Math Case Study Class 8 Factorisation and Identities | CBSE Practice Questions

Math Case Study Class 8 Factorisation and Identities

A math case study class 8 factorisation and identities helps students understand algebra through real-life problems. It builds logical and analytical skills using math case study questions designed around formulas and simplification techniques. Moreover, this topic enhances clarity in algebraic problem-solving.

Understanding Factorisation and Identities

Practicing factorisation and identities strengthens a student’s foundation in algebra and prepares them for Case Study math questions for class 9 and math case study questions class 9. Additionally, it helps in applying algebraic rules effectively while solving complex problems.

Practice and Application

Students should regularly practice math case study questions from NCERT books and worksheets. Furthermore, solving PDF-based examples ensures concept retention and better exam performance.

Case Study 5: School Garden Project

The students of Class 8 at Sunshine Public School decided to create a rectangular garden in their schoolyard. The garden’s length is \((2x + 3y)\) meters, and its width is \((2x – 3y)\) meters. The school also plans to build a square-shaped flower bed inside the garden with each side measuring \((x + y)\) meters.

The students want to calculate the area of the garden and the area of the flower bed to determine how much space will be left for planting vegetables. They also want to explore how the dimensions of the garden and flower bed relate to each other using algebraic identities.

1. What is the area of the garden?
Solution:

The area of a rectangle is given by the product of its length and width. Here, the length is \((2x + 3y)\) and the width is \((2x – 3y)\). Using the identity \((a + b)(a – b) = a^2 – b^2\), the area is:

\((2x + 3y)(2x – 3y) = (2x)^2 – (3y)^2 = 4x^2 – 9y^2\)
2. What is the area of the square-shaped flower bed?
Solution:

The area of a square is given by the square of its side. Here, the side of the flower bed is \((x + y)\). Using the identity \((a + b)^2 = a^2 + 2ab + b^2\), the area is:

\((x + y)^2 = x^2 + 2xy + y^2\)
3. If \(x = 5\) and \(y = 2\), what is the area of the garden?
Solution:

Substitute \(x = 5\) and \(y = 2\) into the area of the garden \(4x^2 – 9y^2\):

\(4(5)^2 – 9(2)^2 = 4(25) – 9(4) = 100 – 36 = 64\)
Note: The correct answer is 64, which is not listed in the options. The closest option is (a) 76, but the correct value is 64.
4. What is the difference between the area of the garden and the area of the flower bed?
Solution:

Subtract the area of the flower bed from the area of the garden:

\((4x^2 – 9y^2) – (x^2 + 2xy + y^2) = 3x^2 – 2xy – 10y^2\)
Note: The correct expression is \(3x^2 – 2xy – 10y^2\), which does not match any of the options. The closest option is (d) \(3x^2 – 6xy – 10y^2\).
5. If the area of the flower bed is 25 square meters, which of the following could be the values of \(x\) and \(y\)?
Solution:

The area of the flower bed is \(x^2 + 2xy + y^2 = 25\). Substitute the options:

  • For \(x = 5, y = 0\): \(5^2 + 2(5)(0) + 0^2 = 25\), which satisfies the equation.

Review your answers and solutions below:

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