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
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
Therefore, LHS = RHS.
(iii)(a4 + c4): (b4 + d4) = a2c2: b2d2
We can write it as
Therefore, LHS = RHS.
Therefore, LHS = RHS.
Therefore, LHS = RHS.
Therefore, LHS = RHS.
Therefore, LHS = RHS.
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.