Maharashtra Board Class 7 Maths Chapter 12 Set 49 Pythagoras Solutions

Question 1. Find the Pythagorean triplets from among the following sets of numbers:
(i) 3,4,5
(ii) 2,4,5
(iii) 4,5,6
(iv) 2,6,7
(v) 9,40,41
(vi) 4,7,8
Answer:
(i) \( 3^2 = 9 \), \( 4^2 = 16 \), \( 5^2 = 25 \) Now, \( 9 + 16 = 25 \)
\( \therefore 3^2 + 4^2 = 5^2 \)
\( \therefore \) 3, 4 and 5 is a Pythagorean triplet.
(ii) \( 2^2 = 4 \), \( 4^2 = 16 \), \( 5^2 = 25 \) But, \( 4 + 16 \neq 25 \)
\( \therefore 2^2 + 4^2 \neq 5^2 \)
\( \therefore \) 2, 4 and 5 is not a Pythagorean triplet.
(iii) \( 4^2 = 16 \), \( 5^2 = 25 \), \( 6^2 = 36 \) But \( 16 + 25 \neq 36 \)
\( \therefore 4^2 + 5^2 \neq 6^2 \)
\( \therefore \) 4, 5 and 6 is not a Pythagorean triplet.
(iv) \( 2^2 = 4 \), \( 6^2 = 36 \), \( 7^2 = 49 \) But, \( 4 + 36 \neq 49 \)
\( \therefore 2^2 + 6^2 \neq 7^2 \)
\( \therefore \) 2, 6 and 7 is not a Pythagorean triplet.
(v) \( 9^2 = 81 \), \( 40^2 = 1600 \), \( 41^2 = 1681 \) Now, \( 81 + 1600 = 1681 \)
\( \therefore 9^2 + 40^2 = 41^2 \)
\( \therefore \) 9, 40 and 41 is a Pythagorean triplet.
(vi) \( 4^2 = 16 \), \( 7^2 = 49 \), \( 8^2 = 64 \) But, \( 16 + 49 \neq 64 \)
\( \therefore 4^2 + 7^2 \neq 8^2 \)
\( \therefore \) 4, 7 and 8 is not a Pythagorean triplet.
In simple words: A Pythagorean triplet is a set of three positive integers a, b, and c, such that \( a^2 + b^2 = c^2 \). To find them, square each number in the set and check if the sum of the squares of the two smaller numbers equals the square of the largest number.

🎯 Exam Tip: Remember to always identify the largest number first, as its square will be compared to the sum of the squares of the other two numbers. Common Pythagorean triplets like (3,4,5) are frequently used in exams.

Question 2. The sides of some triangles are given below. Find out which ones are right-angled triangles?
(i) 8,15,17
(ii) 11,12,15
(iii) 11,60,61
(iv) 1.5, 1.6, 1.7
(v) 40, 20, 30
Answer:
(i) \( 8^2 = 64 \), \( 15^2 = 225 \), \( 17^2 = 289 \) Now, \( 64 + 225 = 289 \)
\( \therefore 8^2 + 15^2 = 17^2 \) The above expression is of the form \( (\text{hypotenuse})^2 = (\text{base})^2 + (\text{height})^2 \)
\( \therefore \) The sides of lengths 8, 15, 17 will form a right-angled triangle.
(ii) \( 11^2 = 121 \), \( 12^2 = 144 \), \( 15^2 = 225 \) But, \( 121 + 144 \neq 225 \)
\( \therefore 11^2 + 12^2 \neq 15^2 \)
\( \therefore \) The above expression is not of the form \( (\text{hypotenuse})^2 = (\text{base})^2 + (\text{height})^2 \)
\( \therefore \) The sides of lengths 11, 12, 15 will not form a right-angled triangle.
(iii) \( 11^2 = 121 \), \( 60^2 = 3600 \), \( 61^2 = 3721 \) Now, \( 121 + 3600 = 3721 \)
\( \therefore 11^2 + 60^2 = 61^2 \)
\( \therefore \) The above expression is of the form \( (\text{hypotenuse})^2 = (\text{base})^2 + (\text{height})^2 \)
\( \therefore \) The sides of lengths 11, 60, 61 will form a right-angled triangle.
(iv) \( 1.5^2 = 2.25 \), \( 1.6^2 = 2.56 \), \( 1.7^2 = 2.89 \) But, \( 2.25 + 2.56 \neq 2.89 \)
\( \therefore 1.5^2 + 1.6^2 \neq 1.7^2 \)
\( \therefore \) The above expression is not of the form \( (\text{hypotenuse})^2 = (\text{base})^2 + (\text{height})^2 \)
\( \therefore \) The sides of lengths 1.5, 1.6, 1.7 will not form a right-angled triangle.
(v) \( 40^2 = 1600 \), \( 20^2 = 400 \), \( 30^2 = 900 \) But, \( 400 + 900 \neq 1600 \)
\( \therefore 20^2 + 30^2 \neq 40^2 \)
\( \therefore \) The above expression is not of the form \( (\text{hypotenuse})^2 = (\text{base})^2 + (\text{height})^2 \)
\( \therefore \) The sides of lengths 40, 20, 30 will not form a right-angled triangle.
In simple words: A triangle is a right-angled triangle if the square of its longest side (hypotenuse) is equal to the sum of the squares of the other two sides. This is based on the Pythagorean theorem.

🎯 Exam Tip: When checking for right-angled triangles, always ensure you correctly identify the longest side to be tested as the hypotenuse. Carefully calculate squares, especially for decimals, to avoid errors.

Maharashtra Board Class 7 Maths Chapter 13 Pythagoras' Theorem Practice Set 49 Intext Questions And Activities

Question 1. From the numbers 1 to 50, pick out the Pythagorean triplets. (Textbook pg. no. 90)
Answer:
(1) 3,4,5
(2) 5,12,13
(3) 7,24,25
(4) 8,15,17
(5) 9,40,41
(6) 12,35,37
(7) 20,21,29
In simple words: Pythagorean triplets are sets of three integers (a, b, c) where \(a^2 + b^2 = c^2\). These triplets represent the sides of a right-angled triangle.

🎯 Exam Tip: Memorizing common Pythagorean triplets can save time during exams, especially when dealing with problems involving right-angled triangles.