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Question
If `a/b = c/d` then prove that `(6a + 7b)/(6c + 7d) = sqrt((5a^2 + 3b^2)/(5c^2 + 3d^2))`
Theorem
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Solution
`a/b = c/d = k`
a = bk
c = dk
L.H.S.
= `(6a + 7b)/(6c + 7d)`
= `(6(bk) + 7b)/(6(dk) + 7d)`
= `(b(6k + 7))/(d(6k + 7))`
= `b/d`
R.H.S.
= `sqrt((5a^2 + 3b^2)/(5c^2 + 3d^2))`
= `sqrt((5(bk)^2 + 3b^2)/(5(dk)^2 + 3d^2))`
= `sqrt((5b^2k^2 + 3b^2)/(5d^2k^2 + 3d^2))`
= `sqrt((b^2(5k^2 + 3))/(d^2(5k^2 + 3)))`
= `sqrt(b^2/d^2)`
= `b/d`
L.H.S. = R.H.S.
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Chapter 7: Ratio and proportion - Exercise 7B [Page 125]
