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The Baudhayana-Pythagoras Theorem - Exercise 2.2 Class 8 Maths - Chapter 2 NCERT Solutions

The second exercise of the chapter The Baudhayana-Pythagoras Theorem contains detailed solutions along with answers, prepared by the ALLEN expert team to support students in their academic development and exam preparation. You can download the PDF for free and study the solutions anytime, anywhere, at your own convenience.The NCERT module uploaded states that Exercise 2.2 is mainly concerned with the application of the Baudhāyana-Pythagoras Theorem to various geometric problems. This includes computing the diagonal of a square, the missing sides of right-angled triangles, the side of a rhombus from its diagonals, and recognising primitive and scaled Baudhāyana triples. The exercise also involves the construction of an equilateral triangle, construction of a square on the difference of the areas of two squares and determination of the height and area of an equilateral triangle.

1.0Download Class 8 Maths Chapter 2 Ex 2.2 NCERT Solutions PDF

Download the Exercise 2.2 detailed solution PDF for free to solve and learn the important questions and practise efficiently for exams.

Chapter 2 : The Baudhayana - The Pythagoras Theorem Exercise 2.2

2.0Detailed NCERT Solutions onTheBaudhayana-Pythagoras Theorem Exercise 2.2 Class8 Maths (Ganita Prakash Term2)

1. If a right-angled triangle has shorter sides of lengths 5 cm and 12 cm , then what is the length of its hypotenuse? First draw the right-angled triangle with these side lengths and measure the hypotenuse, then check your answer using Baudhāyana's Theorem.

Sol. Steps of construction: Step-1: Draw a line segment $AB = 5 cm$ Step-2: At a point A, draw a perpendicular and then cut an arc of length 12 cm such that AC $= 12 cm$ .

Step-3: Join B and C Measure the length of BC.

By measurement, $BC = 13 cm$ Now, using Baudhāyana's Theorem $AC^{2} + AB^{2} = BC^{2}$ $1 2^{2} + 5^{2} = BC^{2}$ $144 + 25 = BC^{2}$ $169 = BC^{2}$ $BC = 13 cm$ Therefore, the hypotenuse of the rightangled triangle is 13 cm , according to Baudhāyana's Theorem.

2. If a right-angled triangle has a short side of length 8 cm and hypotenuse of length 17 cm . What is the length of the third side? Again, try drawing the triangle and measuring, and then check your answer using Baudhāyana's Theorem.

Sol. Step of construction: Step-1: Draw a line segment $PQ = 8 cm$ Step-2: Draw a line segment line passing through point P .

Step-3: With centre Q and radius 17 cm , draw an arc cutting perpendicular line at R. Join Q & R.

Step-4: Required $△ PQR$ is formed. Measure the length of PR.

By measurement PR=15 cm. Now, using Baudhāyana's theorem $PQ^{2} + PR^{2} = QR^{2}$ $8^{2} + PR^{2} = 1 7^{2}$ $PR^{2} = 289 - 64 = 225 = 1 5^{2}$ $∴ PR = 15 cm$ Therefore, the other side of the rightangled triangle is 15 cm , which satisfied Baudhāyana's theorem.

3. Using the constructions, you have now seen, how would you construct a square whose area is triple the area of a given square? Five times the area of a given square. (Baudhayāna's Ṡulba-Ṡutra, Verse 1.10)

Sol. (a) Steps of construction Step-1: Draw a square ABCD of side length $= 2 cm$ Then $AC = 22$ cm

Step-2: Draw $AC ⊥ CE$ such that $CE =$ 2 cm

By Baudhāyana Pythagoras theorem, $AE = 12 cm$ .

Step-3: Now, draw a square AEGH of side length $12 cm$ .

Proof: Here, ar (AEGH) = $12 cm^{2}$ $ar (ABCD) = (2)^{2} = 4 cm^{2}$

From (i) and (ii), we conclude, $ar (AEGH) = 3 \times ar (ABCD)$

(b) Steps of construction:

Step-1: Draw two identical square of side lengths 'a' on keeping side by side.

Step-2: Join D and E to make a rightangled triangle ADE.

Step-3: Draw a square on the hypotenuse DE.

The area of square DEGH formed is 5 times the area of square ABCD .

Proof: The area of square $A BC D$ of side length ' a ' $= a^{2}$

In $△ ADE, AD = a$ and $AE = 2 a$ $DE = a^{2} + (2 a)^{2} = 5 a^{2}$ Area of square DEGH $= 5 a^{2}$ $= 5$ times the area of square ABCD .

4. Let $a, b$ and c denote the length of the sides of a right triangle, with c being the length of the hypotenuse. Find the missing side length in each of the following cases: (i) $a = 5, b = 7$ (ii) $a = 8, b = 12$ (iii) $a = 9, c = 15$ (iv) $a = 7, b = 12$ (v) $a = 1.5, b = 3.5$

Sol. Here $c^{2} = a^{2} + b^{2}$ (i) Now, $c^{2} = 5^{2} + 7^{2} = 25 + 49 = 74$ $\Rightarrow c = 74$ (ii) Now, $c^{2} = 8^{2} + 1 2^{2} = 64 + 144 = 208$

$\Rightarrow c = 208 = 2 \times 2 \times 2 \times 2 \times 13 = 413$

(iii) Here, $1 5^{2} = 9^{2} + b^{2} = b^{2} = 1 5^{2} + 9^{2}$

$= 225 - 81 = 144 \Rightarrow b = 144$ $= 2 \times 2 \times 2 \times 2 \times 3 \times 3 = 12$

(iv) $c^{2} = 7^{2} + 1 2^{2} = 49 + 144 = 193$ $\Rightarrow c = 193$ (v) $c^{2} = 1. 5^{2} + 3. 5^{2} = 2.25 + 12.25 = 14.5$ $\Rightarrow c = 14.5$

3.0Key Concepts of Chapter 2 - The Baudhayana-Pythagoras Theorem Exercise 2.2

Finding the diagonal of a square using the Baudhāyana-Pythagoras Theorem.
Finding a missing side of a right-angled triangle when the other two sides are given.
Finding the side of a rhombus using its diagonals.
Understanding why the hypotenuse is the longest side of a right-angled triangle.
Identifying primitive and scaled Baudhāyana triples using their common factors.
Finding integer-sided rectangles with integer diagonals.
Constructing a square whose area is the difference between the areas of two given squares.
Understanding which integer areas are possible for squares drawn on an integer dot grid.
Using the theorem to find the height and area of an equilateral triangle.

4.0Benefits of NCERT Solutions Class 8 Maths Chapter 2 Exercise 2.2

Helps find the diagonal of a square using the Baudhāyana-Pythagoras Theorem.
Explains how to calculate missing sides in different right-angled triangles.
Makes problems involving rhombus diagonals and side lengths easier to solve.
Helps understand primitive and scaled Baudhāyana triples through examples.
Provides step-by-step methods for square construction and area-based problems.
Explains how to identify possible square areas on an integer dot grid.
Shows the application of the theorem in finding the area of an equilateral triangle.

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

Exercise 2.1