Question

If a, b, c, d are in proportion , then prove that

(i) $\frac{11a^2+9ac}{11b^2+9bd}=\frac{a^2+3ac}{b^2+3bd}$


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



(iii) $\frac{a^2+ab+b^2}{a^2-ab+b^2}=\frac{c^2+cd+d^2}{c^2-cd+d^2}$

Answer 1

It is given that a, b, c, d are in proportion.

$$\begin{gathered} \therefore \frac{a}{b}=\frac{c}{d}=k \\ \Rightarrow a=bk,c=dk \end{gathered}$$
(i)
$$\begin{gathered} \frac{11a^2+9ac}{11b^2+9bd}=\frac{11{\left(bk\right)}^2+9\times bk\times dk}{11b^2+9bd}=\frac{k^2\left(11b^2+9bd\right)}{11b^2+9bd}=k^2.....\left(1\right) \\ \frac{a^2+3ac}{b^2+3bd}=\frac{{\displaystyle {\left(bk\right)}^2+3\times bk\times dk}}{{\displaystyle b^2+3bd}}=\frac{{\displaystyle k^2\left(b^2+3bd\right)}}{{\displaystyle b^2+3bd}}=k^2.....\left(2\right) \end{gathered}$$
From (1) and (2), we get

$\frac{11a^2+9ac}{11b^2+9bd}=\frac{a^2+3ac}{b^2+3bd}$

(ii)
$$\begin{gathered} \sqrt{\frac{a^2+5c^2}{b^2+5d^2}}=\sqrt{\frac{{\left(bk\right)}^2+5{\left(dk\right)}^2}{b^2+5d^2}}=\sqrt{\frac{k^2\left(b^2+5d^2\right)}{b^2+5d^2}}=\sqrt{k^2}=k.....\left(1\right) \\ \frac{a}{b}=\frac{bk}{b}=k.....\left(2\right) \end{gathered}$$
From (1) and (2), we get

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

(iii)
$$\begin{gathered} \frac{a^2+ab+b^2}{a^2-ab+b^2}=\frac{{\displaystyle {\left(bk\right)}^2+bk\times b+b^2}}{{\displaystyle {\left(bk\right)}^2-bk\times b+b^2}}=\frac{b^2\left(k^2+k+1\right)}{b^2\left(k^2-k+1\right)}=\frac{k^2+k+1}{k^2-k+1}.....\left(1\right) \\ \frac{{\displaystyle c^2+cd+d^2}}{{\displaystyle c^2-cd+d^2}}=\frac{{\displaystyle {\left(dk\right)}^2+dk\times d+d^2}}{{\displaystyle {\left(dk\right)}^2-dk\times d+d^2}}=\frac{d^2\left(k^2+k+1\right)}{d^2\left(k^2-k+1\right)}=\frac{k^2+k+1}{k^2-k+1}.....\left(2\right) \end{gathered}$$
From (1) and (2), we get

$\frac{a^2+ab+b^2}{a^2-ab+b^2}=\frac{{\displaystyle c^2+cd+d^2}}{{\displaystyle c^2-cd+d^2}}$