{"id":7325,"date":"2023-05-26T14:52:40","date_gmt":"2023-05-26T09:22:40","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=7325"},"modified":"2023-05-27T17:14:44","modified_gmt":"2023-05-27T11:44:44","slug":"ap-board-8th-class-maths-solutions-chapter-11-ex-11-4","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/ap-board-8th-class-maths-solutions-chapter-11-ex-11-4\/","title":{"rendered":"AP Board 8th Class Maths Solutions Chapter 11 Algebraic Expressions Ex 11.4"},"content":{"rendered":"

AP State Syllabus AP Board 8th Class Maths Solutions<\/a> Chapter 11 Algebraic Expressions Ex 11.4 Textbook Questions and Answers.<\/p>\n

AP State Syllabus 8th Class Maths Solutions 11th Lesson Algebraic Expressions Exercise 11.4<\/h2>\n

\"AP<\/p>\n

Question 1.
\nSelect a suitable identity and find the following products
\n(i) (3k + 4l)(3k + 4l)
\n(ii) (ax2<\/sup> + by2<\/sup>)(ax2<\/sup> + by2<\/sup>)
\n(iii) (7d – 9e)(7d – 9e)
\n(iv) (m2<\/sup> – n2<\/sup>)(m2<\/sup> + n2<\/sup>)
\n(v) (3t + 9s) (3t – 9s)
\n(vi) (kl – mn) (kl + mn)
\n(vii) (6x + 5)(6x + 6)
\n(viii) (2b – a)(2b +c)
\nSolution:
\n(3k + 4l) (3k 4l) = (3k + 4l)2<\/sup> is in the form of (a + b)2<\/sup>.
\n=(3k)2<\/sup> + 2 \u00d7 3k \u00d7 4l+ (4l)2<\/sup> [ (a+ b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup>
\n= 3k \u00d7 3k + 24kl + 4l \u00d7 4l
\n= 9k2<\/sup> + 24kl + 16l2<\/sup><\/p>\n

ii) (ax2<\/sup> + by2<\/sup>) (ax2<\/sup> + by2<\/sup>) = (ax2<\/sup> + by2<\/sup>)2<\/sup> is in the form of (a + b)2<\/sup>.
\n= (ax2<\/sup>)2<\/sup> + 2 \u00d7 ax2<\/sup> \u00d7 by2<\/sup>\u00a0+ (by2<\/sup>)2<\/sup> [ \u2235 (a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup>]
\n= ax2<\/sup> \u00d7 ax2<\/sup> + 2abx2<\/sup>y2<\/sup> + by2<\/sup> \u00d7 by2<\/sup>
\n= a2<\/sup>x4<\/sup> + 2ab x2<\/sup>y2<\/sup> + b2<\/sup>y4<\/sup><\/p>\n

iii) (7d – 9e) (7d – 9e)
\n= (7d – 9e)2<\/sup> is in the form of (a – b)2<\/sup>.
\n= (7d)2<\/sup> – 2 \u00d7 7d \u00d7 9e + (9e)2<\/sup> [ \u2235 (a – b)2<\/sup> = a2<\/sup> – 2ab + b2<\/sup>]
\n= 7d \u00d7 7d – 126de + 9e \u00d7 9e
\n= 49d2<\/sup> – 126de + 81e2<\/sup><\/p>\n

iv) (m2<\/sup> – n2<\/sup>) (m2<\/sup> + n2<\/sup>) is in the form of (a + b) (a – b).
\n\u2234 (a + b) (a – b) = a2<\/sup> – b2<\/sup>
\n\u2234 (m2<\/sup> + n2<\/sup>) (m2<\/sup> – n2<\/sup>) = (m2<\/sup>)2<\/sup> – (n2<\/sup>)2<\/sup> = m4<\/sup> – n4<\/sup><\/p>\n

v) (3t + 9s) (3t – 9s) = (3t)2<\/sup> – (9s)2<\/sup> [ \u2235\u00a0(a + b) (a – b) = a2<\/sup> – b2<\/sup> ]
\n= 3t \u00d7 3t – 9s \u00d7 9s
\n= 9t2<\/sup> – 81s2<\/sup><\/p>\n

vi) (kl – mn) (kl + mn) = (kl)2<\/sup> – (mn)2<\/sup> [ \u2235(a + b) (a – b) = a2<\/sup> – b2<\/sup> ]
\n= kl \u00d7 kl – mn \u00d7 mn
\n= k2<\/sup>l2<\/sup> – m2<\/sup>n2<\/sup><\/p>\n

vii) (6x + 5) (6x + 6) is in the form of
\n(ax + b) (ax + c).
\n(ax + b) (ax + c) = a2<\/sup>x2<\/sup> + ax(b + c) + bc
\n(6x + 5) (6x + 6) = (6)2<\/sup>x2<\/sup> + 6x (5 + 6) + 5 \u00d7 6
\n= 36x2<\/sup> + 6x \u00d7 11 + 30
\n= 36x2<\/sup> + 66x + 30<\/p>\n

viii) (2b – a) (2b + c) is in the form of (ax – b) (ax + c).
\n(ax – b) (ax + c) = a2<\/sup>x2<\/sup> + ax(c – b) – cb
\n(2b – a) (2b + c) = (2)2<\/sup>(b)2<\/sup> + 2b (c – a) – ca
\n= 4b2<\/sup> + 2bc – 2ab – ca<\/p>\n

Question 2.
\nEvaluate the following by using suitable identities:
\n(i) 3042<\/sup>
\n(ii) 5092<\/sup>
\n(iii) 9922<\/sup>
\n(iv) 7992<\/sup>
\n(v) 304 \u00d7 296
\n(vi) 83 \u00d7 77
\n(vii) 109 \u00d7 108
\n(viii) 204 \u00d7 206
\nSolution:
\ni) 3042<\/sup> = (300 + 4)2<\/sup> is in the form of (a + b)2<\/sup>.
\n\u2235 (a+b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup>
\na = 300, b = 4
\n(300 + 4)2<\/sup> = (300)2<\/sup> + 2 \u00d7 300 \u00d7 4 + (4)2<\/sup>
\n= 300 \u00d7 300+ 2400 + 4 \u00d7 4
\n= 90,000 + 2400 + 16
\n= 92,416<\/p>\n

ii) 5092<\/sup> = (500 + 9)2<\/sup>
\na\u00a0 = 500, b = 9
\n= (500)2<\/sup> + 2 \u00d7 500 \u00d7 9 + (9)2<\/sup>
\n[ \u2235 (a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup>]
\n= 500 \u00d7 500 + 9000 + 9 \u00d7 9
\n= 2,50,000 + 9000 + 81
\n= 2,59,081<\/p>\n

iii) 9922<\/sup> = (1000 – 8)2<\/sup>
\na = 1000, b = 8
\n= (1000)2<\/sup> – 2 \u00d7 1000 \u00d7 8 + (8)2<\/sup> [\u2235 (a-b)2<\/sup> = a2<\/sup> – 2ab + b2<\/sup>]
\n= 1000 \u00d7 1000 – 16,000 + 8 \u00d7 8
\n= 10,00,000 – 16000 + 64
\n= 10,00,064 – 1600
\n= 9,98,464<\/p>\n

iv) 7992<\/sup> = (800 – 1)2<\/sup>
\na = 800, b = 1
\n= (800)2<\/sup> – 2 \u00d7 800 \u00d7 1 + (1)2<\/sup>
\n= 800 \u00d7 800 – 1600 + 1
\n= 6,40,000 – 1600 + 1
\n= 6,40,001 – 1600
\n= 6,38,401<\/p>\n

v) 304 \u00d7 296 = (300 + 4) (300 – 4) is in the form of (a + b) (a – b).
\n(a + b) (a – b) = a2<\/sup> – b2<\/sup>
\n\u2234 (300 + 4) (300 – 4) = (300)2<\/sup> – (4)2<\/sup>
\n= 300 \u00d7 300 – 4 \u00d7 4
\n= 90,000 – 16
\n= 89,984<\/p>\n

vi) 83 \u00d7 77 = (80 + 3) (80 – 3)
\n= (80)2<\/sup> – (3)2<\/sup> [ \u2235 (a + b) (a – b) = a2<\/sup> – b2<\/sup>]
\n= 80 \u00d7 80 – 3 \u00d7 3
\n= 6400 – 9
\n= 6391<\/p>\n

vii) 109 \u00d7 108 = (100 + 9) (100 + 8)
\n= (100)2<\/sup> + (9 + 8)100 + 9 \u00d7 8
\n= 10,000 + 1700 + 72
\n= 11,772<\/p>\n

viii) 204 \u00d7 206 = (205 – 1) (205 + 1)
\n= (205)2<\/sup> – (1)2<\/sup> [\u2235 (a + b)(a-b) = a2<\/sup> – b2<\/sup>]
\n= 205 \u00d7 205 – 1 \u00d7 1
\n= 42,025 -1
\n= 42,024<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 11 Algebraic Expressions Ex 11.4 Textbook Questions and Answers. AP State Syllabus 8th Class Maths Solutions 11th Lesson Algebraic Expressions Exercise 11.4 Question 1. Select a suitable identity and find the following products (i) (3k + 4l)(3k + 4l) (ii) (ax2 + by2)(ax2 + by2) … Read more<\/a><\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[6],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/7325"}],"collection":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/users\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/comments?post=7325"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/7325\/revisions"}],"predecessor-version":[{"id":40704,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/7325\/revisions\/40704"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=7325"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=7325"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=7325"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}