KSEEB Solutions for Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Ex 5.2

In this chapter, we provide KSEEB SSLC Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Ex 5.2 for English medium students, Which will very helpful for every student in their exams. Students can download the latest KSEEB SSLC Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Ex 5.2 pdf, free KSEEB SSLC Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Ex 5.2 pdf download. Now you will get step by step solution to each question.

Karnataka State Syllabus Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Ex 5.2

Karnataka Board Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Ex 5.2

Question 1.
Find the sum 1 + 3 + 5 + _____ +51 (the sum fo all odd numbers from 1 to 51) without actually adding them.
Answer:
The sum of first n odd numbers is n2
∴ 1 +3 + 5+ _____ + 51 = 262 = 676.

Question 2.
Express 144 as a sum of 12 odd numbers.
Answer:
144 = 122 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23.

Question 3.
Find the 14th and 15th triangular numbers and find their sum verify the statement 8 for this sum.
Answer:
14 th triangular number =1 + 2 + 3+4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 = 105.
15 th triangular number = 1 + 2 + 3 + 4 + ___ + 15 = 120
Sum of 105 + 120 = 225
According to statement 8, the sum of 14th and 15th triangular numbers in equal to
n + (n + 1) = (n + 1)2
105 + 120 = (14 + 1)2
225 = 225.

Question 4.
What are the remainders of a perfect square when divided by 5?
Answer:
When perfect squares 9, 16, 25, 36, 49, 64 are divided by 5 the remainders are 4, 1, 0, 1, 4, 4 respectively.
Therefore the remainders of a perfect square when divided by 5 are 0, 1, and 4.

All Chapter KSEEB Solutions For Class 8 maths

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