{"id":108685,"date":"2022-02-14T20:55:10","date_gmt":"2022-02-14T15:25:10","guid":{"rendered":"https:\/\/infinitylearn.com\/surge\/?p=108685"},"modified":"2025-01-27T17:40:05","modified_gmt":"2025-01-27T12:10:05","slug":"ncert-solutions-for-class-8-maths-chapter-9-algebraic-expressions-and-identities-ex-9-2","status":"publish","type":"post","link":"https:\/\/infinitylearn.com\/surge\/study-material\/ncert-solution\/class-8\/maths\/ncert-solutions-for-class-8-maths-chapter-9-algebraic-expressions-and-identities-ex-9-2\/","title":{"rendered":"NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.2"},"content":{"rendered":"

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.2<\/h2>\n

 <\/p>\n

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.2<\/strong><\/p>\n

Ex 9.2 Class 8 Maths Question 1.
\nFind the product of the following pairs of monomials.
\n(i) 4, 7p
\n(ii) -4p, 7p
\n(iii) -4p, 7pq
\n(iv) 4p3<\/sup>, -3p
\n(v) 4p, 0
\nSolution:
\n(i) 4 \u00d7 7p = (4 \u00d7 7) \u00d7 p = 28p
\n(ii) -4p \u00d7 7p = (-4 \u00d7 7) \u00d7 p \u00d7 p = -28p2<\/sup>
\n(iii) -4p \u00d7 7pq = (-4 \u00d7 7) \u00d7 p \u00d7 pq = -28p2<\/sup>q
\n(iv) 4p3<\/sup> \u00d7 -3p = (4 \u00d7 -3) \u00d7 p3<\/sup> \u00d7 p = -12p4<\/sup>
\n(v) 4p x 0 = (4 \u00d7 0) \u00d7 p = 0 \u00d7 p = 0<\/p>\n

Ex 9.2 Class 8 Maths Question 2.
\nFind the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.
\n(p, q); (10m, 5n); (20x2<\/sup>, 5y2<\/sup>); (4x, 3x2<\/sup>); (3mn, 4np)
\nSolution:
\n(i) Length = p units and breadth = q units
\nArea of the rectangle = length \u00d7 breadth = p \u00d7 q = pq sq units
\n(ii) Length = 10 m units, breadth = 5n units
\nArea of the rectangle = length \u00d7 breadth = 10 m \u00d7 5 n = (10 \u00d7 5) \u00d7 m \u00d7 n = 50 mn sq units
\n(iii) Length = 20x2<\/sup> units, breadth = 5y2<\/sup> units
\nArea of the rectangle = length \u00d7 breadth = 20x2<\/sup> \u00d7 5y2<\/sup> = (20 \u00d7 5) \u00d7 x2<\/sup> \u00d7 y2<\/sup> = 100x2<\/sup>y2<\/sup> sq units
\n(iv) Length = 4x units, breadth = 3x2<\/sup> units
\nArea of the rectangle = length \u00d7 breadth = 4x \u00d7 3x2<\/sup> = (4 \u00d7 3) \u00d7 x \u00d7 x2<\/sup> = 12x3<\/sup> sq units
\n(v) Length = 3mn units, breadth = 4np units
\nArea of the rectangle = length \u00d7 breadth = 3mn \u00d7 4np = (3 \u00d7 4) \u00d7 mn \u00d7 np = 12mn2<\/sup>p sq units<\/p>\n

Ex 9.2 Class 8 Maths Question 3.
\nComplete the table of Products.
\n\"NCERT
\nSolution:
\nCompleted Table
\n\"NCERT<\/p>\n

Ex 9.2 Class 8 Maths Question 4.
\nObtain the volume of rectangular boxes with the following length, breadth and height respectively.
\n(i) 5a, 3a2<\/sup>, 7a4<\/sup>
\n(ii) 2p, 4q, 8r
\n(iii) xy, 2x2<\/sup>y, 2xy2<\/sup>
\n(iv) a, 2b, 3c
\nSolution:
\n(i) Here, length = 5a, breadth = 3a2<\/sup>, height = 7a4<\/sup>
\nVolume of the box = l \u00d7 b \u00d7 h = 5a \u00d7 3a2<\/sup> \u00d7 7a4<\/sup> = 105 a7<\/sup> cu. units
\n(ii) Here, length = 2p, breadth = 4q, height = 8r
\nVolume of the box = l \u00d7 b \u00d7 h = 2p \u00d7 4q \u00d7 8r = 64pqr cu. units
\n(iii) Here, length = xy, breadth = 2x2<\/sup>y, height = 2xy2<\/sup>
\nVolume of the box = l \u00d7 b \u00d7 h = xy \u00d7 2x2<\/sup>y \u00d7 2xy2<\/sup> = (1 \u00d7 2 \u00d7 2) \u00d7 xy \u00d7 x2<\/sup>y \u00d7 xy2<\/sup> = 4x4<\/sup>y4<\/sup> cu. units
\n(iv) Here, length = a, breadth = 2b, height = 3c
\nVolume of the box = length \u00d7 breadth \u00d7 height = a \u00d7 2b \u00d7 3c = (1 \u00d7 2 \u00d7 3)abc = 6 abc cu. units<\/p>\n

Ex 9.2 Class 8 Maths Question 5.
\nObtain the product of
\n(i) xy, yz, zx
\n(ii) a, -a2<\/sup>, a3<\/sup>
\n(iii) 2, 4y, 8y2<\/sup>, 16y3<\/sup>
\n(iv) a, 2b, 3c, 6abc
\n(v) m, -mn, mnp
\nSolution:
\n(i) xy \u00d7 yz \u00d7 zx = x2<\/sup>y2<\/sup>z2<\/sup>
\n(ii) a \u00d7 (-a2<\/sup>) \u00d7 a3<\/sup> = -a6<\/sup>
\n(iii) 2 \u00d7 4y \u00d7 8y2<\/sup> \u00d7 16y3<\/sup> = (2 \u00d7 4 \u00d7 8 \u00d7 16) \u00d7 y \u00d7 y2<\/sup> \u00d7 y3<\/sup> = 1024y6<\/sup>
\n(iv) a \u00d7 2b \u00d7 3c \u00d7 6abc = (1 \u00d7 2 \u00d7 3 \u00d7 6) \u00d7 a \u00d7 b \u00d7 c \u00d7 abc = 36 a2<\/sup>b2<\/sup>c2<\/sup>
\n(v) m \u00d7 (-mn) \u00d7 mnp = [1 \u00d7 (-1) \u00d7 1 ]m \u00d7 mn \u00d7 mnp = -m3<\/sup>n2<\/sup>p<\/p>\n

\"NCERT<\/p>\n

\"NCERT<\/p>\n

\"NCERT<\/p>\n

 <\/p>\n","protected":false},"excerpt":{"rendered":"

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.2   NCERT Solutions for Class 8 […]<\/p>\n","protected":false},"author":7,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_yoast_wpseo_focuskw":"","_yoast_wpseo_title":"","_yoast_wpseo_metadesc":"Get NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.2 Infinity Learn.","custom_permalink":"study-material\/ncert-solution\/class-8\/maths\/ncert-solutions-for-class-8-maths-chapter-9-algebraic-expressions-and-identities-ex-9-2\/"},"categories":[13,19,21],"tags":[],"table_tags":[],"acf":[],"yoast_head":"\nNCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.2 - Infinity Learn by Sri Chaitanya<\/title>\n<meta name=\"description\" content=\"Get NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.2 Infinity Learn.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/infinitylearn.com\/surge\/study-material\/ncert-solution\/class-8\/maths\/ncert-solutions-for-class-8-maths-chapter-9-algebraic-expressions-and-identities-ex-9-2\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.2 - 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