What topics are covered in Exercise 6.5 of Chapter 6 (Triangles)?

Exercise 6.5 mainly focuses on the application of the Pythagoras theorem and its converse. It includes problems related to right-angled triangles and their properties.

Why is Exercise 6.5 important for Class 10 students?

This exercise helps students understand how to apply the Pythagoras theorem in different scenarios, which is crucial for board exams and competitive exams like JEE and NTSE.

How does the NCERT solution help in solving Exercise 6.5?

NCERT solutions provide step-by-step explanations of each question, ensuring that students grasp the logic behind the theorem and its applications.

Are NCERT solutions enough to score full marks in this chapter?

Yes, NCERT solutions provide accurate and exam-oriented answers, but practicing additional problems from reference books can enhance understanding.

Exercise 6.5

NCERT Solutions Class 10 Maths Chapter 6 Triangles : Exercise 6.5

NCERT Solutions Class-10 Maths Chapter 6 Triangles Exercise 6.5 will help you explore and understand the problems related to Pythagoras' Theorem and other forms of this theorem. This is a foundational concept of the geometry of triangles, explained and proved here by the concept of similarity of triangles. These concepts also help in understanding other complex topics of mathematics, such as trigonometry or other geometric figures. So, let’s start exploring this foundational topic of class 10th, which is important for both the higher classes and examination purposes.

1.0Download NCERT Solutions Class 10 Maths Chapter 6 Triangles Exercise 6.5 : Free PDF

NCERT Solutions Class 10 Maths Chapter 6 Triangles Exercise 6.5

2.0Introduction to Pythagoras Theorem and Similarity of Triangles:

The Pythagoras Theorem is a key concept in geometry, and it is used for right-angled triangles. According to the theorem, the square of the hypotenuse (the side on the opposite side of the right angle) is always equal to the sum of the squares of the perpendicular & base (other two sides). This can be expressed mathematically as:

$Hypotenuse (H)^{2} = Perpendicular (P)^{2} + Base (B)^{2}$

$A C^{2} = A B^{2} + B C^{2}$

Understanding Similarity of Triangles:

The similarity is the backbone of proving Pythagoras' Theorem. Similarity of triangles refers to two triangles sharing the same form but potentially different dimensions. Two triangles are required to meet two conditions to be similar.

Any two corresponding angles of these two triangles are equal.
The corresponding sides of the two triangles are proportional (i.e., the ratio of the corresponding sides' lengths is the same).

Access our NCERT Solutions Class 10 Maths Chapter 6 Triangle Exercise 6.5 PDF, crafted with the guidance of our expert faculty to ensure accurate solutions.

3.0Exercise 6.5 Key Concepts – Important Theorems

Theorem: If a perpendicular is drawn from the right-angled vertex of any given right triangle to its hypotenuse, then triangles on both sides of the perpendicular are always similar to the whole triangle & each other.

To Prove:

$△ B D A \sim △ B A C$

$△ A D C \sim △ B A C$

$△ B D A \sim △ A D C$

Given: $(A D ⊥ BC), (A BC)$ is a right-angle triangle. Right angled at A.

Proof: In

$△ B D A and △ B A C$

∠B = ∠B (Common)

∠BDA = ∠BAC (Right angle)

$△ B D A \sim △ B A C) (AA)$

Similarly in

$△ A D C and △ B A C$

$△ A D C \sim △ B A C (AA)$

$△ B D A \sim △ A D C$

(A similar triangle shows transitive properties, meaning if a triangle is similar to a 2nd triangle and the 2nd triangle is similar to a 3rd triangle, then triangles one and third are also similar.)

Theorem: In any given right-angled triangle, the square of the hypotenuse (side opposite to right angle) is always equal to the sum of the squares of perpendicular and base. (Standard form of Pythagoras theorem)

To prove:

$Hy p o t e n u s e^{2} = P er p e n d i c u l a r^{2} + B a s e^{2}$

$A C^{2} = B C^{2} + B A^{2}$

In any given right-angled triangle, the square of the hypotenuse (side opposite to right angle) is always equal to the sum of the squares of perpendicular and base. (Standard form of Pythagoras theorem)

Construction: $B D ⊥ A C$

Given: ABC is a right-angle triangle. Right angled at B.

Proof: In $△ A D B and △ A BC$

∠A = ∠A

∠ADB = ∠ABC (Right Angle)

$△ A D B \sim △ A BC (AA)$

$\frac{A D}{A B} = \frac{A B}{A C} (CPST)$

$A B^{2} = A D \times A C \dots (1)$

In $△ B D C and △ A BC$

∠C = ∠C (Common)

∠BDC = ∠ABC (Right angle)

$△ B D C \sim △ A BC (AA)$

$\frac{D C}{BC} = \frac{BC}{A C} (CPCT)$

$B C^{2} = D C \times A C \dots (2)$

Now, add equation 1 and 2

$A B^{2} + B C^{2} = A D \times A C + D C \times A C$

$A B^{2} + B C^{2} = A C (A D + D C)$

$A B^{2} + B C^{2} = A C \times A C$

$A C^{2} = A B^{2} + B C^{2}$

Theorem: In a triangle, if the square of any one side is equal to the sum of the squares of its other two sides, then the angle opposite the first side is always a right angle. (Converse of Pythagoras theorem)

To Prove: ∠B = 90°

Given:

$A C^{2} = A B^{2} + B C^{2}$

Construction: Construct a right-angled triangle PQR, such that AB = PQ and BC = QR. Also, ∠Q = 90°

In a triangle, if the square of any one side is equal to the sum of the squares of its other two sides, then the angle opposite the first side is always a right angle. (Converse of Pythagoras theorem)

Proof: In $△ A BC$

$A C^{2} = A B^{2} + B C^{2} \dots (1)$

In $△ PQR$

$P R^{2} = P Q^{2} + R Q^{2}$

As AB = PQ and QR = BC (By construction)

$P R^{2} = A B^{2} + B C^{2} \dots (2)$

From equation (1) and (2)

$A C^{2} = P R^{2}$

$A C = PR \dots (3)$

In $△ A BC and △ PQR$

AB = PQ (By construction)

BC = QR (By construction)

AC = PR (equation 3)

$△ A BC ≅ △ PQR (SSS)$

∠B = ∠Q (CPCT)

∠B = 90°

4.0NCERT Class 10 Maths Chapter 6 Triangles Exercise 6.5 : Detailed Solutions

1. Sides of some 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

Solution:

(i) (7)² + (24)² = 49 + 576 = 625 = (25)²

Therefore, given sides 7 cm, 24 cm, 25 cm make a right triangle.

(ii) (6)² + (3)² = 36 + 9 = 45

(8)² = 64

(6)² + (3)² ≠ (8)²

Therefore, given sides 3 cm, 8 cm, 6 cm does not make a right triangle.

(iii) (50)² + (80)² = 2500 + 6400 = 8900

(100)² = 10000

(50)² + (80)² ≠ 100²

Therefore, given sides 50 cm, 80 cm, 100 cm does not make a right triangle.

(iv) (12)² + (5)² = 144 + 25 = 169 = (13)²

Therefore, given sides 13 cm, 12 cm, 5 cm make a right triangle.

2. PQR is a triangle with ∠PQR = ∠PRQ, and M is a point on QR such that PM ⊥ QR.

Show that: PM² = QM × MR.

Solution:

∠1 + ∠2 = ∠2 + ∠4 (Each is 90 degrees)

PQR is a triangle with ∠PQR = ∠PRQ, and M is a point on QR such that PM ⊥ QR. Show that: PM² = QM × MR.

⇒ ∠1 = ∠4

Similarly, ∠2 = ∠3. Now, this gives

ΔQPM ~ ΔPRM (AA similarity)

PM/MR = QM/PM

⇒ PM² = QM × MR

3. In figure, ABD is a right triangle right angled at A and AC ⊥ BD. Show that

(i) AB² = BC ⋅ BD

(ii) AC² = BC ⋅ DC

(iii) AD² = BD ⋅ CD

Solution:

In the given figure, we have

△ABC ∼ △DAC ∼ △DBA

(i) △ABC ∼ △DBA

⇒ ar(△ABC) / ar(△DBA) = AB² / DB²

⇒ (1/2)(BC)(AC) / (1/2)(BD)(AC) = AB² / DB²

⇒ AB² = BC ⋅ BD

(ii) △ABC ∼ △DAC

⇒ ar(△ABC) / ar(△DAC) = AC² / DC²

⇒ (1/2)(BC)(AC) / (1/2)(DC)(AC) = AC² / DC²

⇒ AC² = BC ⋅ DC

(iii) △DAC ∼ △DBA

⇒ ar(△DAC) / ar(△DBA) = DA² / DB²

⇒ (1/2)(CD)(AC) / (1/2)(BD)(AC) = AD² / BD²

⇒ AD² = BD ⋅ CD

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

Solution:

In △ABC, ∠ACB = 90°. We are given that △ABC is an isosceles triangle.

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

=> ∠A = ∠B = 45°

=> AC = BC

By Pythagoras theorem, we have:

AB² = AC² + BC²

= AC² + AC² [∵ BC = AC]

= 2AC²

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

Solution:

As, AB² = 2AC²

AB² = AC² + AC²

= AC² + BC²

As it satisfies the Pythagorean triplet,

So, ∆ABC is a right triangle, right angled at ∠C.

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

Solution:

Altitude of equilateral triangle = (√3 / 2) × Side = (√3 / 2) × 2a = √3a

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.

Solution:

ABCD is a rhombus in which AB = BC = CD = DA = a (say). Its diagonals AC and BD are right bisectors of each other at O.

In ∆OAB, ∠AOB = 90°, OA = (1/2)AC and OB = (1/2)BD.

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

By Pythagoras theorem, we have:

OA² + OB² = AB²

=> (1/2 AC)² + (1/2 BD)² = AB²

=> AC² + BD² = 4AB²

or 4AB² = AC² + BD²

=> AB² + BC² + CD² + DA² = AC² + BD²

Hence proved.

8. In the given 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

In the given figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC, and OF ⊥ AB. Show that:

Solution:

(i) In right angled △OFA,

O A² = O F² + A F² (Pythagoras theorem)

=> OA² - OF² = AF² ...(1)

Similarly, OB² - OD² = BD² ...(2)

And OC² - OE² = CE² ...(3)

Adding (1), (2) and (3), we get:

OA² + OB² + OC² - OD² - OE² - OF² = AF² + BD² + CE² ...(4)

(ii) We have proved that:

OA² + OB² + OC² - OD² - OE² - OF² = AF² + BD² + CE²

Similarly, we can prove that:

OA² + OB² + OC² - OD² - OE² - OF² = BF² + CD² + AE² ...(5)

From (4) and (5), we have:

AF² + BD² + CE² = AE² + CD² + BF²

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.

Solution:

Let AC = x metres be the distance of the foot of the ladder from the base of the wall.

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.

AB = 8 m (Height of window)

BC = 10 m (length of ladder)

Now, x² + (8)² = (10)²

=> x² = 100 - 64 = 36

=> x = 6, i.e., AC = 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:

Let AB be the vertical pole of 18 m and AC be the wire of 24 m.

In triangle ABC, by Pythagoras theorem:

. 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?

AC² = AB² + BC²

24² = 18² + BC²

BC² = 252

BC = 6√7 m

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 1 ½ hours?

Solution:

The first plane travels the distance BC in the direction of north in 1 ½ hours at a speed of 1000 km/hr.

Therefore, BC = 1000 × (3/2) km = 1500 km.

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 1 ½ hours?

(Plane II)

The second plane travels distance BA in the direction of west in 1 ½ hours at a speed of 1200 km/hr.

Therefore, BA = 1200 × (3/2) km = 1800 km.

From right angled triangle ABC,

AC² = AB² + BC²

= (1800)² + (1500)²

= 3240000 + 2250000 = 5490000

=> AC = √5490000 m

=> AC = 300√61 m

12. Two poles of height 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:

Let AD and BE be two poles of height 6 m and 11 m and AB = 12 m.

In triangle DEC, by Pythagoras theorem:

DE² = CD² + CE²

DE² = 12² + 5² (DC = AB = 12 m)

DE = √(144 + 25) = √169 = 13 m

Thus, 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:

In right angled triangle ACE,

AE² = CA² + CE²

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.

and in right angled triangle BCD,

BD² = BC² + CD² (2)

Adding (1) and (2), we get

AE² + BD² = (CA² + CE²) + (BC² + CD²)

= (BC² + CA²) + (CD² + CE²)

= BA² + DE

Therefore, AE² + BD² = AB² + DE²

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

Solution:

DB = 3CD

=> CD = (1/4)BC

And DB = (3/4)BC

In triangle ABD, AB² = DB² + AD² ...(1)

In triangle ACD, AC² = CD² + AD² ...(2)

Subtracting (2) from (1), we get:

AB² - AC² = DB² - CD²

= ((3/4)BC)² - ((1/4)BC)²

= (9/16)BC² - (1/16)BC²

= (8/16)BC² = (1/2)BC²

=> 2AB² - 2AC² = BC²

=> 2AB² = 2AC² + BC²

Hence proved.

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

Solution:

AB = BC = CA = a (say)

BD = (1/3)BC = (1/3)a

=> CD = (2/3)BC = (2/3)a

Let AE be perpendicular to BC.

=> BE = EC = (1/2)a

DE = BE - BD = (1/2)a - (1/3)a = (1/6)a

AD² = AE² + DE²

We know AE² = AB² - BE² (Pythagoras Theorem in triangle ABE)

AD² = AB² - BE² + DE²

= a² - ((1/2)a)² + ((1/6)a)²

= a² - (1/4)a² + (1/36)a²

= (36a² - 9a² + a²)/36

= (28a²)/36

= (7/9)a²

= (7/9)AB²

=> 9AD² = 7AB²

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:

Altitude of equilateral triangle = (√3 / 2) * side

h = (√3 / 2) * a

h² = (3 / 4) * a²

4h² = 3a²

17. Tick the correct answer and justify : In triangle ABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm. The angle B is:

(1) 120°

(2) 60°

(3) 90°

(4) 45°

Solution:

Tick the correct answer and justify : In triangle ABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm. The angle B is:

AB² = (6√3)² = 108

BC² = 6² = 36

AC² = 12² = 144

So, AB² + BC² = AC²

△ABC is a right triangle, right angled at B.

∠B = 90°.

5.0Benefits of NCERT Solutions Class 10 Maths Chapter 6 Triangles Exercise 6.5

Provides detailed solutions, making it easier to grasp problem-solving techniques.
Encourages analytical thinking and logical reasoning.
Aligns with CBSE syllabus, ensuring thorough preparation for board exams.
Covers essential theorems like Pythagoras and similarity theorems with practical applications.
Builds a strong foundation for exams like JEE, NTSE, and Olympiads.

NCERT Class 10 Maths Ch 6 Triangles Other Exercises:

Exercise 6.5

NCERT Solutions Class 10 Maths All Chapters:-

Chapter 1 - Real Numbers

Chapter 2 - Polynomials

Chapter 3 - Linear Equations in Two Variables

Chapter 4 - Quadratic Equations

Chapter 5 - Arithmetic Progressions

Chapter 6 - Triangles

Chapter 7 - Coordinate Geometery

Chapter 8 - Introdction to Trigonometry

Chapter 9 - Some Applications of Trigonometry

Chapter 10 - Circles

Chapter 11 - Areas Related to Circles

Chapter 12 - Surface Areas and Volumes

Chapter 13 - Statistics

Chapter 14 - Probability

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