NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities (EX 9.2) Exercise 9.2

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities (EX 9.2) Exercise 9.2

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NCERT Solutions for Class 8 Maths Chapter 9 part-1

Access NCERT solutions for Class 8 Chapter 9-Algebraic Expressions and Identities

Exercise 9.2

Refer to page 5-7 for Exercise 9.2 in the PDF

1. Find the product of the following pairs of monomials.

i. $4$ and $7p$

Ans: The required product is,

$4 \times 7p = 4 \times 7 \times p$

$4 \times 7p = 28p$


iii. $ - 4p$ and $7p$

Ans: The required product is,

$ - 4p \times 7p = \left( { - 4} \right) \times p \times 7 \times p$

$ - 4p \times 7p = \left( { - 4 \times 7} \right) \times \left( {p \times p} \right)$

$ - 4p \times 7p =  - 28{p^2}$


iii. $ - 4p$ and $7pq$

Ans: The required product is,

$ - 4p \times 7pq = \left( { - 4} \right) \times p \times 7 \times p \times q$

$ - 4p \times 7pq = \left( { - 4 \times 7} \right) \times \left( {p \times p} \right) \times q$

$ - 4p \times 7pq =  - 28{p^2}q$


iv. $4{p^3}$ and $ - 3p$

Ans: The required product is,

\[4{p^3} \times \left( { - 3p} \right) = 4 \times {p^3} \times \left( { - 3} \right) \times p\]

$4{p^3} \times \left( { - 3p} \right) = \left( {4 \times  - 3} \right) \times \left( {{p^3} \times p} \right)$

$4{p^3} \times \left( { - 3p} \right) =  - 12{p^4}$


v. $4p$ and $0$

Ans: The required product is,

$4p \times \left( 0 \right) = 4 \times p \times 0$

$4p \times \left( 0 \right) = 0$


2. Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.

$\left( {p,q} \right)$;$\left( {10m,5n} \right)$;$\left( {20{x^2},5{y^2}} \right)$;$\left( {4x,3{x^2}} \right)$;$\left( {3mn,4np} \right)$.

Ans: The area of a rectangle is the product of length and breadth.

The first rectangle has dimensions, $\left( {p,q} \right)$. Let the area be ${A_1}$. Thus, ${A_1} = pq$.

The second rectangle has dimensions, $\left( {10m,5n} \right)$. Let the area be ${A_2}$. Thus, ${A_2} = 10m \times 5n$

${A_2} = 10 \times 5 \times m \times n$

${A_2} = 50mn$

The third rectangle has dimensions, $\left( {20{x^2},5{y^2}} \right)$. Let the area be ${A_3}$. Thus, ${A_3} = 20{x^2} \times 5{y^2}$

${A_3} = \left( {20 \times 5} \right) \times \left( {{x^2} \times {y^2}} \right)$

${A_3} = 100{x^2}{y^2}$

The third rectangle has dimensions, $\left( {4x,3{x^2}} \right)$. Let the area be ${A_4}$. Thus, ${A_4} = 4x \times 3{x^2}$

${A_4} = \left( {4 \times 3} \right) \times \left( {x \times {x^2}} \right)$

${A_4} = 12{x^3}$

The third rectangle has dimensions, $\left( {3mn,4np} \right)$. Let the area be ${A_5}$. Thus, ${A_5} = 3mn \times 4np$

${A_5} = \left( {3 \times 4} \right) \times \left( {m \times n \times n \times p} \right)$

${A_5} = 12m{n^2}p$


3. Complete the table of products.

$\frac{{{\text{First monomial}} \to }}{{{\text{Second monomial}} \downarrow }}$

$2x$

$ - 5y$

$3{x^2}$

$ - 4xy$

$7{x^2}y$

$ - 9{x^2}{y^2}$

$2x$

$4{x^2}$






$ - 5y$



$ - 15{x^2}y$




$3{x^2}$







$ - 4xy$







$7{x^2}y$







$ - 9{x^2}{y^2}$








Ans:  Multiply the term in particular row with respective column to complete the table.

$\frac{{{\text{First monomial}} \to }}{{{\text{Second monomial}} \downarrow }}$

$2x$

$ - 5y$

$3{x^2}$

$ - 4xy$

$7{x^2}y$

$ - 9{x^2}{y^2}$

$2x$

$4{x^2}$

$ - 10xy$

$6{x^2}$

$ - 8{x^2}y$

$14{x^3}y$

$ - 18{x^2}{y^2}$

$ - 5y$

$ - 10xy$

$25{y^2}$

$ - 15{x^2}y$

$20x{y^2}$

$ - 35{x^2}{y^2}$

$45{x^2}{y^3}$

$3{x^2}$

$6{x^2}$

$ - 15{x^2}y$

$9{x^4}$

$ - 12{x^3}y$

$21{x^4}y$

$ - 27{x^4}{y^2}$

$ - 4xy$

$ - 8{x^2}y$

$20x{y^2}$

$ - 12{x^3}y$

$16{x^2}{y^2}$

$ - 28{x^3}{y^2}$

$36{x^3}{y^3}$

$7{x^2}y$

$14{x^3}y$

$ - 35{x^2}{y^2}$

$21{x^4}y$

$ - 28{x^3}{y^3}$

$49{x^4}{y^2}$

$ - 63{x^3}{y^3}$

$ - 9{x^2}{y^2}$

$ - 18{x^2}{y^2}$

$45{x^2}{y^3}$

$ - 27{x^4}{y^2}$

$36{x^3}{y^3}$

$ - 63{x^3}{y^3}$

\[81{x^4}{y^4}\]


4. Find the volume of rectangular boxes with the following length, and breadth, and height respectively.

i. $5a,3{a^2},7{a^4}$

Ans: The volume of a rectangle is the product of length, breadth and height.

The rectangular box has dimensions, $5a$, $7{a^4}$, and $3{a^2}$. Let the volume be ${V_1}$. Thus,${V_1} = 5a \times 3{a^2} \times 7{a^4}$

${V_1} = 5 \times 3 \times 7 \times a \times {a^2} \times {a^4}$

${V_1} = 105{a^7}$


ii. $2p$,$4q$,$8r$

Ans: The rectangular box has dimensions, $2p$,$4q$,$8r$. Let the volume be ${V_2}$. Thus, 

${V_2} = 2p \times 4q \times 8r$

${V_2} = 2 \times 4 \times 8 \times p \times q \times r$

${V_2} = 64pqr$


iii. $xy$,$2{x^2}y$,$2x{y^2}$

Ans: The rectangular box has dimensions, $xy$,$2{x^2}y$,$2x{y^2}$. Let the volume be ${V_3}$. Thus, ${V_3} = xy \times 2{x^2}y \times 2x{y^2}$

${V_3} = 2 \times 2 \times x \times {x^2} \times x \times y \times y \times {y^2}$

${V_3} = 4{x^4}{y^4}$


iv. $a$,$2b$,$3c$

Ans: The rectangular box has dimensions, $a$,$2b$,$3c$. Let the volume be ${V_3}$. Thus, \[{V_4} = a \times 2b \times 3c\]

\[{V_4} = 2 \times 3 \times a \times b \times c\]

\[{V_4} = 6abc\]


5. Obtain the product of the following:

i. $xy$,$yz$,$zx$

Ans: Group the like terms and multiply.

$xy \times yz \times zx = {x^2}{y^2}{z^2}$


ii. $a$,$ - {a^2}$,${a^3}$

Ans: Group the like terms and multiply.

$a \times \left( { - {a^2}} \right) \times {a^3} =  - {a^6}$


iii. $2$,$4y$,$8{y^2}$,$16{y^3}$

Ans: Group the like terms and multiply.

$2 \times 4y \times 8{y^2} \times 16{y^3} = 2 \times 4 \times 8 \times 16 \times y \times {y^2} \times {y^3}$

$2 \times 4y \times 8{y^2} \times 16{y^3} = 1024{y^6}$


iv. $a$,$2b$,$3c$,$6abc$

Ans: Group the like terms and multiply.

$a \times 2b \times 3c \times 6abc = 2 \times 3 \times 6 \times a \times b \times c \times abc$

$a \times 2b \times 3c \times 6abc = 36{a^2}{b^2}{c^2}$


v. $m$,\[ - mn\],$mnp$

Ans: Group the like terms and multiply.

$m \times \left( { - mn} \right) \times mnp = m \times \left( { - m} \right) \times m \times n \times p$

$m \times \left( { - mn} \right) \times mnp =  - {m^3}{n^2}p$


NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.2

Opting for the NCERT solutions for Ex 9.2 Class 8 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 9.2 Class 8 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 8 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 8 Maths Chapter 9 Exercise 9.2 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 8 Maths Chapter 9 Exercise 9.2, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it. 

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