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

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

(A) $18$

(B) $10$

(C) $15$

(D) $36$

(A) $ \pm 11\sqrt {17} $

(B) $ \pm 13\sqrt {17} $

(C) $ \pm 16\sqrt {17} $

(D) $ \pm 17\sqrt {17} $

A) 0.9664

B) 0.9604

C) 0.9864

D) 0.9964

A) 1062608

B) 1061208

C) 1082908

D) None of these

A) 35

B) 65

C) 145

D) 95

## 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.

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