Question

If $a,b,c,d$ are in proportion, then prove that
$\sqrt{\frac{a^2+{5c}^2}{b^2+{5d}^2}}=\frac{a}{b}$

Answer 1

Given $a,b,c$ and $d$ are in proportion

$\frac{a}{b}=\frac{c}{d}-------\left(1\right)$

$ad=bc$

$a=\frac{bc}{d}$

$\sqrt{\frac{a^2+5c^2}{b^2+5d^2}}=\frac{a}{b}$

$\Rightarrow \sqrt{\frac{(\frac{bc}{d}{)}^2+5c^2}{b^2+5d^2}}$

$\Rightarrow \sqrt{\frac{\frac{b^2{c}^2}{d^2}+5c^2}{b^2+5d^2}}$

$\Rightarrow \sqrt{\frac{b^2{c}^2+5c^2{d}^2}{b^2{d}^2+5d^4}}$

$\Rightarrow \sqrt{\frac{c^2(b^2+5d^2)}{d^2(b^2+5d^2)}}$

$\Rightarrow \sqrt{\frac{c^2}{d^2}}=\frac{c}{d}$

$\Rightarrow \sqrt{\frac{a^2+5c^2}{b^2+5d^2}}=\frac{c}{d}$ From (1)

$\Rightarrow \frac{c}{d}=\frac{a}{b}$

$\therefore \sqrt{\frac{a^2+5x^2}{b^2+5d^2}}=\frac{a}{b}$