CBSE Class 10 Mathematics Triangles Assignment Set 11

Exercise 5: Pythagoras Theorem

Question 1. 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
Answer:
Check whether the square of the longest side equals the sum of the squares of the other two (converse of the Pythagoras theorem).
(i) \( 7^2+24^2=49+576=625=25^2 \). Right triangle, hypotenuse 25 cm.
(ii) \( 3^2+6^2=45\neq 8^2=64 \). Not a right triangle.
(iii) \( 50^2+80^2=8900\neq 100^2=10000 \). Not a right triangle.
(iv) \( 12^2+5^2=169=13^2 \). Right triangle, hypotenuse 13 cm.
In simple words: Test the longest side first.

Teacher's Note:
1. Identify the hypotenuse (the longest side, opposite the right angle) before writing the Pythagoras relation.
2. In proofs, name the right-angled triangle in which you apply the theorem and write each relation as a numbered line.

Question 2. PQR is a triangle right-angled at P and M is a point on QR such that PM ⊥ QR. Show that PM² = QM × MR.

Answer:
In ΔPQM (right-angled at M), PQ² = PM² + QM², so PM² = PQ² − QM² ...(i).
In ΔPMR, PR² = PM² + MR², so PM² = PR² − MR² ...(ii).
Adding (i) and (ii): 2PM² = (PQ² + PR²) − (QM² + MR²) = QR² − QM² − MR² (as ∠P = 90°, QR² = PQ² + PR²).
With QR = QM + MR: 2PM² = (QM + MR)² − QM² − MR² = 2 QM × MR.
Hence PM² = QM × MR.

Teacher's Note:
1. Identify the hypotenuse (the longest side, opposite the right angle) before writing the Pythagoras relation.
2. In proofs, name the right-angled triangle in which you apply the theorem and write each relation as a numbered line.

Question 3. In the given figure, ABD is a triangle right-angled at A and AC ⊥ BD. Show that
(i) AB² = BC × BD   (ii) AC² = BC × DC   (iii) AD² = BD × CD

Answer:
(i) In ΔADB and ΔCAB: ∠DAB = ∠ACB = 90° and ∠B is common, so ΔADB ~ ΔCAB (AA). Hence \( \frac{AB}{CB}=\frac{BD}{AB} \), that is AB² = BC × BD.
(ii) Let ∠CAB = θ. In ΔCBA, ∠CBA = 90° − θ. In ΔCAD, ∠CAD = 90° − θ (since ∠A = 90°), so ∠CAD = ∠CBA; also ∠ACB = ∠DCA = 90°. Hence ΔCBA ~ ΔCAD (AA) and \( \frac{AC}{DC}=\frac{BC}{AC} \), that is AC² = BC × DC.
(iii) In ΔDCA and ΔDAB: ∠DCA = ∠DAB = 90° and ∠D is common, so ΔDCA ~ ΔDAB (AA). Hence \( \frac{DC}{DA}=\frac{DA}{DB} \), that is AD² = BD × CD.

Teacher's Note:
1. Identify the hypotenuse (the longest side, opposite the right angle) before writing the Pythagoras relation.
2. In proofs, name the right-angled triangle in which you apply the theorem and write each relation as a numbered line.

Question 4. ABC is an isosceles triangle right-angled at C. Prove that AB² = 2AC².

Answer:
Since AC = BC and ∠C = 90°, by the Pythagoras theorem AB² = AC² + BC² = AC² + AC² = 2AC².

Teacher's Note:
1. Identify the hypotenuse (the longest side, opposite the right angle) before writing the Pythagoras relation.
2. In proofs, name the right-angled triangle in which you apply the theorem and write each relation as a numbered line.

Question 5. ABC is an isosceles triangle with AC = BC. If AB² = 2AC², prove that ABC is a right triangle.

Answer:
AB² = 2AC² = AC² + AC² = AC² + BC² (as AC = BC).
By the converse of the Pythagoras theorem, ∠C = 90°, so ABC is a right triangle.

Teacher's Note:
1. Identify the hypotenuse (the longest side, opposite the right angle) before writing the Pythagoras relation.
2. In proofs, name the right-angled triangle in which you apply the theorem and write each relation as a numbered line.

Question 6. ABC is an equilateral triangle of side 2a. Find each of its altitudes.

Answer:
Draw AD ⊥ BC. In ΔADB and ΔADC: AB = AC, AD is common and ∠ADB = ∠ADC = 90°, so ΔADB ≅ ΔADC (RHS) and BD = DC = a.
In ΔADB: \( AD^2=AB^2-BD^2=4a^2-a^2=3a^2 \), so AD = √3 a.
Each altitude is √3 a.

Teacher's Note:
1. Identify the hypotenuse (the longest side, opposite the right angle) before writing the Pythagoras relation.
2. In proofs, name the right-angled triangle in which you apply the theorem and write each relation as a numbered line.

Question 7. Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

Answer:
The diagonals of a rhombus bisect each other at right angles: AO = CO, BO = DO and ∠AOB = 90°.
In right ΔAOB, AB² = AO² + BO². Similarly BC² = BO² + CO², CD² = CO² + DO², DA² = DO² + AO².
Adding: AB² + BC² + CD² + DA² = 2(AO² + BO² + CO² + DO²) = 4AO² + 4BO² (as CO = AO, DO = BO) = (2AO)² + (2BO)² = AC² + BD².
Hence proved.

Teacher's Note:
1. Identify the hypotenuse (the longest side, opposite the right angle) before writing the Pythagoras relation.
2. In proofs, name the right-angled triangle in which you apply the theorem and write each relation as a numbered line.

Question 8. In the figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that
(i) OA² + OB² + OC² − OD² − OE² − OF² = AF² + BD² + CE²
(ii) AF² + BD² + CE² = AE² + CD² + BF²

Answer:
Join OA, OB, OC. Apply the Pythagoras theorem in the six right triangles:
OA² = OF² + AF², OB² = OD² + BD², OC² = OE² + EC² (from ΔAOF, ΔBOD, ΔCOE).
(i) Adding the three: OA² + OB² + OC² = OF² + OD² + OE² + AF² + BD² + CE², so OA² + OB² + OC² − OD² − OE² − OF² = AF² + BD² + CE² ...(iv).
(ii) Similarly, from ΔAOE, ΔBOF, ΔCOD: OA² = OE² + AE², OB² = OF² + BF², OC² = OD² + CD². Hence OA² + OB² + OC² − OD² − OE² − OF² = AE² + BF² + CD² ...(v).
From (iv) and (v): AF² + BD² + CE² = AE² + CD² + BF².

Teacher's Note:
1. Identify the hypotenuse (the longest side, opposite the right angle) before writing the Pythagoras relation.
2. In proofs, name the right-angled triangle in which you apply the theorem and write each relation as a numbered line.

Question 9. 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.

Answer:
Let PQ be the wall and PR the ladder. In right ΔPQR: PR² = PQ² + QR², so 10² = 8² + QR².
\[ QR^2=100-64=36\Rightarrow\ QR=6 \]
The foot of the ladder is 6 m from the base of the wall.

Teacher's Note:
1. Identify the hypotenuse (the longest side, opposite the right angle) before writing the Pythagoras relation.
2. In proofs, name the right-angled triangle in which you apply the theorem and write each relation as a numbered line.

Question 10. 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?

Answer:
A guy wire is a wire used to give stability to a free-standing structure such as a pole. The pole AB = 18 m, the wire AC = 24 m and BC is the distance of the stake from the base.
In right ΔABC: \( BC^2=AC^2-AB^2=576-324=252 \), so \( BC=\sqrt{252}=6\sqrt7 \) m.
The stake should be driven 6√7 m (about 15.87 m) from the base of the pole.

Teacher's Note:
1. Identify the hypotenuse (the longest side, opposite the right angle) before writing the Pythagoras relation.
2. In proofs, name the right-angled triangle in which you apply the theorem and write each relation as a numbered line.

Question 11. An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after 1½ hours?

Answer:
Let P be the airport. Distance flown north after 1½ hours: PQ = 1000 × 3/2 = 1500 km. Distance flown west: PR = 1200 × 3/2 = 1800 km.
Since north and west are perpendicular, ΔPQR is right-angled at P:
\[ QR^2=1500^2+1800^2=2250000+3240000=5490000\Rightarrow\ QR=\sqrt{5490000}=300\sqrt{61} \]
The planes are 300√61 km (about 2343 km) apart.

Teacher's Note:
1. Identify the hypotenuse (the longest side, opposite the right angle) before writing the Pythagoras relation.
2. In proofs, name the right-angled triangle in which you apply the theorem and write each relation as a numbered line.

Question 12. 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.

Answer:
Let AB = 6 m and CD = 11 m be the poles with AC = 12 m. Draw BE ∥ AC, meeting CD at E. Then BE = AC = 12 m and DE = 11 − 6 = 5 m.
In right ΔBED: \( BD^2=BE^2+DE^2=12^2+5^2=169 \), so BD = 13 m.
The distance between the tops of the poles is 13 m.

Teacher's Note:
1. Identify the hypotenuse (the longest side, opposite the right angle) before writing the Pythagoras relation.
2. In proofs, name the right-angled triangle in which you apply the theorem and write each relation as a numbered line.

Question 13. D and E are points on the sides CA and CB respectively of a triangle ABC right-angled at C. Prove that AE² + BD² = AB² + DE².

Answer:
Right angle at C: apply the Pythagoras theorem in four right triangles.
ΔACE: AE² = AC² + CE² ...(i). ΔBCD: BD² = BC² + CD² ...(ii).
Adding: AE² + BD² = (AC² + BC²) + (CE² + CD²) ...(iii).
ΔCDE: DE² = CD² + CE² and ΔABC: AB² = AC² + BC².
Substituting in (iii): AE² + BD² = AB² + DE².

Teacher's Note:
1. Identify the hypotenuse (the longest side, opposite the right angle) before writing the Pythagoras relation.
2. In proofs, name the right-angled triangle in which you apply the theorem and write each relation as a numbered line.

Question 14. The perpendicular from A on side BC of a ΔABC intersects BC at D such that DB = 3CD. Prove that 2AB² = 2AC² + BC².

Answer:
In right triangles ADB and ADC: AB² = AD² + BD² and AC² = AD² + CD².
Subtracting: AB² − AC² = BD² − CD² = 9CD² − CD² = 8CD².
Since BC = BD + CD = 4CD, CD = BC/4, so 8CD² = BC²/2.
Hence AB² − AC² = BC²/2, that is 2AB² = 2AC² + BC².

Teacher's Note:
1. Draw the perpendicular as stated in the construction and apply the Pythagoras theorem in both right triangles.
2. Subtract or add the two relations so that the common term cancels.

Question 15. In an equilateral triangle ABC, D is a point on side BC such that BD = ⅓BC. Prove that 9AD² = 7AB².

Answer:
Let each side be a. Draw AE ⊥ BC; then BE = EC = a/2 and AE = (√3/2)a. Also BD = a/3, so DE = a/2 − a/3 = a/6.
In right ΔADE: \( AD^2=AE^2+DE^2=\frac{3a^2}{4}+\frac{a^2}{36}=\frac{28a^2}{36}=\frac{7a^2}{9} \).
So 9AD² = 7a² = 7AB². Hence proved.

Teacher's Note:
1. Draw the perpendicular as stated in the construction and apply the Pythagoras theorem in both right triangles.
2. Subtract or add the two relations so that the common term cancels.

Question 16. 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.

Answer:
Let ABC be equilateral with side a and AD ⊥ BC. Then BD = DC = a/2.
In ΔABD: \( AB^2=AD^2+BD^2\Rightarrow\ a^2=AD^2+\frac{a^2}{4}\Rightarrow\ AD^2=\frac{3a^2}{4} \).
Hence 4AD² = 3a² = 3AB². Hence proved.

Teacher's Note:
1. Identify the hypotenuse (the longest side, opposite the right angle) before writing the Pythagoras relation.
2. In proofs, name the right-angled triangle in which you apply the theorem and write each relation as a numbered line.

Question 17. Tick the correct answer and justify: In ΔABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm. The angle B is
(a) 120° (b) 60° (c) 90° (d) 45°

Answer:
Correct option: (c)
AB² = 108, AC² = 144 and BC² = 36. Since 108 + 36 = 144, AB² + BC² = AC².
By the converse of the Pythagoras theorem, ∠B = 90°.

Teacher's Note:
1. Write the similarity statement with corresponding vertices in the same order, then use the ratio of areas equal to the square of the ratio of corresponding sides.
2. Give a reason (AA, SAS or SSS) for every similarity you claim.