NCERT Solutions for Class 10 Maths Chapter 6 - Triangles Exercise 6.5

Chapter 6, Triangles, Exercise 6.5 begins on page number 150 of the NCERT textbook. The exercise covers the portions in which theorems related to right angles are present. In other words, the exercise-wise NCERT Class 10 Solutions contains questions that could be solved using the theorems that deal with right angles.

The subject experts at BYJU’S make sure that the questions present in the NCERT textbooks are solved in the easiest way possible. The NCERT solutions for Class 10 Maths can help you not only with your board examination preparation but can also be used to check if the answers you gave to the questions are correct or not.

NCERT Solutions for Class 10 Maths Chapter 6- Triangles Exercise 6.5

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Access other exercise solutions of Class 10 Maths Chapter 6 – Triangles

Exercise 6.1 Solutions 3 Questions (3 Short Answer Questions)

Exercise 6.2 Solutions 10 Questions (9 Short Answer Questions, 1 Long Answer Question)

Exercise 6.3 Solutions 16 Questions (1 Main Question with 6 Sub-questions, 12 Short Answer Questions, 3 Long Answer Questions)

Exercise 6.4 Solutions 9 Questions (2 Short Answer with Reasoning Questions, 5 Short Answer Questions, 2 Long Answer Questions)

Exercise 6.6 Solutions 10 Questions (5 Short Answer Questions, 5 Long Answer Questions)

Access Answers to Maths NCERT Class 10 Chapter 6 – Triangles Exercise 6.5

1.  Sides of 4 triangles are given below. Determine which of them are right triangles. In the 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

Solution:

(i) Given, the sides of the triangle are 7 cm, 24 cm, and 25 cm.

Squaring the lengths of the sides of the, we will get 49, 576, and 625.

49 + 576 = 625

(7)² + (24)² = (25)²

Therefore, the above equation satisfies Pythagoras theorem. Hence, it is a right-angled triangle.

Length of Hypotenuse = 25 cm

(ii) Given, the sides of the triangle are 3 cm, 8 cm, and 6 cm.

Squaring the lengths of these sides, we will get 9, 64, and 36.

Clearly, 9 + 36 ≠ 64

Or, 3² + 6² ≠ 8²

Therefore, the sum of the squares of the lengths of two sides is not equal to the square of the length of the hypotenuse.

Hence, the given triangle does not satisfy Pythagoras theorem.

(iii) Given, the sides of triangle’s are 50 cm, 80 cm, and 100 cm.

Squaring the lengths of these sides, we will get 2500, 6400, and 10000.

However, 2500 + 6400 ≠ 10000

Or, 50² + 80² ≠ 100²

As you can see, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.

Therefore, the given triangle does not satisfy Pythagoras theorem.

Hence, it is not a right triangle.

(iv) Given, the sides are 13 cm, 12 cm, and 5 cm.

Squaring the lengths of these sides, we will get 169, 144, and 25.

Thus, 144 +25 = 169

Or, 12² + 5² = 13²

The sides of the given triangle satisfy Pythagoras theorem.

Therefore, it is a right triangle.

Hence, the length of the hypotenuse of this triangle is 13 cm.

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.

Solution:

Given, ΔPQR is right angled at P is a point on QR such that PM ⊥QR

We have to prove, PM² = QM × MR

In ΔPQM, by Pythagoras theorem

PQ² = PM² + QM²

Or, PM² = PQ² – QM² ……………………………..(i)

In ΔPMR, by Pythagoras theorem

PR² = PM² + MR²

Or, PM² = PR² – MR² ………………………………………..(ii)

Adding equations (i) and (ii), we get,

2PM² = (PQ² + PM²) – (QM² + MR²)

= QR² – QM² – MR²        [∴ QR² = PQ² + PR²]

= (QM + MR)² – QM² – MR²

= 2QM × MR

∴ PM² = QM × MR

3. In 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

Solution:

(i) In ΔADB and ΔCAB,

∠DAB = ∠ACB (Each 90°)

∠ABD = ∠CBA (Common angles)

∴ ΔADB ~ ΔCAB [AA similarity criterion]

⇒ AB/CB = BD/AB

⇒ AB² = CB × BD

(ii) Let ∠CAB = x

In ΔCBA,

∠CBA = 180° – 90° – x

∠CBA = 90° – x

Similarly, in ΔCAD

∠CAD = 90° – ∠CBA

= 90° – x

∠CDA = 180° – 90° – (90° – x)

∠CDA = x

In ΔCBA and ΔCAD, we have

∠CBA = ∠CAD

∠CAB = ∠CDA

∠ACB = ∠DCA (Each 90°)

∴ ΔCBA ~ ΔCAD [AAA similarity criterion]

⇒ AC/DC = BC/AC

⇒ AC² =  DC × BC

(iii) In ΔDCA and ΔDAB,

∠DCA = ∠DAB (Each 90°)

∠CDA = ∠ADB (common angles)

∴ ΔDCA ~ ΔDAB [AA similarity criterion]

⇒ DC/DA = DA/DA

⇒ AD² = BD × CD

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

Solution:

Given, ΔABC is an isosceles triangle right angled at C.

In ΔACB, ∠C = 90°

AC = BC (By isosceles triangle property)

AB² = AC² + BC² [By Pythagoras theorem]

= AC² + AC² [Since, AC = BC]

AB² = 2AC²

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

Solution:

Given, ΔABC is an isosceles triangle having AC = BC and AB² = 2AC²

In ΔACB,

AC = BC

AB² = 2AC²

AB² = AC² + AC²

= AC² + BC² [Since, AC = BC]

Hence, by Pythagoras theorem, ΔABC is a right angle triangle.

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

Solution:

Given, ABC is an equilateral triangle of side 2a.

Draw, AD ⊥ BC

In ΔADB and ΔADC,

AB = AC

AD = AD

∠ADB = ∠ADC [Both are 90°]

Therefore, ΔADB ≅ ΔADC by RHS congruence.

Hence, BD = DC [by CPCT]

In the right angled ΔADB,

AB² = AD² + BD²

(2a)² = AD² + a² 

⇒ AD^{2 =} 4a² – a²

⇒ AD^{2 =} 3a²

⇒ AD ⁼ √3a

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

Solution:

Given, ABCD is a rhombus whose diagonals AC and BD intersect at O.

We have to prove, as per the question,

AB² + BC² + CD² + AD² = AC² + BD²

Since the diagonals of a rhombus bisect each other at right angles.

Therefore, AO = CO and BO = DO

In ΔAOB,

∠AOB = 90°

AB² = AO² + BO² …………………….. (i) [By Pythagoras theorem]

Similarly,

AD² = AO² + DO² …………………….. (ii)

DC² = DO² + CO² …………………….. (iii)

BC² = CO² + BO² …………………….. (iv)

Adding equations (i) + (ii) + (iii) + (iv), we get,

AB² + AD² + DC² + BC² = 2(AO² + BO² + DO² + CO²)

= 4AO² + 4BO² [Since, AO = CO and BO =DO]

= (2AO)² + (2BO)² = AC² + BD²

AB² + AD² + DC² + BC² = AC² + BD²

Hence, proved.

8. In Fig. 6.54, 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.

Solution:

Given, in ΔABC, O is a point in the interior of a triangle.

And OD ⊥ BC, OE ⊥ AC and OF ⊥ AB.

Join OA, OB and OC

(i) By Pythagoras theorem in ΔAOF, we have

OA² = OF² + AF²

Similarly, in ΔBOD

OB² = OD² + BD²

Similarly, in ΔCOE

OC² = OE² + EC²

Adding these equations,

OA² + OB² + OC² = OF² + AF² + OD² + BD² + OE² + EC²

OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + CE².

(ii) AF² + BD² + EC² = (OA² – OE²) + (OC² – OD²) + (OB² – OF²)

∴ AF² + BD² + CE² = AE² + CD² + BF².

9. A ladder 10 m long reaches a window 8 m above the ground. Find the distance between the foot of the ladder from the base of the wall.

Solution:

Given, a ladder 10 m long reaches a window 8 m above the ground.

Let BA be the wall and AC be the ladder,

Therefore, by Pythagoras theorem,

AC² = AB² + BC²

10² = 8² + BC²

BC² = 100 – 64

BC² = 36

BC = 6m

Therefore, the distance between the foot of the ladder from the base of the wall is 6 m.

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?

Solution:

Given, 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.

Let AB be the pole and AC be the wire.

By Pythagoras theorem,

AC² = AB² + BC²

24² = 18² + BC²

BC² = 576 – 324

BC² = 252

BC = 6√7m

Therefore, the distance from the base is 6√7m.

11. An aeroplane leaves an airport and flies due north at a speed of 1,000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1,200 km per hour. How far apart will be the two planes after
 hours?

Solution:

Given,

Speed of first aeroplane = 1000 km/hr

Distance covered by the first aeroplane flying due north in
  hours (OA) = 1000 × 3/2 km = 1500 km

Speed of second aeroplane = 1200 km/hr

Distance covered by the second aeroplane flying due west in
  hours (OB) = 1200 × 3/2 km = 1800 km

In right angle ΔAOB, by Pythagoras Theorem,

AB² = AO² + OB²

⇒ AB² = (1500)² + (1800)²

⇒ AB = √(2250000 + 3240000)

= √5490000

⇒ AB = 300√61 km

Hence, the distance between two aeroplanes will be 300√61 km.

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.

Solution:

Given, Two poles of heights 6 m and 11 m stand on a plane ground.

And the distance between the feet of the poles is 12 m.

Let AB and CD be the poles of height 6m and 11m.

Therefore, CP = 11 – 6 = 5m

From the figure, it can be observed that AP = 12m

By Pythagoras theorem for ΔAPC, we get,

AP² = PC² + AC²

(12m)² + (5m)² = (AC)²

AC² = (144+25) m² = 169 m²

AC = 13m

Therefore, the distance between their tops is 13 m.

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.

Solution:

Given, D and E are points on the sides CA and CB, respectively, of a triangle ABC right angled at C.

By Pythagoras theorem in ΔACE, we get

AC² + CE² = AE² ………………………………………….(i)

In ΔBCD, by Pythagoras theorem, we get

BC² + CD² = BD² ………………………………..(ii)

From equations (i) and (ii), we get

AC² + CE² + BC² + CD² = AE² + BD² …………..(iii)

In ΔCDE, by Pythagoras theorem, we get

DE² = CD² + CE²

In ΔABC, by Pythagoras theorem, we get

AB² = AC² + CB²

Putting the above two values in equation (iii), we get

DE² + AB² = AE² + BD².

14. The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3CD (see figure). Prove that 2AB² = 2AC² + BC².

Solution:

Given, the perpendicular from A on side BC of a Δ ABC intersects BC at D such that;

DB = 3CD.

In Δ ABC,

AD ⊥BC and BD = 3CD

In right angle triangle, ADB and ADC, by Pythagoras theorem,

AB² = AD² + BD² ……………………….(i)

AC² = AD² + DC² ……………………………..(ii)

Subtracting equation (ii) from equation (i), we get

AB² – AC² = BD² – DC²

= 9CD² – CD² [Since, BD = 3CD]

= 8CD²

= 8(BC/4)² [Since, BC = DB + CD = 3CD + CD = 4CD]

Therefore, AB² – AC² = BC²/2

⇒ 2(AB² – AC²) = BC²

⇒ 2AB² – 2AC² = BC²

∴ 2AB² = 2AC² + BC².

15.  In an equilateral triangle, ABC, D is a point on side BC such that BD = 1/3BC. Prove that 9AD² = 7AB².

Solution:

Given, ABC is an equilateral triangle.

And D is a point on side BC such that BD = 1/3BC.

Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC.

∴ BE = EC = BC/2 = a/2

And, AE = a√3/2

Given, BD = 1/3BC

∴ BD = a/3

DE = BE – BD = a/2 – a/3 = a/6

In ΔADE, by Pythagoras theorem,

AD² = AE² + DE² 

⇒ 9 AD² = 7 AB²

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.

Solution:

Given, an equilateral triangle, say ABC,

Let the sides of the equilateral triangle be of length a, and AE be the altitude of ΔABC.

∴ BE = EC = BC/2 = a/2

In ΔABE, by Pythagoras Theorem, we get

AB² = AE² + BE²

4AE² = 3a²

⇒ 4 × (Square of altitude) = 3 × (Square of one side)

Hence, proved.

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°

Solution:

Given, in ΔABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm.

We can observe that,

AB² = 108

AC² = 144

And, BC² = 36

AB² + BC² = AC²

The given triangle, ΔABC, satisfies Pythagoras theorem.

Therefore, the triangle is a right triangle, right-angled at B.

∴ ∠B = 90°

Hence, the correct answer is (C).

The Class 10 Maths Chapter 6, Triangles, covers a wide range of topics with the six exercises present in the chapter. Exercise 6.5 deals with the theorems related to Pythagoras theorem. 17 questions are present in this exercise, out of which 15 questions are short answer type questions, and 2 questions are long answer type questions. There are 3 theorems that are covered in this exercise. The three theorems that form the base of Exercise 6.5 is given below:

• Theorem 6.7: If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.
• Theorem 6.8: In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. It is also known as Pythagoras theorem.
• 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.

The best way to get acquainted with a mathematical concept is to solve as many problems as possible from that particular topic. To familiarise yourself with all the concepts of Maths present in Class 10, practising with the solutions for the questions given in the NCERT textbook is mandatory. These NCERT Solutions will help the students in scoring excellent marks in the board examination.