Math Problem Statement

Find the square of the given numbers and write a Pythagorean triplet with one member being the specified number.

Solution

The exercise in the image contains problems related to square roots, squares, and Pythagorean triplets. Let’s go through each section you might need help with.

Exercise Details

  1. Find the square of the following numbers:

    • (i) 6
    • (ii) 14
    • (iii) 16
    • (iv) 18
    • (v) 32
    • (vi) 35
    • (vii) 71
    • (viii) 46

    To solve these, simply square each number. For example, 62=366^2 = 36, 142=19614^2 = 196, and so on.

  2. Write a Pythagorean triplet whose one member is:

    • (i) 6
    • (ii) 14
    • (iii) 16
    • (iv) 18
    • (v) 86

    To create Pythagorean triplets, you can use the formulas a=m2n2a = m^2 - n^2, b=2mnb = 2mn, and c=m2+n2c = m^2 + n^2, where mm and nn are positive integers with m>nm > n. For each given number, you’d need to find suitable values for mm and nn such that one of the generated triplet numbers matches the given value.

  3. Square Roots Section:

    • There’s a problem about finding the side of a square with an area of 144 cm².
      • Since the area of a square is given by side2\text{side}^2, you can find the side by taking the square root: 144=12\sqrt{144} = 12.

Let me know if you want the full solutions for each of these or have any specific questions.


Additional Questions

  1. What are the steps to find the square of a number?
  2. How do you identify Pythagorean triplets, and why do the formulas work?
  3. What is the relationship between area and side length in a square?
  4. How do you simplify square roots?
  5. Can you generate a list of other common Pythagorean triplets?

Tip: Remember that for perfect squares, the square root is a whole number, making calculations straightforward.

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Math Problem Analysis

Mathematical Concepts

Square Roots
Squares
Pythagorean Triplets

Formulas

Square of a number: n^2
Pythagorean triplet formulas: a = m^2 - n^2, b = 2mn, c = m^2 + n^2

Theorems

Pythagorean theorem

Suitable Grade Level

Grade 8