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If a/b = c/d = e/f prove that: \[\frac{{{a}^{2}}}{{{b}^{2}}}+\frac{{{c}^{2}}}{{{d}^{2}}}+\frac{{{e}^{2}}}{{{f}^{2}}}=\frac{ac}{bd}+\frac{ce}{df}+\frac{ae}{df}\] (iv) \[bdf{{\left[ \frac{a+b}{b}+\frac{c+d}{d}+\frac{c+f}{f} \right]}^{3}}=27(a+b)(c+d)(e+f)\]
If a/b = c/d = e/f prove that: \[\frac{{{a}^{2}}}{{{b}^{2}}}+\frac{{{c}^{2}}}{{{d}^{2}}}+\frac{{{e}^{2}}}{{{f}^{2}}}=\frac{ac}{bd}+\frac{ce}{df}+\frac{ae}{df}\] (iv) \[bdf{{\left[ \frac{a+b}{b}+\frac{c+d}{d}+\frac{c+f}{f} \right]}^{3}}=27(a+b)(c+d)(e+f)\]
Class 10 | Exercise 1 | ICSE | Maths | ML Aggarwal | Ratio and Proportion
If a/b = c/d = e/f prove that: \[\frac{{{a}^{2}}}{{{b}^{2}}}+\frac{{{c}^{2}}}{{{d}^{2}}}+\frac{{{e}^{2}}}{{{f}^{2}}}=\frac{ac}{bd}+\frac{ce}{df}+\frac{ae}{df}\] (iv) \[bdf{{\left[ \frac{a+b}{b}+\frac{c+d}{d}+\frac{c+f}{f} \right]}^{3}}=27(a+b)(c+d)(e+f)\]

Consider

a/b = c/d = e/f = k

So we get

a = bk, c = dk, e = fk

Therefore, LHS = RHS.

So we get

\[=\text{ }bdf\text{ }{{\left( k\text{ }+\text{ }1\text{ }+\text{ }k\text{ }+\text{ }1\text{ }+\text{ }k\text{ }+\text{ }1 \right)}^{3}}\]

By further calculation

\[\begin{array}{*{35}{l}}

=\text{ }bdf\text{ }{{\left( 3k\text{ }+\text{ }3 \right)}^{3}}  \\

=\text{ }27\text{ }bdf\text{ }{{\left( k\text{ }+\text{ }1 \right)}^{3}}  \\

\end{array}\]

RHS = \[27\text{ }\left( a\text{ }+\text{ }b \right)\text{ }\left( c\text{ }+\text{ }d \right)\text{ }\left( e\text{ }+\text{ }f \right)\]

It can be written as

\[=\text{ }27\text{ }\left( bk\text{ }+\text{ }b \right)\text{ }\left( dk\text{ }+\text{ }d \right)\text{ }\left( fk\text{ }+\text{ }f \right)\]

Taking out the common terms

\[=\text{ }27\text{ }b\text{ }\left( k\text{ }+\text{ }1 \right)\text{ }d\text{ }\left( k\text{ }+\text{ }1 \right)\text{ }f\text{ }\left( k\text{ }+\text{ }1 \right)\]

So we get

\[=\text{ }27\text{ }bdf\text{ }{{\left( k\text{ }+\text{ }1 \right)}^{3}}\]

Therefore, LHS = RHS.

i

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