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The Baudhayana-Pythagoras Theorem - Exercise 2.4 Class 8 Maths - Chapter 2 NCERT Solutions

NCERT detailed solutions for Class 8 Maths Term 2 Chapter 2, The Baudhayana-Pythagoras Theorem, Exercise 2.4 are provided on this webpage. Prepared by the experienced ALLEN faculty, these solutions give clear explanations for each question to support students in their exam preparation.In Exercise 2.4, we use the Baudhāyana-Pythagoras Theorem to solve geometric problems involving the diagonal of a square, the unknown sides of right-angled triangles, the side of a rhombus with given diagonals, and the height and area of an equilateral triangle.

1.0Download Class 8 Maths Chapter 2 Ex 2.4 NCERT Solutions PDF

Download the Class 8 Maths Chapter 2 Exercise 2.4 NCERT Solutions PDF for step-by-step solutions and convenient revision. The PDF can be used to practise the questions and review the methods while preparing for exams.

Chapter 2 : The Baudhayana - The Pythagoras Theorem Exercise 2.4

2.0Detailed NCERT Solutions for The Baudhayana-Pythagoras Theorem Exercise 2.4 Class 8 Maths (Ganita Prakash Term 2)

1. Find the diagonal of a square with side length 5 cm.

Sol.

BD^{2} = 5^{2} + 5^{2} = 25 + 25 = 50

BD^{2} = 50 = 5 \times 5 \times 2

= 52 = 5 \times 1.414 = 7.07 (\sim 7.1)

Hence, the length of the diagonal is

52

( 7.1 cm ) approx.

2. Find the missing side lengths in the following right triangles:

(a)

(b)

(f)

(d)

(e)

Sol. (a) $a^{2} = 7^{2} + 9^{2} = 49 + 81 = 130$

$\Rightarrow a = 130$

(b) $b^{2} = 4^{2} + 1 0^{2} = 16 + 100 = 116$

$\Rightarrow b = \Rightarrow b = 116 2 \times 2 \times 29 = 229 = 116$

(c) $4 0^{2} = c^{2} + 4 1^{2}$ $\Rightarrow 1600 + c^{2} = 1681$ $\Rightarrow c^{2} = 1681 - 1600$ $\Rightarrow c^{2} = 81$ $\Rightarrow c = 81$ $\Rightarrow c = 9$ (d) $d^{2} + 1 0^{2} = (200)^{2}$ $d^{2} + 100 = 200$ $d^{2} = 200 - 100$ $d^{2} = 100$ d = 10 (e) $e^{2} = 1 0^{2} + (150)^{2}$ $\Rightarrow e^{2} = 100 + 150$ $\Rightarrow e^{2} = 250$ $\Rightarrow e = 250$ $\Rightarrow e = 5 \times 5 \times 5 \times 2$ $\Rightarrow d = 510$ (f) $2 7^{2} = f^{2} = 4 5^{2}$ $\Rightarrow 729 = f^{2} = 2025$ $\Rightarrow f^{2} = 2025 - 729$ $\Rightarrow f^{2} = 1296$ $\Rightarrow f = 1296$ $= 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3$ $= 2 \times 2 \times 3 \times 3 = 36$

3. Find the side length of a rhombus whose diagonals are of length 24 units and 70 units. Sol.

Since diagonals of Rhombus bisect each other at $9 0^{\circ}$ . $OA = \frac{1}{2} \times 24 = 12$ OD $= \frac{1}{2} \times 70 = 35$ In $△ AOD, a^{2} = 1 2^{2} + 3 5^{2}$ $= 144 + 1225 = 1369$ $a = 1369 = 37$ $∴$ The side of the rhombus is 37 units.

4. Is the hypotenuse the longest side of a right triangle? Justify your answer. Sol. $c^{2} = a^{2} + b^{2}$ $∴ c^{2} > a^{2}$ and $c^{2} > b^{2}$ Or $c > a$ and $c > b$ Hence, 'c' is the longest side of the right triangle.

5. True or False: Every Buadhayana triple is either a primitive triple or a scaled version of a primitive triple. Sol. Let $(a, b, c)$ be a Baudhayana triple. Then $HCF (a, b, c) 1$ or $HCF (a, b, c) \neq = 1$ $(3, 4, 5) : HCF$ is 1 $(10, 24, 26) : HCF = 2$ If HCF is 1 , then the triple is primitive If HCF is a number other than 1 , then the triple is scaled.

6. Give 5 examples of rectangles whose side lengths and diagonals are all integers.
Sol. (i)

(ii)

(iii)

(iv)

(v)

7. Construct a square whose area is equal to the difference of the areas of squares of side-lengths 5 units and 7 units.

Sol. To form a square whose area should be $\Rightarrow 7^{2} - 5^{2} = 49 - 25 = 24$ sq. units. Now, draw a right angles triangle XYZ such that $XY = 5 cm$ and $YZ = 7 cm$ .

Here, $XZ^{2} - XY^{2} = ZY^{2}$

$X Z^{2} = 7^{2} - 5^{2} = 49 - 25 = 24 XZ = 24 cm$

Then draw a square XZUV of side length $24 cm$ as shown.

Hence, the square XZUN is formed whose area is equal to difference of area of squares of side lengths 5 units and 7 units.

8. (i) Using the dots of a grid as the vertices, can you create a square that has an area of (a) 2 sq. units, (b) 3 sq. units, (c) 4 sq-units, (d) 5 unit (ii) Suppose the grid extends indefinitely. What are the possible integer-values areas of squares you can create in this manner?

Sol. (i) Since, one side of square formed will be hypotenuse 'c' of a right angled triangle formed by the grid lines.

The legs of this triangle will be 'a' units and 'b' units.

Now, $c^{2} = a^{2} + b^{2}$ Because, the dots are on a grid, a and b must be whole number ( $0, 1, 2, 3, \dots$ ). This means that the area of a square on a dot grid must be equal to sum of two perfect square. (a) Area of 2 square units

Since $2 = (1)^{2} + (1)^{2}$ So, $AP = PD = 1$ units

$\Rightarrow AD = AP^{2} + PD^{2} = 1 - 11 = 2 units$

Area of square $ABCD = 2 sq$ . units

Tilted square ABCD is formed.

(b) Area of 3 sq. units

There is no combination of two perfect square that add up to 3 .

It is impossible to draw a square with an area of 3 sq. units on a standard integer dot grid. (c) Area of 4 sq. units $∵ 4 = (2)^{2} + 0^{2} = 4 + 0 = 4$

Non tilted square ABCD is formed.

(d) Area of 5 sq. units

5 = (2)^{2} + (1)^{2}

Area of square

ABCD = 5 sq

. units Tilted square ABCD is formed.

(ii) If the grid extends indefinitely, the possible integer area are exactly the numbers that can be written as the sum of two perfect squares.

$\Rightarrow \Rightarrow 0^{2} + 1^{2} = 1 1^{2} + 1^{2} = 2 2^{2} + 0^{2} = 4 2^{2} + 1^{2} = 5 2^{2} + 2^{2} = 8 3^{2} + 0 = 9 3^{2} + 1^{2} = 10 3^{2} + 2^{2} = 13 4^{2} + 0^{2} = 16 4^{2} + 1^{2} = 17$

... and so on Here, 3 , 6 , 7 , 11 , 12 , 14 , 15 are shifted as it is impossible to draw a square on a standard integer dot grid.

9. Find the area of an equilateral triangle with sidelength 6 units. [Hint: Show that an altitude bisects the opposite side. Use this to find the height.] Sol. Let $△ ABC$ be an equilateral triangle. $AB = BC = CA = 6 cm$

Let AD be perpendicular to BC . Then $∠1 = ∠2$ (each $9 0^{\circ}$ ) $AB = AC$ (each 6 cm ) $AD = AD$ (common) $△ ADB ≅ △ ADC$ (by RHS congruency) $BD = DC$ (CPCT)

$∴ BD = DC = \frac{1}{2} \times 6 cm = 3 cm$

In $△ ADC, h^{2} + 3^{2}$

$\Rightarrow \Rightarrow h = 6^{2} (Baudhayana’s triple) h^{2} = 36 - 9 = 27 = ar (Δ ABC) = \frac{1}{2} \times BC \times AD = \frac{1}{2} \times 6 \times 33 sq. units = 93 sq. units$

3.0Key Concepts of Chapter 2 - The Baudhayana-Pythagoras Theorem Exercise 2.4

Calculating the diagonal of a square with side length 5 cm.
Determining unknown sides in the given right-angled triangles.
Using the diagonals of a rhombus measuring 24 units and 70 units to calculate its side.
Identifying the hypotenuse as the longest side of a right-angled triangle.
Working with an isosceles right triangle with a hypotenuse of 10 units.
Calculating the height and area of an equilateral triangle with side length 6 units.

4.0Benefits of NCERT Solutions Class 8 Maths Chapter 2 Exercise 2.4

Shows the steps for calculating the diagonal of a square.
Makes the calculation of unknown sides in right-angled triangles easier to follow.
Explains how rhombus diagonals are used to calculate its side.
Helps understand the position of the hypotenuse in a right-angled triangle.
Provides the method used for the isosceles right triangle problem.
Explains the calculation of the height and area of the given equilateral triangle.

NCERT Class 8 Maths Ch. 2 The Baudhayana - The Pythagoras Theorem Other Exercises:-

Exercise 2.1

Exercise 2.2

Exercise 2.3

Exercise 2.4

NCERT Solutions Class 8 Maths All Chapters:-

Chapter 1 : Fractions in Disguise

Chapter 2 : The Baudhayana - The Pythagoras Theorem

Chapter 3 : Proportional Reasoning - 2

Chapter 4 : Exploring Some Geometric Themes

Chapter 5 : Tales by Dots and Lines

Chapter 6 : Algebra Play

Chapter 7 : Area

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The Baudhayana-Pythagoras Theorem - Exercise 2.4 Class 8 Maths - Chapter 2 NCERT Solutions

NCERT detailed solutions for Class 8 Maths Term 2 Chapter 2, The Baudhayana-Pythagoras Theorem, Exercise 2.4 are provided on this webpage. Prepared by the experienced ALLEN faculty, these solutions give clear explanations for each question to support students in their exam preparation.In Exercise 2.4, we use the Baudhāyana-Pythagoras Theorem to solve geometric problems involving the diagonal of a square, the unknown sides of right-angled triangles, the side of a rhombus with given diagonals, and the height and area of an equilateral triangle.

1.0Download Class 8 Maths Chapter 2 Ex 2.4 NCERT Solutions PDF

Download the Class 8 Maths Chapter 2 Exercise 2.4 NCERT Solutions PDF for step-by-step solutions and convenient revision. The PDF can be used to practise the questions and review the methods while preparing for exams.

Chapter 2 : The Baudhayana - The Pythagoras Theorem Exercise 2.4

2.0Detailed NCERT Solutions for The Baudhayana-Pythagoras Theorem Exercise 2.4 Class 8 Maths (Ganita Prakash Term 2)

1. Find the diagonal of a square with side length 5 cm.

Sol.

BD^{2} = 5^{2} + 5^{2} = 25 + 25 = 50

BD^{2} = 50 = 5 \times 5 \times 2

= 52 = 5 \times 1.414 = 7.07 (\sim 7.1)

Hence, the length of the diagonal is

52

( 7.1 cm ) approx.

2. Find the missing side lengths in the following right triangles:

(a)

(b)

(f)

(d)

(e)

Sol. (a) $a^{2} = 7^{2} + 9^{2} = 49 + 81 = 130$

$\Rightarrow a = 130$

(b) $b^{2} = 4^{2} + 1 0^{2} = 16 + 100 = 116$

$\Rightarrow b = \Rightarrow b = 116 2 \times 2 \times 29 = 229 = 116$

(c) $4 0^{2} = c^{2} + 4 1^{2}$ $\Rightarrow 1600 + c^{2} = 1681$ $\Rightarrow c^{2} = 1681 - 1600$ $\Rightarrow c^{2} = 81$ $\Rightarrow c = 81$ $\Rightarrow c = 9$ (d) $d^{2} + 1 0^{2} = (200)^{2}$ $d^{2} + 100 = 200$ $d^{2} = 200 - 100$ $d^{2} = 100$ d = 10 (e) $e^{2} = 1 0^{2} + (150)^{2}$ $\Rightarrow e^{2} = 100 + 150$ $\Rightarrow e^{2} = 250$ $\Rightarrow e = 250$ $\Rightarrow e = 5 \times 5 \times 5 \times 2$ $\Rightarrow d = 510$ (f) $2 7^{2} = f^{2} = 4 5^{2}$ $\Rightarrow 729 = f^{2} = 2025$ $\Rightarrow f^{2} = 2025 - 729$ $\Rightarrow f^{2} = 1296$ $\Rightarrow f = 1296$ $= 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3$ $= 2 \times 2 \times 3 \times 3 = 36$

3. Find the side length of a rhombus whose diagonals are of length 24 units and 70 units. Sol.

Since diagonals of Rhombus bisect each other at $9 0^{\circ}$ . $OA = \frac{1}{2} \times 24 = 12$ OD $= \frac{1}{2} \times 70 = 35$ In $△ AOD, a^{2} = 1 2^{2} + 3 5^{2}$ $= 144 + 1225 = 1369$ $a = 1369 = 37$ $∴$ The side of the rhombus is 37 units.

4. Is the hypotenuse the longest side of a right triangle? Justify your answer. Sol. $c^{2} = a^{2} + b^{2}$ $∴ c^{2} > a^{2}$ and $c^{2} > b^{2}$ Or $c > a$ and $c > b$ Hence, 'c' is the longest side of the right triangle.

5. True or False: Every Buadhayana triple is either a primitive triple or a scaled version of a primitive triple. Sol. Let $(a, b, c)$ be a Baudhayana triple. Then $HCF (a, b, c) 1$ or $HCF (a, b, c) \neq = 1$ $(3, 4, 5) : HCF$ is 1 $(10, 24, 26) : HCF = 2$ If HCF is 1 , then the triple is primitive If HCF is a number other than 1 , then the triple is scaled.

6. Give 5 examples of rectangles whose side lengths and diagonals are all integers.
Sol. (i)

(ii)

(iii)

(iv)

(v)

7. Construct a square whose area is equal to the difference of the areas of squares of side-lengths 5 units and 7 units.

Sol. To form a square whose area should be $\Rightarrow 7^{2} - 5^{2} = 49 - 25 = 24$ sq. units. Now, draw a right angles triangle XYZ such that $XY = 5 cm$ and $YZ = 7 cm$ .

Here, $XZ^{2} - XY^{2} = ZY^{2}$

$X Z^{2} = 7^{2} - 5^{2} = 49 - 25 = 24 XZ = 24 cm$

Then draw a square XZUV of side length $24 cm$ as shown.

Hence, the square XZUN is formed whose area is equal to difference of area of squares of side lengths 5 units and 7 units.

8. (i) Using the dots of a grid as the vertices, can you create a square that has an area of (a) 2 sq. units, (b) 3 sq. units, (c) 4 sq-units, (d) 5 unit (ii) Suppose the grid extends indefinitely. What are the possible integer-values areas of squares you can create in this manner?

Sol. (i) Since, one side of square formed will be hypotenuse 'c' of a right angled triangle formed by the grid lines.

The legs of this triangle will be 'a' units and 'b' units.

Now, $c^{2} = a^{2} + b^{2}$ Because, the dots are on a grid, a and b must be whole number ( $0, 1, 2, 3, \dots$ ). This means that the area of a square on a dot grid must be equal to sum of two perfect square. (a) Area of 2 square units

Since $2 = (1)^{2} + (1)^{2}$ So, $AP = PD = 1$ units

$\Rightarrow AD = AP^{2} + PD^{2} = 1 - 11 = 2 units$

Area of square $ABCD = 2 sq$ . units

Tilted square ABCD is formed.

(b) Area of 3 sq. units

There is no combination of two perfect square that add up to 3 .

It is impossible to draw a square with an area of 3 sq. units on a standard integer dot grid. (c) Area of 4 sq. units $∵ 4 = (2)^{2} + 0^{2} = 4 + 0 = 4$

Non tilted square ABCD is formed.

(d) Area of 5 sq. units

5 = (2)^{2} + (1)^{2}

Area of square

ABCD = 5 sq

. units Tilted square ABCD is formed.

(ii) If the grid extends indefinitely, the possible integer area are exactly the numbers that can be written as the sum of two perfect squares.

$\Rightarrow \Rightarrow 0^{2} + 1^{2} = 1 1^{2} + 1^{2} = 2 2^{2} + 0^{2} = 4 2^{2} + 1^{2} = 5 2^{2} + 2^{2} = 8 3^{2} + 0 = 9 3^{2} + 1^{2} = 10 3^{2} + 2^{2} = 13 4^{2} + 0^{2} = 16 4^{2} + 1^{2} = 17$

... and so on Here, 3 , 6 , 7 , 11 , 12 , 14 , 15 are shifted as it is impossible to draw a square on a standard integer dot grid.

9. Find the area of an equilateral triangle with sidelength 6 units. [Hint: Show that an altitude bisects the opposite side. Use this to find the height.] Sol. Let $△ ABC$ be an equilateral triangle. $AB = BC = CA = 6 cm$

Let AD be perpendicular to BC . Then $∠1 = ∠2$ (each $9 0^{\circ}$ ) $AB = AC$ (each 6 cm ) $AD = AD$ (common) $△ ADB ≅ △ ADC$ (by RHS congruency) $BD = DC$ (CPCT)

$∴ BD = DC = \frac{1}{2} \times 6 cm = 3 cm$

In $△ ADC, h^{2} + 3^{2}$

$\Rightarrow \Rightarrow h = 6^{2} (Baudhayana’s triple) h^{2} = 36 - 9 = 27 = ar (Δ ABC) = \frac{1}{2} \times BC \times AD = \frac{1}{2} \times 6 \times 33 sq. units = 93 sq. units$

3.0Key Concepts of Chapter 2 - The Baudhayana-Pythagoras Theorem Exercise 2.4

Calculating the diagonal of a square with side length 5 cm.
Determining unknown sides in the given right-angled triangles.
Using the diagonals of a rhombus measuring 24 units and 70 units to calculate its side.
Identifying the hypotenuse as the longest side of a right-angled triangle.
Working with an isosceles right triangle with a hypotenuse of 10 units.
Calculating the height and area of an equilateral triangle with side length 6 units.

4.0Benefits of NCERT Solutions Class 8 Maths Chapter 2 Exercise 2.4

Shows the steps for calculating the diagonal of a square.
Makes the calculation of unknown sides in right-angled triangles easier to follow.
Explains how rhombus diagonals are used to calculate its side.
Helps understand the position of the hypotenuse in a right-angled triangle.
Provides the method used for the isosceles right triangle problem.
Explains the calculation of the height and area of the given equilateral triangle.

Frequently Asked Questions

What is covered in Exercise 2.4?

How is the square diagonal calculated?

How are the rhombus diagonals used?

Which side is the longest in a right-angled triangle?

What is the side length of the equilateral triangle?

Where can I find Exercise 2.4 solutions?

Join ALLEN!

The Baudhayana-Pythagoras Theorem - Exercise 2.4 Class 8 Maths - Chapter 2 NCERT Solutions

1.0Download Class 8 Maths Chapter 2 Ex 2.4 NCERT Solutions PDF

2.0Detailed NCERT Solutions for The Baudhayana-Pythagoras Theorem Exercise 2.4 Class 8 Maths (Ganita Prakash Term 2)

3.0Key Concepts of Chapter 2 - The Baudhayana-Pythagoras Theorem Exercise 2.4

4.0Benefits of NCERT Solutions Class 8 Maths Chapter 2 Exercise 2.4

Frequently Asked Questions

What is covered in Exercise 2.4?

How is the square diagonal calculated?

How are the rhombus diagonals used?

Which side is the longest in a right-angled triangle?

What is the side length of the equilateral triangle?

Where can I find Exercise 2.4 solutions?

Join ALLEN!

The Baudhayana-Pythagoras Theorem - Exercise 2.4 Class 8 Maths - Chapter 2 NCERT Solutions

1.0Download Class 8 Maths Chapter 2 Ex 2.4 NCERT Solutions PDF

2.0Detailed NCERT Solutions for The Baudhayana-Pythagoras Theorem Exercise 2.4 Class 8 Maths (Ganita Prakash Term 2)

3.0Key Concepts of Chapter 2 - The Baudhayana-Pythagoras Theorem Exercise 2.4

4.0Benefits of NCERT Solutions Class 8 Maths Chapter 2 Exercise 2.4