NCERT Solutions for Class 9 Maths Exercise 2.4 Question 15
Understanding the Question 🧐
This question connects algebra with geometry. We know that the area of a rectangle is calculated by multiplying its length and breadth (Area = Length × Breadth). Here, we are given the area as a quadratic polynomial and asked to find possible algebraic expressions for the length and breadth. To do this, we need to factorise the given quadratic polynomial. The resulting factors will be the expressions for the length and breadth. This page offers detailed ncert solutions using the ‘splitting the middle term’ method.
Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:
(i) Area: &&25a^2 – 35a + 12&&
(ii) Area: &&35y^2 + 13y – 12&&
Part (i): Area = &&25a^2 – 35a + 12&& 📝
To find the length and breadth, we need to factorise the quadratic trinomial &&25a^2 – 35a + 12&&.
Step 1: Identify the coefficients and find the product AC.
Comparing with &&Ax^2 + Bx + C&&, we have &&A=25, B=-35, C=12&&.
Product AC = &&25 \times 12 = 300&&.
Sum = B = &&-35&&.
Step 2: Find two numbers that multiply to 300 and add to -35.
Since the product is positive (&&+300&&) and the sum is negative (&&-35&&), both numbers must be negative. Let’s find factors of 300.
After checking pairs like (&&-10, -30&&), (&&-12, -25&&), we find that &&-15&& and &&-20&& work:
&&(-15) \times (-20) = 300&& (Product)
&&(-15) + (-20) = -35&& (Sum)
Step 3: Split the middle term.
We rewrite &&-35a&& as &&-15a – 20a&&.
Area = &&25a^2 – 15a – 20a + 12&&
Step 4: Factor by grouping.
Group the first two and the last two terms:
&&= (25a^2 – 15a) + (-20a + 12)&&
Factor out the greatest common factor from each group:
&&= 5a(5a – 3) – 4(5a – 3)&&
Step 5: Write the final factors.
Factor out the common binomial &&(5a – 3)&&:
Area = &&(5a – 3)(5a – 4)&&
Answer: Since Area = Length × Breadth, the possible expressions are:
Length = &&(5a – 3)&& and Breadth = &&(5a – 4)&& (or vice-versa).
Part (ii): Area = &&35y^2 + 13y – 12&& 📝
Let’s factorise the quadratic expression &&35y^2 + 13y – 12&&.
Step 1: Identify coefficients and find the product AC.
Here, &&A=35, B=13, C=-12&&.
Product AC = &&35 \times (-12) = -420&&.
Sum = B = &&13&&.
Step 2: Find two numbers that multiply to -420 and add to 13.
Since the product is negative, one number will be positive and the other negative. Since the sum is positive, the larger number must be positive. Let’s check factors of 420.
After checking pairs, we find &&28&& and &&-15&&:
&&28 \times (-15) = -420&& (Product)
&&28 + (-15) = 13&& (Sum)
Step 3: Split the middle term.
We rewrite &&+13y&& as &&+28y – 15y&&.
Area = &&35y^2 + 28y – 15y – 12&&
Step 4: Factor by grouping.
&&= (35y^2 + 28y) + (-15y – 12)&&
Factor out the GCF from each group:
&&= 7y(5y + 4) – 3(5y + 4)&&
Step 5: Write the final factors.
Factor out the common binomial &&(5y + 4)&&:
Area = &&(5y + 4)(7y – 3)&&
Answer: The possible expressions for the dimensions are:
Length = &&(5y + 4)&& and Breadth = &&(7y – 3)&& (or vice-versa).
- If &&AC&& is positive and &&B&& is positive, both numbers are positive.
- If &&AC&& is positive and &&B&& is negative, both numbers are negative.
- If &&AC&& is negative, one number is positive and one is negative.
- Factorising a quadratic expression representing area gives the dimensions (length and breadth).
- The “splitting the middle term” method is a reliable way to factorise trinomials.
- The choice of which factor is length and which is breadth is arbitrary.
FAQ
Q: How is the area of a rectangle related to its length and breadth?
A: The area of a rectangle is the product of its length and breadth (Area = Length × Breadth).
Q: What mathematical process is used to find the length and breadth from the area’s polynomial?
A: The process used is the factorisation of the quadratic polynomial. When we factor the polynomial representing the area, we get two expressions multiplied together. These two expressions represent the possible length and breadth.
Q: What is the ‘splitting the middle term’ method?
A: For a quadratic polynomial &&Ax^2 + Bx + C&&, this method involves finding two numbers that multiply to give &&A \times C&& and add up to give &&B&&. We then use these two numbers to ‘split’ the middle term &&Bx&& into two terms and then factor the expression by grouping.
Q: For the area &&25a^2 – 35a + 12&&, what two numbers were used to split the middle term?
A: We needed two numbers that multiply to &&300&& (&&25 \times 12&&) and add to &&-35&&. The numbers are &&-15&& and &&-20&&.
Q: For the area &&35y^2 + 13y – 12&&, what two numbers were used?
A: We needed two numbers that multiply to &&-420&& (&&35 \times -12&&) and add to &&13&&. The numbers are &&28&& and &&-15&&.
Q: Are the answers for length and breadth interchangeable?
A: Yes. Since multiplication is commutative (Length × Breadth = Breadth × Length), either factor can represent the length and the other will represent the breadth.
Further Reading
For more practice on factorising quadratic polynomials, you can refer to the official NCERT textbook and resources available on their website: https://ncert.nic.in/
