This MCQ module is based on: 6.5 Pythagoras Theorem (via Similarity)
TOPIC 16 OF 35
6.5 Pythagoras Theorem (via Similarity)
🎓 Class 10
Mathematics
CBSE
Theory
Ch 6 — Triangles
⏱ ~28 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: 6.5 Pythagoras Theorem (via Similarity)
Targeting Class 10 level in Geometry, with Intermediate difficulty.
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6.5 Pythagoras Theorem (via Similarity)
The Pythagoras theoremⓘ is one of the oldest and most celebrated results in mathematics. Having established the similarity of the triangles produced when an altitude is dropped from the right-angle vertex onto the hypotenuse (Example 9 of Part 2), we can now give a clean, similarity-based proof. Before stating the main theorem, we first prove a lemma that is its essential ingredient.
Theorem 6.6 — Altitude on Hypotenuse Lemma
If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, then the triangles on both sides of the perpendicular are similar to the whole triangle and to each other.Let \(\triangle ABC\) be right-angled at B, and let BD be perpendicular to AC (D on AC). Then \(\triangle ADB \sim \triangle ABC\), \(\triangle BDC \sim \triangle ABC\), and consequently \(\triangle ADB \sim \triangle BDC\).
Fig 6.7: Right triangle ABC (right angle at B); altitude BD drops to hypotenuse AC.
Proof of Theorem 6.6
In \(\triangle ADB\) and \(\triangle ABC\): \(\angle A\) is common and \(\angle ADB = \angle ABC = 90°\). By AA similarity, \(\triangle ADB \sim \triangle ABC\).In \(\triangle BDC\) and \(\triangle ABC\): \(\angle C\) is common and \(\angle BDC = \angle ABC = 90°\). By AA, \(\triangle BDC \sim \triangle ABC\).
Since similarity is transitive, \(\triangle ADB \sim \triangle BDC\). ∎
Theorem 6.7 — Pythagoras' Theorem
Theorem 6.7
In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.That is, if \(\triangle ABC\) is right-angled at B, then \(AC^2 = AB^2 + BC^2\).
Proof (using Similarity)
Drop perpendicular BD from B onto AC. By Theorem 6.6, \(\triangle ADB \sim \triangle ABC\). So
\[\frac{AD}{AB}=\frac{AB}{AC}\quad\Longrightarrow\quad AB^2 = AD\cdot AC \quad (*)\]
Similarly, \(\triangle BDC \sim \triangle ABC\) gives
\[\frac{CD}{BC}=\frac{BC}{AC}\quad\Longrightarrow\quad BC^2 = CD\cdot AC \quad (**)\]
Adding (*) and (**):
\[AB^2 + BC^2 = (AD + CD)\cdot AC = AC\cdot AC = AC^2.\]
Hence \(AC^2 = AB^2 + BC^2\). ∎
Theorem 6.8 — Converse of Pythagoras' Theorem
Theorem 6.8 — Converse
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.If in \(\triangle ABC\), \(AC^2 = AB^2 + BC^2\), then \(\angle B = 90°\).
Proof (Converse)
Construct a right triangle \(\triangle PQR\) right-angled at Q with PQ = AB and QR = BC. By Theorem 6.7, \(PR^2 = PQ^2 + QR^2 = AB^2 + BC^2 = AC^2\), so PR = AC.Now in \(\triangle ABC\) and \(\triangle PQR\), the three sides are equal (AB = PQ, BC = QR, AC = PR). By SSS congruence, \(\triangle ABC \cong \triangle PQR\). Therefore \(\angle B = \angle Q = 90°\). ∎
Fig 6.8: Converse — building a right triangle PQR with PQ = AB, QR = BC forces PR = AC, hence congruence and \(\angle B = 90°\).
Activity: Verify Pythagoras by Paper Cut-out
L3 ApplyMaterials: Graph paper, ruler, scissors, pencil.
Predict: For a right triangle with legs 3 cm and 4 cm, can you cover the square on the hypotenuse using only the squares built on the two legs?
- On graph paper, draw a right triangle ABC with right angle at B, AB = 3 cm, BC = 4 cm. Measure AC.
- Construct squares externally on each of the three sides. Count the unit squares inside each.
- Record: square on AB = 9, square on BC = 16, square on AC = ?
- Cut out the two smaller squares, rearrange the 9 + 16 = 25 unit pieces to fit exactly inside the hypotenuse square.
- Conclude: \(AB^2 + BC^2 = AC^2\), so AC = 5 cm.
The unit squares fill the hypotenuse-square exactly. This geometric "dissection proof" was known to Indian mathematicians in the Sulbasutras (c. 800 BCE), long before Pythagoras. The similarity-based proof we gave is a more modern, algebraic route to the same eternal truth.
Worked Examples (9–14)
Example 9 — Altitude on hypotenuse (foundational)
In \(\triangle PQR\), right-angled at Q, QM is drawn perpendicular to PR. Prove that \(QM^2 = PM \cdot MR\).
Solution. By Theorem 6.6, \(\triangle PMQ \sim \triangle QMR\). So \(\dfrac{PM}{QM}=\dfrac{QM}{MR}\), giving \(QM^2 = PM \cdot MR\). ∎
Example 10 — Ladder against a wall
A ladder 10 m long reaches a window 8 m above the ground on one side of a street. Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window 6 m high. Find the width of the street.
Solution. In the first right triangle, horizontal distance \(= \sqrt{10^2 - 8^2} = \sqrt{36} = 6\) m. In the second right triangle, horizontal distance \(= \sqrt{10^2 - 6^2} = \sqrt{64} = 8\) m. Width of street \(= 6 + 8 = 14\) m.
Fig 6.9: Ladder pivoting at one foot; reaches 8 m on left wall and 6 m on right wall.
Example 11 — Diagonal of a rectangle
Find the length of the diagonal of a rectangle of sides 24 cm and 10 cm.
Solution. The diagonal is the hypotenuse of a right triangle with legs 24 and 10. \(d = \sqrt{24^2 + 10^2} = \sqrt{576+100} = \sqrt{676} = 26\) cm.
Example 12 — Is it a right triangle?
Sides of a triangle are 7 cm, 24 cm, 25 cm. Is it a right triangle? If yes, name the right-angle vertex opposite the longest side.
Solution. Longest side = 25. Check: \(7^2 + 24^2 = 49 + 576 = 625 = 25^2\). By the converse of Pythagoras' theorem, the triangle is right-angled at the vertex opposite the side of length 25.
Example 13 — Aeroplane problem
Two aeroplanes leave the same airport at the same time. One flies due north at 1000 km/h and the other due west at 1200 km/h. How far apart are they after 1.5 hours?
Solution. Distance north = 1000 × 1.5 = 1500 km. Distance west = 1200 × 1.5 = 1800 km. The paths are perpendicular, so the separation is \(\sqrt{1500^2 + 1800^2} = \sqrt{2{,}250{,}000 + 3{,}240{,}000} = \sqrt{5{,}490{,}000} = 300\sqrt{61}\) km ≈ 2343.1 km.
Example 14 — Equilateral triangle altitude
Prove that in an equilateral triangle of side \(a\), the square of the altitude equals three times the square of half the side; i.e., \(h^2 = 3(a/2)^2\).
Solution. In equilateral \(\triangle ABC\) with side \(a\), drop altitude AD from A onto BC. Then D is the midpoint of BC (altitude = median in equilateral), so BD = \(a/2\). In right \(\triangle ABD\), \(AB^2 = AD^2 + BD^2 \Rightarrow a^2 = h^2 + (a/2)^2 \Rightarrow h^2 = a^2 - a^2/4 = 3a^2/4 = 3(a/2)^2\). Hence \(h = \tfrac{\sqrt 3}{2}a\). ∎
Fig 6.10: Equilateral triangle with altitude AD; D is the midpoint of BC.
Exercise 6.5
Q1. Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse. (i) 7 cm, 24 cm, 25 cm (ii) 3 cm, 8 cm, 6 cm (iii) 50 cm, 80 cm, 100 cm (iv) 13 cm, 12 cm, 5 cm.
(i) \(7^2+24^2 = 49+576=625=25^2\). Right triangle, hypotenuse 25 cm.
(ii) \(3^2+6^2=9+36=45\ne 64=8^2\). Not a right triangle.
(iii) \(50^2+80^2=2500+6400=8900\ne 10000=100^2\). Not a right triangle.
(iv) \(5^2+12^2=25+144=169=13^2\). Right triangle, hypotenuse 13 cm.
(ii) \(3^2+6^2=9+36=45\ne 64=8^2\). Not a right triangle.
(iii) \(50^2+80^2=2500+6400=8900\ne 10000=100^2\). Not a right triangle.
(iv) \(5^2+12^2=25+144=169=13^2\). Right triangle, hypotenuse 13 cm.
Q2. PQR is a triangle right-angled at P and M is a point on QR such that PM ⊥ QR. Show that \(PM^2 = QM \cdot MR\).
By Theorem 6.6, \(\triangle QMP \sim \triangle PMR\). So \(\dfrac{QM}{PM}=\dfrac{PM}{MR}\Rightarrow PM^2 = QM\cdot MR\). ∎
Q3. In the figure, ABD is a triangle right-angled at A and AC ⊥ BD. Show that (i) \(AB^2 = BC\cdot BD\) (ii) \(AC^2 = BC\cdot DC\) (iii) \(AD^2 = BD\cdot CD\).
All three follow from the altitude-on-hypotenuse lemma applied to right \(\triangle ABD\) with altitude AC.
(i) \(\triangle ABC \sim \triangle DBA \Rightarrow AB/BD = BC/AB \Rightarrow AB^2 = BC\cdot BD\).
(ii) \(\triangle ACB \sim \triangle DCA \Rightarrow AC/DC = BC/AC \Rightarrow AC^2 = BC\cdot DC\).
(iii) \(\triangle ACD \sim \triangle BAD \Rightarrow AD/BD = CD/AD \Rightarrow AD^2 = BD\cdot CD\).
(i) \(\triangle ABC \sim \triangle DBA \Rightarrow AB/BD = BC/AB \Rightarrow AB^2 = BC\cdot BD\).
(ii) \(\triangle ACB \sim \triangle DCA \Rightarrow AC/DC = BC/AC \Rightarrow AC^2 = BC\cdot DC\).
(iii) \(\triangle ACD \sim \triangle BAD \Rightarrow AD/BD = CD/AD \Rightarrow AD^2 = BD\cdot CD\).
Q4. ABC is an isosceles triangle right-angled at C. Prove that \(AB^2 = 2 AC^2\).
Right angle at C, so \(AB^2 = AC^2 + BC^2\). Isosceles with \(AC = BC\) (legs equal): \(AB^2 = AC^2 + AC^2 = 2AC^2\). ∎
Q5. ABC is an isosceles triangle with \(AC = BC\). If \(AB^2 = 2 AC^2\), prove that ABC is right-angled at C.
\(AB^2 = 2 AC^2 = AC^2 + AC^2 = AC^2 + BC^2\) (since AC = BC). By the converse of Pythagoras (Theorem 6.8), the right angle is opposite AB, i.e., at C. ∎
Q6. ABC is an equilateral triangle of side \(2a\). Find each of its altitudes.
Altitude \(h = \tfrac{\sqrt 3}{2}\cdot(\text{side}) = \tfrac{\sqrt 3}{2}\cdot 2a = \sqrt 3\, a\).
Q7. Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
In rhombus ABCD, diagonals AC and BD bisect each other at right angles at O. So \(AO = AC/2,\ BO = BD/2\). In right \(\triangle AOB\): \(AB^2 = AO^2 + BO^2 = \tfrac14 AC^2 + \tfrac14 BD^2\). Multiplying by 4: \(4 AB^2 = AC^2 + BD^2\). Since all four sides equal AB: \(AB^2 + BC^2 + CD^2 + DA^2 = 4 AB^2 = AC^2 + BD^2\). ∎
Q8. In the figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that (i) \(OA^2 + OB^2 + OC^2 - OD^2 - OE^2 - OF^2 = AF^2 + BD^2 + CE^2\) (ii) \(AF^2 + BD^2 + CE^2 = AE^2 + CD^2 + BF^2\).
Apply Pythagoras in the six right triangles formed by the feet of perpendiculars:
\(OA^2 = OF^2 + AF^2 = OE^2 + AE^2\); \(OB^2 = OF^2 + BF^2 = OD^2 + BD^2\); \(OC^2 = OD^2 + CD^2 = OE^2 + CE^2\).
(i) Add: \(OA^2+OB^2+OC^2 = (OF^2+AF^2)+(OD^2+BD^2)+(OE^2+CE^2)\). Rearrange ⇒ required identity.
(ii) From the two expressions for each of \(OA^2, OB^2, OC^2\): \(AF^2 + BD^2 + CE^2 = AE^2 + BF^2 + CD^2\). ∎
\(OA^2 = OF^2 + AF^2 = OE^2 + AE^2\); \(OB^2 = OF^2 + BF^2 = OD^2 + BD^2\); \(OC^2 = OD^2 + CD^2 = OE^2 + CE^2\).
(i) Add: \(OA^2+OB^2+OC^2 = (OF^2+AF^2)+(OD^2+BD^2)+(OE^2+CE^2)\). Rearrange ⇒ required identity.
(ii) From the two expressions for each of \(OA^2, OB^2, OC^2\): \(AF^2 + BD^2 + CE^2 = AE^2 + BF^2 + CD^2\). ∎
Q9. A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from the base of the wall.
\(d = \sqrt{10^2 - 8^2} = \sqrt{36} = 6\) m.
Q10. A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?
\(d = \sqrt{24^2 - 18^2} = \sqrt{576 - 324} = \sqrt{252} = 6\sqrt{7}\) m ≈ 15.87 m.
Q11. An aeroplane leaves an airport and flies due north at 1000 km/h. At the same time, another aeroplane leaves the same airport and flies due west at 1200 km/h. How far apart will be the two planes after \(1\tfrac12\) hours?
Distances: north 1500 km, west 1800 km. Paths are perpendicular. Separation \(= \sqrt{1500^2 + 1800^2} = \sqrt{5{,}490{,}000} = 300\sqrt{61}\) km.
Q12. Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.
Vertical difference = 11 − 6 = 5 m; horizontal = 12 m. Distance between tops = \(\sqrt{5^2+12^2} = \sqrt{169} = 13\) m.
Q13. D and E are points on the sides CA and CB respectively of a triangle ABC right-angled at C. Prove that \(AE^2 + BD^2 = AB^2 + DE^2\).
Right angle at C. By Pythagoras in the four right triangles: \(AE^2 = AC^2 + CE^2\); \(BD^2 = BC^2 + CD^2\); \(AB^2 = AC^2 + BC^2\); \(DE^2 = CD^2 + CE^2\). Adding the first two: \(AE^2 + BD^2 = AC^2 + CE^2 + BC^2 + CD^2 = (AC^2 + BC^2) + (CE^2 + CD^2) = AB^2 + DE^2\). ∎
Q14. The perpendicular from A on side BC of a \(\triangle ABC\) intersects BC at D such that DB = 3 CD. Prove that \(2 AB^2 = 2 AC^2 + BC^2\).
Let CD = x, so DB = 3x, BC = 4x. AD = h (altitude).
\(AB^2 = h^2 + (3x)^2 = h^2 + 9x^2\); \(AC^2 = h^2 + x^2\).
\(AB^2 - AC^2 = 8x^2 = 8\cdot(BC/4)^2 = \tfrac{BC^2}{2}\).
So \(2(AB^2 - AC^2) = BC^2\), i.e., \(2AB^2 = 2AC^2 + BC^2\). ∎
\(AB^2 = h^2 + (3x)^2 = h^2 + 9x^2\); \(AC^2 = h^2 + x^2\).
\(AB^2 - AC^2 = 8x^2 = 8\cdot(BC/4)^2 = \tfrac{BC^2}{2}\).
So \(2(AB^2 - AC^2) = BC^2\), i.e., \(2AB^2 = 2AC^2 + BC^2\). ∎
Q15. In an equilateral triangle ABC, D is a point on side BC such that BD = \(\tfrac13\) BC. Prove that \(9 AD^2 = 7 AB^2\).
Let AB = a. Drop altitude AM to BC; M is midpoint, BM = a/2, AM = \(\tfrac{\sqrt3}{2}a\). BD = a/3, so DM = BM − BD = a/2 − a/3 = a/6. In right \(\triangle ADM\): \(AD^2 = AM^2 + DM^2 = \tfrac{3a^2}{4} + \tfrac{a^2}{36} = \tfrac{27a^2 + a^2}{36} = \tfrac{28 a^2}{36} = \tfrac{7 a^2}{9}\). Hence \(9 AD^2 = 7 a^2 = 7 AB^2\). ∎
Q16. 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.
Altitude \(h = \tfrac{\sqrt 3}{2} a\Rightarrow h^2 = \tfrac{3 a^2}{4}\). Hence \(4 h^2 = 3 a^2\), i.e., \(3\cdot\text{side}^2 = 4\cdot\text{altitude}^2\). ∎
Q17. Tick the correct answer and justify: In \(\triangle ABC\), AB = \(6\sqrt 3\) cm, AC = 12 cm and BC = 6 cm. The angle B is: (A) 120° (B) 60° (C) 90° (D) 45°.
\(AB^2 + BC^2 = (6\sqrt 3)^2 + 6^2 = 108 + 36 = 144 = 12^2 = AC^2\). By the converse of Pythagoras, the right angle is opposite AC, i.e., at vertex B. (C) 90°.
Exercise 6.6 (Optional — for enrichment)
Q1. In the figure, PS is the bisector of \(\angle QPR\) of \(\triangle PQR\). Prove that \(\dfrac{QS}{SR}=\dfrac{PQ}{PR}\) (Angle-Bisector Theorem).
Through R, draw a line parallel to PS meeting QP produced at T. Then \(\angle RPS = \angle PRT\) (alternate) and \(\angle QPS = \angle PTR\) (corresponding). Since PS bisects \(\angle QPR\), \(\angle PRT = \angle PTR\), hence PR = PT. In \(\triangle QRT\), PS ∥ RT, so by BPT \(\dfrac{QS}{SR} = \dfrac{QP}{PT} = \dfrac{PQ}{PR}\). ∎
Q2. In the figure, D is a point on hypotenuse AC of \(\triangle ABC\), DM ⊥ BC and DN ⊥ AB. Prove that (i) \(DM^2 = DN \cdot MC\) (ii) \(DN^2 = DM \cdot AN\).
BDMN is a rectangle (three right angles), so DN = BM and DM = BN. In right \(\triangle DMC\) with common angle at C, \(\triangle DMC \sim \triangle BMD\) (AA); hence \(\dfrac{DM}{BM} = \dfrac{MC}{DM}\Rightarrow DM^2 = BM\cdot MC = DN\cdot MC\).
(ii) Similarly using \(\triangle DNA \sim \triangle BND\) gives \(DN^2 = AN\cdot BN = AN\cdot DM\). ∎
(ii) Similarly using \(\triangle DNA \sim \triangle BND\) gives \(DN^2 = AN\cdot BN = AN\cdot DM\). ∎
Q3. In the figure, ABC is a triangle in which \(\angle ABC > 90°\) and AD ⊥ CB produced. Prove that \(AC^2 = AB^2 + BC^2 + 2\, BC\cdot BD\).
In right \(\triangle ADB\), \(AB^2 = AD^2 + BD^2\). In right \(\triangle ADC\), \(AC^2 = AD^2 + DC^2 = AD^2 + (BD+BC)^2 = (AD^2+BD^2) + BC^2 + 2BC\cdot BD = AB^2 + BC^2 + 2 BC\cdot BD\). ∎
Q4. In the figure, ABC is a triangle in which \(\angle ABC < 90°\) and AD ⊥ BC. Prove that \(AC^2 = AB^2 + BC^2 - 2\, BC\cdot BD\).
Analogous to Q3: \(AC^2 = AD^2 + DC^2 = AD^2 + (BC - BD)^2 = (AD^2 + BD^2) + BC^2 - 2BC\cdot BD = AB^2 + BC^2 - 2 BC\cdot BD\). ∎
Q5. In the figure, AD is a median of \(\triangle ABC\) and AM ⊥ BC. Prove that (i) \(AC^2 = AD^2 + BC\cdot DM + \bigl(\tfrac{BC}{2}\bigr)^2\) (ii) \(AB^2 = AD^2 - BC\cdot DM + \bigl(\tfrac{BC}{2}\bigr)^2\) (iii) \(AC^2 + AB^2 = 2 AD^2 + \tfrac12 BC^2\) (Apollonius' theorem).
Using Q3 on \(\triangle ADC\) (obtuse at D side with C) and Q4 on \(\triangle ADB\): (i) and (ii) follow directly. Adding: the \(BC\cdot DM\) terms cancel and the \((BC/2)^2\) terms add, giving (iii). ∎
Q6. Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.
In parallelogram ABCD, diagonals bisect each other at O. In \(\triangle ABC\), BO is a median, so by Apollonius (Q5): \(AB^2 + BC^2 = 2 BO^2 + \tfrac12 AC^2\). Similarly for \(\triangle ACD\) with median DO: \(AD^2 + CD^2 = 2 DO^2 + \tfrac12 AC^2\). Adding and using BO = DO = BD/2: \(AB^2 + BC^2 + CD^2 + DA^2 = 4\cdot(BD/2)^2 + AC^2 = BD^2 + AC^2\). ∎
Q7. In the figure, two chords AB and CD of a circle intersect at P. Prove that (i) \(\triangle APC \sim \triangle DPB\) (ii) \(AP\cdot PB = CP\cdot DP\).
\(\angle APC = \angle DPB\) (vertically opposite). \(\angle PAC = \angle PDB\) (angles in same segment on arc CB). By AA, \(\triangle APC \sim \triangle DPB\). Hence \(\dfrac{AP}{DP} = \dfrac{CP}{PB}\Rightarrow AP\cdot PB = CP\cdot DP\). ∎
Q8. In the figure, two chords AB and CD of a circle intersect each other at point P (when produced) outside the circle. Prove (i) \(\triangle PAC \sim \triangle PDB\) (ii) \(PA\cdot PB = PC\cdot PD\).
\(\angle P\) is common. \(\angle PAC = \angle PDB\) (exterior angle of cyclic quadrilateral = interior opposite). By AA similar. Ratio gives the product result. ∎
Q9. In the figure, D is a point on side BC of \(\triangle ABC\) such that \(\dfrac{BD}{CD} = \dfrac{AB}{AC}\). Prove that AD is the bisector of \(\angle BAC\).
This is the converse of the Angle-Bisector theorem (Q1). Produce AD-like construction: through C draw line parallel to the would-be bisector; using BPT in \(\triangle ABE\) with E on extension, show \(\angle BAD = \angle DAC\). ∎
Q10. Nazima is fly-fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away from a point on the water directly below the tip. Assuming her string (from the tip of the rod to the fly) is taut, how much string does she have out? If she pulls in the string at the rate of 5 cm/s, what will be the horizontal distance of the fly from her after 12 seconds?
String length \(= \sqrt{1.8^2 + 3.6^2} = \sqrt{3.24 + 12.96} = \sqrt{16.2} \approx 4.025\) m.
After 12 s at 5 cm/s, she pulls in 60 cm = 0.6 m. New string length = 4.025 − 0.6 ≈ 3.425 m. New horizontal distance = \(\sqrt{3.425^2 - 1.8^2} = \sqrt{11.73 - 3.24} = \sqrt{8.49} \approx 2.914\) m.
After 12 s at 5 cm/s, she pulls in 60 cm = 0.6 m. New string length = 4.025 − 0.6 ≈ 3.425 m. New horizontal distance = \(\sqrt{3.425^2 - 1.8^2} = \sqrt{11.73 - 3.24} = \sqrt{8.49} \approx 2.914\) m.
Competency-Based Questions
Scenario: A municipal corporation plans to install a straight zip-line from the top of a 24 m tall clock-tower to a ground point 7 m away from the tower's base. A safety inspector must determine the zip-line's length, its angle, and whether small design changes compromise the structure's geometry. She applies Pythagoras' theorem and its converse to verify the setup.
Q1. Compute the length of the zip-line from the top of the tower to the ground anchor.
L3 Apply(b) 25 m (and equivalently (d) \(\sqrt{625}\)). \(\text{length} = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25\) m.
Q2. Analyse: the engineer claims a triangle with sides 9 m, 40 m and 41 m is right-angled. Which side is the hypotenuse and which theorem justifies the claim?
L4 Analyse\(9^2 + 40^2 = 81 + 1600 = 1681 = 41^2\). The side of length 41 m is the hypotenuse. The converse of Pythagoras' theorem (Theorem 6.8) certifies the right angle (opposite the 41 m side).
Q3. Evaluate: if the tower's height is measured as 24.2 m instead of 24 m (1% error), by what percent does the zip-line length change?
L5 EvaluateNew length \(= \sqrt{24.2^2 + 7^2} = \sqrt{585.64 + 49} = \sqrt{634.64} \approx 25.19\) m. Change = 0.19 m on 25 m ≈ 0.77%. Pythagoras' theorem amplifies errors less than linearly for the dominant leg here — the vertical leg dominates the sum of squares, so small absolute errors propagate modestly.
Q4. Design a verification procedure, using only a long measuring tape, to confirm that a corner of a newly-laid rectangular floor is exactly 90°. Justify which theorem powers your check.
L6 CreateUse the 3–4–5 method: mark 3 m along one edge from the corner and 4 m along the adjacent edge. Measure the straight-line distance between these two marks; if it is exactly 5 m, the corner is 90° (since \(3^2 + 4^2 = 5^2\), by the converse of Pythagoras' theorem). For greater accuracy, use 6–8–10 or 9–12–15 (same ratio, larger scale). Record the reading; if ≠ 5 m, the corner is not square and must be re-laid.
Assertion–Reason Questions
Assertion (A): A triangle with sides 5 cm, 12 cm, 13 cm is right-angled.
Reason (R): If \(a^2 + b^2 = c^2\) then the angle opposite \(c\) is 90° (converse of Pythagoras).
Reason (R): If \(a^2 + b^2 = c^2\) then the angle opposite \(c\) is 90° (converse of Pythagoras).
Answer: (a). \(5^2 + 12^2 = 169 = 13^2\), and the converse gives the right angle at the vertex opposite 13.
Assertion (A): In any triangle, the square of the longest side always equals the sum of the squares of the other two.
Reason (R): Pythagoras' theorem holds for every triangle.
Reason (R): Pythagoras' theorem holds for every triangle.
Answer: (d) — actually both are false. Pythagoras' theorem holds only for right triangles; in general triangles the relation \(c^2 = a^2 + b^2 - 2ab\cos C\) (Law of Cosines) holds, with equality to \(a^2+b^2\) only when \(C = 90°\). Strictly per the four options, the best match is (d) treating R as ambiguously "true in the right-triangle context" — but a careful student marks both false.
Assertion (A): In a right triangle, the altitude from the right-angle vertex to the hypotenuse creates two smaller triangles similar to the original.
Reason (R): Each smaller triangle shares an acute angle with the original triangle and has a right angle.
Reason (R): Each smaller triangle shares an acute angle with the original triangle and has a right angle.
Answer: (a). This is exactly Theorem 6.6, proved by AA similarity.
6.6 Summary
This chapter connected geometric intuition (shape) with algebraic ratio (proportion). Below are the twelve ideas that every Class 10 student should carry forward.
- Two figures with the same shape but (possibly) different sizes are called similar.
- All congruent figures are similar, but similar figures need not be congruent.
- Two polygons of the same number of sides are similar if (a) corresponding angles are equal and (b) corresponding sides are in the same ratio (proportion).
- BPT (Thales' Theorem): If a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, it divides those two sides in the same ratio.
- Converse of BPT: If a line divides any two sides of a triangle in the same ratio, the line is parallel to the third side.
- Two triangles are similar if their corresponding angles are equal (AAA), and consequently if two pairs of angles are equal (AA).
- Two triangles are similar if their corresponding sides are in the same ratio (SSS similarity).
- Two triangles are similar if one pair of corresponding angles is equal and the sides including those angles are proportional (SAS similarity).
- If a perpendicular is drawn from the vertex of the right angle to the hypotenuse, the two triangles formed are similar to the whole triangle and to each other.
- Pythagoras' Theorem: In a right triangle, the square of the hypotenuse equals the sum of the squares of the other two sides.
- Converse of Pythagoras' Theorem: If the square of one side of a triangle equals the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.
- RHS Similarity (A Note to the Reader): If in two right triangles, the hypotenuse and one side of one triangle are proportional to the hypotenuse and one side of the other, the triangles are similar.
Frequently Asked Questions — Triangles
What is Part 3 — Pythagoras Theorem & Final Exercises | Class 10 Maths Ch 6 | MyAiSchool in NCERT Class 10 Mathematics?
Part 3 — Pythagoras Theorem & Final Exercises | Class 10 Maths Ch 6 | MyAiSchool is a key concept covered in NCERT Class 10 Mathematics, Chapter 6: Triangles. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.
How do I solve problems on Part 3 — Pythagoras Theorem & Final Exercises | Class 10 Maths Ch 6 | MyAiSchool step by step?
To solve problems on Part 3 — Pythagoras Theorem & Final Exercises | Class 10 Maths Ch 6 | MyAiSchool, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 10 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.
What are the most important formulas for Chapter 6: Triangles?
The essential formulas of Chapter 6 (Triangles) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.
Is Part 3 — Pythagoras Theorem & Final Exercises | Class 10 Maths Ch 6 | MyAiSchool important for the Class 10 board exam?
Part 3 — Pythagoras Theorem & Final Exercises | Class 10 Maths Ch 6 | MyAiSchool is part of the NCERT Class 10 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.
What mistakes should students avoid in Part 3 — Pythagoras Theorem & Final Exercises | Class 10 Maths Ch 6 | MyAiSchool?
Common mistakes in Part 3 — Pythagoras Theorem & Final Exercises | Class 10 Maths Ch 6 | MyAiSchool include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.
Where can I find more NCERT practice questions on Part 3 — Pythagoras Theorem & Final Exercises | Class 10 Maths Ch 6 | MyAiSchool?
End-of-chapter NCERT exercises for Part 3 — Pythagoras Theorem & Final Exercises | Class 10 Maths Ch 6 | MyAiSchool cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 6, and solve at least one previous-year board paper to consolidate your understanding.
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