDC ~ ΔABC (AA). Therefore\[ \frac{DC}{BC}=\frac{BC}{AC}\Rightarrow\ BC^2=AC\times DC\qquad\dots(ii) \]Adding (i) and (ii),\[ AB^2+BC^2=AC\,(AD+DC)=AC\times AC=AC^2 \]Hence AC² = AB² + BC². Hence proved.

Converse of Pythagoras Theorem

Theorem 8. In a triangle, if the square of the longest side is equal to the sum of the squares of the other two sides, then the angle opposite to the longest side is a right angle.

Given: A triangle ABC such that AC² = AB² + BC².
To prove: ∠B = 90°.
Construction: Construct a triangle XYZ with XY = AB, YZ = BC and ∠Y = 90°.

Proof: ΔXYZ is right-angled at Y, so by the Pythagoras theorem XZ² = XY² + YZ² = AB² + BC². But AB² + BC² = AC² (given). So XZ² = AC², that is XZ = AC. Now in ΔABC and ΔXYZ: AB = XY, BC = YZ and AC = XZ, so ΔABC ≅ ΔXYZ (SSS). Hence ∠B = ∠Y = 90° (CPCT), that is ABC is a triangle right-angled at B. Hence proved.

Knowledge Hub
(i) If ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced, then AC² = AB² + BC² + 2 BC × BD.
(ii) If ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC, then AC² = AB² + BC² − 2 BC × BD.

Example 1. The sides of a triangle are \( (2a-1) \) cm, \( 2\sqrt{2a} \) cm and \( (2a+1) \) cm. Determine whether it is a right triangle or not.
Answer:
Let p = \( (2a-1) \) cm, q = \( 2\sqrt{2a} \) cm and r = \( (2a+1) \) cm. The longest side is r.
\[ p^2+q^2=(2a-1)^2+\left(2\sqrt{2a}\right)^2=4a^2-4a+1+8a=4a^2+4a+1=(2a+1)^2=r^2 \]
By the converse of the Pythagoras theorem, the triangle is a right-angled triangle (right angle opposite the side (2a + 1) cm).

Example 2. A man goes 7 m due east and then 24 m due north. How far is he from the starting point?

Answer:
Let the starting point be O. The man walks OA = 7 m east and then AB = 24 m north, so ΔOAB is right-angled at A.
\[ OB^2=OA^2+AB^2=7^2+24^2=49+576=625\Rightarrow\ OB=\sqrt{625}=25 \]
He is 25 m from the starting point.

Example 3. If PQR is an equilateral triangle of side a, prove that its altitude = \( \frac{\sqrt3}{2}a \).

Answer:
Let PM ⊥ QR. In ΔPQM and ΔPRM: PQ = PR (given), PM is common and ∠PMQ = ∠PMR = 90°, so ΔPQM ≅ ΔPRM (RHS). Hence QM = RM = a/2 (CPCT).
In right ΔPQM, by the Pythagoras theorem PQ² = PM² + QM²:
\[ a^2=PM^2+\left(\frac a2\right)^2\Rightarrow\ PM^2=a^2-\frac{a^2}{4}=\frac{3a^2}{4}\Rightarrow\ PM=\frac{\sqrt3}{2}a \]
Hence proved.

Example 4. P and Q are the mid-points of the sides CA and CB respectively of ΔABC, right-angled at C. Prove that (i) 4AQ² = 4AC² + BC² (ii) 4BP² = 4BC² + AC².

Answer:
Since ∠C = 90°, ΔAQC and ΔBPC are right-angled at C. Also BC = 2QC and AC = 2CP.
(i) In ΔAQC, AQ² = AC² + QC². Multiplying by 4: 4AQ² = 4AC² + (2QC)² = 4AC² + BC².
(ii) In ΔBPC, BP² = BC² + CP². Multiplying by 4: 4BP² = 4BC² + (2CP)² = 4BC² + AC².
Hence proved.