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If a, b, c and d are in proportion, prove that: (i) (5a + 7v) (2c – 3d) = (5c + 7d) (2a – 3b) (ii) (ma + nb): b = (mc + nd): d (iii)(a4 + c4): (b4 + d4) = a2c2: b2d2 (iv)(a2+ab) /(c2+cd) =(b2-2ab)/(d2-2cd) (v) (a+c)3 /(b+d) 3=a(a-c) 2/(b(b-d)2 (vi) a2+ab+b2/a2-ab+b2=c2+cd+d2/c2-cd+d2 (vii) (a2+b2) /(c2+d2) =ab+ad-bc/bc+cd-ad) (viii) abcd(1/a2+1/b2+1/c2+1/d2)=a2+b2+b2+c2)

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This is a basic question from ML aggarwal book of class 10th, chapter 7, ratio and proportion., ICSE board

If a, b, c and d are in proportion,

We have to prove that:

(i) (5a + 7v) (2c – 3d) = (5c + 7d) (2a – 3b)

(ii) (ma + nb): b = (mc + nd): d

(iii)(a4 + c4): (b4 + d4) = a2c2: b2d2

(iv)(a2+ab) /(c2+cd) =(b2-2ab)/(d2-2cd)

(v) (a+c)3 /(b+d) 3=a(a-c) 2/(b(b-d)2 (vi) a2+ab+b2/a2-ab+b2=c2+cd+d2/c2-cd+d2

(vii) (a2+b2) /(c2+d2) =ab+ad-bc/bc+cd-ad) (viii)abcd(1/a2+1/b2+1/c2+1/d2)=a2+b2+b2+c2)

Question 19, 7.2

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  1. Solution:

    It is given that

    a, b, c, d are in proportion

    Consider a/b = c/d = k

    a = b, c = dk

    (i) LHS = (5a + 7b) (2c – 3d)

    Substituting the values

    = (5bk + 7b) (2dk – 3d)

    Taking out the common terms

    = k (5b + 7b) k (2d – 3d)

    So we get

    = k2 (12b) (-d)

    = – 12 bd k2

    RHS = (5c + 7d) (2a – 3b)

    Substituting the values

    = (5dk + 7d) (2kb – 3b)

    Taking out the common terms

    = k (5d + 7d) k (2b – 3b)

    So we get

    = k2 (12d) (-b)

    = – 12 bd k2

    Therefore, LHS = RHS.

    (ii) (ma + nb): b = (mc + nd): d

    We can write it as

    ML Aggarwal Solutions for Class 10 Chapter 7 Image 15

    ML Aggarwal Solutions for Class 10 Chapter 7 Image 16

    Therefore, LHS = RHS.

    (iii)(a4 + c4): (b4 + d4) = a2c2: b2d2

    We can write it as

    ML Aggarwal Solutions for Class 10 Chapter 7 Image 17

    Therefore, LHS = RHS.

    ML Aggarwal Solutions for Class 10 Chapter 7 Image 18

    Therefore, LHS = RHS.

    ML Aggarwal Solutions for Class 10 Chapter 7 Image 19

    ML Aggarwal Solutions for Class 10 Chapter 7 Image 20

    Therefore, LHS = RHS.

    ML Aggarwal Solutions for Class 10 Chapter 7 Image 21

    ML Aggarwal Solutions for Class 10 Chapter 7 Image 22

    Therefore, LHS = RHS.

    ML Aggarwal Solutions for Class 10 Chapter 7 Image 23

    Therefore, LHS = RHS.

    ML Aggarwal Solutions for Class 10 Chapter 7 Image 24

    So we get

    = d2 (1 + k2) + b2 (1 + k2)

    = (1 + k2) (b2 + d2)

    RHS = a2 + b2 + c2 + d2

    We can write it as

    = b2k2 + b2 + d2k2 + d2

    Taking out the common terms

    = b2 (k2 + 1) + d2 (k2 + 1)

    = (k2 + 1) (b2 + d2)

    Therefore, LHS = RHS.

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