6. Solutions for Proportions¶

1. Let the number is x. Then, we have,
$\frac{3}{4}=\frac{9}{x}~ \Rightarrow~x=12$

Second and third part are left as exercise to the reader.

1. Let mean proportional be x so we have,
$x^2=ab~\Rightarrow~x=\sqrt{ab}$

Second part is left as an exercise.

1. Let third proportional be x. Then, we have,
$9^2=7x~\Rightarrow~x=\frac{81}{7}$

Second part is left as an exercise.

1. Let us equate the given proportion’s ratios to k.
$\frac{a}{b}=\frac{c}{d}~\Rightarrow~a=bk,~c=dk$

Now let us evaluate L.H.S. by substituting above value for a and c.

$\frac{a^c+ac^2}{b^d+bd^2}=\frac{b^2k^2dk+bkd^2k^2}{b^d+db^2}=k^3$

Now let us evaluate R.H.S. by substituting above value for a and c.

$\frac{(bk+dk)^3}{(b+d)^3}=k^3$
$\text{Hence,} L.H.S.= R.H.S.$
1. Proceeding the same way as previous problem we have L.H.S. as
$\frac{pb^2k^2+qb^2}{p^2k^2-qb^2}=\frac{pk^2+q}{pk^2-q}$

and R.H.S. as

$\frac{pd^2k^2+qd^2}{pd^2k^2-qd^2}=\frac{pk^2+q}{pk^2-q}$
$\text{Hence,} L.H.S.= R.H.S.$
1. Processing similarly L.H.S. is
$\frac{bk-dk}{b-d}=k$

and R.H.S. is

$\frac{\sqrt{b^2k^2+d^2k^2}}{\sqrt{b^2+d^2}} = k$
$\text{Hence,} L.H.S.= R.H.S.$

7. We have computed R.H.S. to be k in last problem’s R.H.S. So let us try to calculate R.H.S.

$\frac{\sqrt{bdk^2+\frac{d^3k^3}{bk}}}{\sqrt{bd+\frac{d^3}{b}}}=k$
1. Since a, b, c and d are in continued proportion we can write them as
$\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k$
$\Rightarrow~a=dk^3, b=dk^2,~\text{and}~c=dk$

First we do L.H.S.

$\frac{a}{b+d}=\frac{dk^3}{dk^2+d}=\frac{k^3}{k^2+1}$

now we do R.H.S.

$\frac{c^3}{c^2d+d^3}=\frac{d^3k^3}{d^2k^2d+d^2}=\frac{k^3}{k^2+1}.$

Problem no. 9 and 10 are left as exercises to the reader as they are similar.

11. Since b is mean proportional between a and b we can write $$b^2=ac$$. Now let us evaluate the expression.

$L.H.S.=\frac{a^2-ac+c^2}{\frac{1}{a^2}-\frac{1}{ac}+\frac{1}{c^2}}$
$\Rightarrow~\frac{(a^2-ac-c^2)a^2c^2}{c^2-ac+a^2}$
$\Rightarrow~a^2c^2 = b^4$

12. Equating first set of ratio to $$k$$ and second set to $$l$$ we have following:

$a=bk\text{ and }c=dk;~e=fl\text{ and }g=hl$

Substituting these value for $$a, c, e \text{ and } g$$ we get both left hand side and right hand side equal to

$\frac{kl+1}{kl-1}$

13. Proceeding similarly as previous problems we can say that $$a=bk$$ and $$c=dk$$. Now substituting for $$a$$ and $$c$$ we get L.H.S. as

$L.H.S. = \{bk+b+dk+d\}\{bk-b-dk+d\} = \{(b+d)(k+1)\}\{(b-d)(k-1)\}$
$\Rightarrow~(b^2-d^2)(k^2-1)$

Similarly for R.H.S. we have,

$R.H.S. = \{bk-b+dk-d\}\{bk+b-dk-d\} = \{(b-d)(k+1)\}\{(b+d)(k+1)\}$
$\Rightarrow~(b^2-d^2)(k^2-1).$

Remaining problems are left exercise to the reader.