Let, the radius of the circle be \(r\).

Then, the diameter of the circle is \(2r\).

∴ The ratio of the radius to the diameter is

= \(\displaystyle \frac {r}{2r}\)

= \(\displaystyle \frac {1}{2}\)

= \(\displaystyle 1 : 2\)

∴ The ratio of the radius to the diameter of a circle is 1 : 2.

Let’s find the diagonal of the rectangle.
Length of the rectangle = 4 cm
Breadth of the rectangle = 3 cm

Using the Pythagoras’ theorem, the diagonal of the rectangle is

= \(\displaystyle \sqrt{(4)^2 + (3)^2}\)

= \(\displaystyle \sqrt{16 + 9}\)

= \(\displaystyle \sqrt{25}\)

= \(\displaystyle 5\) ... (i)

∴ The ratio of the diagonal to the length is

= \(\displaystyle \frac {5}{4}\)

= \(\displaystyle 5 : 4\)

∴ The ratio of the diagonal to the length of that rectangle is 5 : 4.

Side of the square = 4 cm
Perimeter of the square = \(4 \times \text{side} = 4 \times 4 = 16\) cm
Area of the square = \((\text{side})^2 = (4)^2 = 16\) cm\(^2\)

∴ The ratio of the perimeter to the area is

= \(\displaystyle \frac {16}{16}\)

= \(\displaystyle 1 : 1\)

∴ The ratio of the perimeter to the area of that square is 1 : 1.

If a, b, c are in continued proportion, then
\(\displaystyle b^2 = a \times c\)

Let, a = 2, b = 4, c = 8

Now, \(\displaystyle b^2 = (4)^2 = 16\)

And \(\displaystyle a \times c = 2 \times 8 = 16\)

∴ \(b^2 = a \times c\)

∴ The numbers 2, 4, 8 are in continued proportion.

If a, b, c are in continued proportion, then
\(\displaystyle b^2 = a \times c\)

Let, a = 1, b = 2, c = 3

Now, \(\displaystyle b^2 = (2)^2 = 4\)

And \(\displaystyle a \times c = 1 \times 3 = 3\)

∴ \(b^2 \neq a \times c\)

∴ The numbers 1, 2, 3 are not in continued proportion.

If a, b, c are in continued proportion, then
\(\displaystyle b^2 = a \times c\)

Let, a = 9, b = 12, c = 16

Now, \(\displaystyle b^2 = (12)^2 = 144\)

And \(\displaystyle a \times c = 9 \times 16 = 144\)

∴ \(b^2 = a \times c\)

∴ The numbers 9, 12, 16 are in continued proportion.

If a, b, c are in continued proportion, then
\(\displaystyle b^2 = a \times c\)

Let, a = 3, b = 5, c = 8

Now, \(\displaystyle b^2 = (5)^2 = 25\)

And \(\displaystyle a \times c = 3 \times 8 = 24\)

∴ \(b^2 \neq a \times c\)

∴ The numbers 3, 5, 8 are not in continued proportion.

Since a, b, c are in continued proportion, we have:

\(\displaystyle \frac{a}{b} = \frac{b}{c}\)

∴ \(\displaystyle b^2 = a \times c\)

Given that a = 3 and c = 27:

 \(\displaystyle b^2 = 3 \times 27 = 81\)

∴ \(\displaystyle b = \sqrt{81} = 9\)

∴ The value of b is 9.

 \(\displaystyle 37 : 500\)

= \(\displaystyle \frac{37}{500}\)

= \(\displaystyle \frac{37 \div 5}{500 \div 5}\)

= \(\displaystyle \frac{7.4}{100}\)

= \(\displaystyle 7.4\%\)

 \(\displaystyle \frac{5}{8}\)

= \(\displaystyle \frac{5 \times \displaystyle \frac {100}{8}}{8 \times \displaystyle \frac {100}{8}}\)

= \(\displaystyle \frac{\displaystyle\frac{500}{8}}{100}\)

= \(\displaystyle \frac{62.5}{100}\)

= \(\displaystyle 62.5\%\)

 \(\displaystyle \frac{22}{30}\)

= \(\displaystyle \frac{11}{15}\)

= \(\displaystyle \frac{11 \times \displaystyle \frac {100}{15}}{15 \times \displaystyle \frac {100}{15}}\)

= \(\displaystyle \frac{\displaystyle\frac{1100}{15}}{100}\)

= \(\displaystyle \frac{73.33...}{100}\)

= \(\displaystyle 73.33 \%\)

 \(\displaystyle \frac{5}{16}\)

= \(\displaystyle \frac{5 \times \displaystyle \frac {100}{16}}{16 \times \displaystyle \frac {100}{16}}\)

= \(\displaystyle \frac{\displaystyle\frac{500}{16}}{100}\)

= \(\displaystyle \frac{31.25}{100}\)

= \(\displaystyle 31.25 \%\)

 \(\displaystyle \frac{144}{1200}\)

= \(\displaystyle \frac{144 \div 12}{1200 \div 12}\)

= \(\displaystyle \frac{12}{100}\)

= \(\displaystyle 12 \%\)

 \(\displaystyle \frac{a}{b} = \frac{2}{3}\) ... (Given)

Multiplying both sides by \(\displaystyle \frac{4}{3}\),

 \(\displaystyle \frac{a}{b} \times \frac{4}{3} = \frac{2}{3} \times \frac{4}{3}\)

∴ \(\displaystyle \frac{4a}{3b} = \frac{8}{9}\)

By Componendo,

 \(\displaystyle \frac{4a + 3b}{3b} = \frac{8 + 9}{9}\)

∴ \(\displaystyle \frac{4a + 3b}{3b} = \frac{17}{9}\) ... (i)

 \(\displaystyle \frac{a}{b} = \frac{2}{3}\) ... (Given)

Squaring both sides,

 \(\displaystyle \left(\frac{a}{b}\right)^2 = \left(\frac{2}{3}\right)^2\)

∴ \(\displaystyle \frac{a^2}{b^2} = \frac{4}{9}\)

Multiplying both sides by \(\displaystyle \frac{5}{2}\),

 \(\displaystyle \frac{a}{b} \times \frac{5}{2} = \frac{4}{9} \times \frac{5}{2}\)

∴ \(\displaystyle \frac{5a^2}{2b^2} = \frac{10}{9}\)

By Componendo – Dividendo,

 \(\displaystyle \frac{5a^2 + 2b^2}{5a^2 - 2b^2} = \frac{10 + 9}{10 - 9}\)

∴ \(\displaystyle \frac{5a^2 + 2b^2}{5a^2 - 2b^2}\)

∴ \(\displaystyle \frac{5a^2 + 2b^2}{5a^2 - 2b^2} = 19\) ... (ii)

 \(\displaystyle \frac{a}{b} = \frac{2}{3}\) ... (Given)

Taking cubes of both sides,

 \(\displaystyle \left(\frac{a}{b}\right)^3 = \left(\frac{2}{3}\right)^3\)

∴ \(\displaystyle \frac{a^3}{b^3} = \frac{8}{27}\)

By Componendo,

 \(\displaystyle \frac{a^3 + b^3}{b^3} = \frac{8 + 27}{27}\)

∴ \(\displaystyle \frac{a^3 + b^3}{b^3} = \frac{35}{27}\) ... (iii)

 \(\displaystyle \frac{a}{b} = \frac{2}{3}\) ... (Given)

By Invertendo,

 \(\displaystyle \frac{b}{a} = \frac{3}{2}\)

Multiplying both sides by \(\displaystyle \frac{7}{4}\),

 \(\displaystyle \frac{b}{a} \times \frac{7}{4} = \frac{3}{2} \times \frac{7}{4}\)

∴ \(\displaystyle \frac{7b}{4a} = \frac{21}{8}\)

By Componendo – Dividendo,

 \(\displaystyle \frac{7b + 4a}{7b - 4a} = \frac{21 + 8}{21 - 8}\)

∴ \(\displaystyle \frac{7b + 4a}{7b - 4a} = \frac{29}{13}\)

By Invertendo,

 \(\displaystyle \frac{7b - 4a}{7b + 4a} = \frac{13}{29}\) ... (iv)

a, b, c, d are in proportion ... (Given)

∴ \(\displaystyle \frac{a}{b}\) = \(\displaystyle \frac{c}{d} = k \text{ (say)}\)

∴ \(\displaystyle a = bk\) and \(\displaystyle c = dk\)

Now,

 LHS = \(\displaystyle \frac{11a^2 + 9ac}{11b^2 + 9bd}\)

∴ LHS = \(\displaystyle \frac{11(bk)^2 + 9 \times {bk} \times{dk}}{11b^2 + 9bd}\)

∴ LHS = \(\displaystyle \frac{11b^2k^2 + 9bdk^2}{11b^2 + 9bd}\)

∴ LHS = \(\displaystyle \frac{k^2(\enclose{downdiagonalstrike}{11b^2 + 9bd})}{\enclose{downdiagonalstrike}{11b^2 + 9bd}}\)

∴ LHS = \(\displaystyle k^2\) ... (i)

Also,

 RHS = \(\displaystyle \frac{a^2 + 3ac}{b^2 + 3bd}\)

∴ RHS = \(\displaystyle \frac{(bk)^2 + 3 \times {bk} \times{dk}}{b^2 + 3bd}\)

∴ RHS = \(\displaystyle \frac{b^2k^2 + 3bdk^2}{b^2 + 3bd}\)

∴ RHS = \(\displaystyle \frac{k^2(\enclose{downdiagonalstrike}{b^2 + 3bd})}{\enclose{downdiagonalstrike}{b^2 + 3bd}}\)

∴ RHS = \(\displaystyle k^2\) ... (ii)

From (i) and (ii),

 LHS = RHS

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

a, b, c, d are in proportion ... (Given)

∴ \(\displaystyle \frac{a}{b}\) = \(\displaystyle \frac{c}{d} = k \text{ (say)}\)

∴ \(\displaystyle a = bk\) and \(\displaystyle c = dk\)

Now,

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

∴ LHS = \(\displaystyle \sqrt{\frac{(bk)^2 + 5(dk)^2}{b^2 + 5d^2}}\)

∴ LHS = \(\displaystyle \sqrt{\frac{b^2k^2 + 5d^2k^2}{b^2 + 5d^2}}\)

∴ LHS = \(\displaystyle \sqrt{\frac{k^2(\enclose{downdiagonalstrike}{b^2 + 5d^2})}{\enclose{downdiagonalstrike}{b^2 + 5d^2}}}\)

∴ LHS = \(\displaystyle \sqrt{k^2}\)

∴ LHS = \(\displaystyle k\) ... (i)

Also,

 RHS = \(\displaystyle \frac{a}{b}\)

∴ RHS = \(\displaystyle \frac{bk}{b}\)

∴ RHS = \(\displaystyle k\) ... (ii)

From (i) and (ii),

 LHS = RHS

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

a, b, c are in continued proportion. ... (Given)

∴ \(\displaystyle \frac{a}{b}\) = \(\displaystyle \frac{b}{c} = k \text{ (say)}\)

∴ \(\displaystyle b = ck\) and
\(\displaystyle a = bk = ck \times k = k^2\)

Now,

 LHS = \(\displaystyle \frac{a}{a + 2b}\)

∴ LHS = \(\displaystyle \frac{ck^2}{ck^2 + 2ck}\)

∴ LHS = \(\displaystyle \frac{ck \times k}{ck(k + 2)}\)

∴ LHS = \(\displaystyle \frac{\enclose{downdiagonalstrike}{ck} \times k}{\enclose{downdiagonalstrike}{ck}(k + 2)}\)

∴ LHS = \(\displaystyle \frac{k}{k + 2}\) ... (i)

Also,

 RHS = \(\displaystyle \frac{a - 2b}{a - 4c}\)

∴ RHS = \(\displaystyle \frac{ck^2 - 2ck}{ck^2 - 4c}\)

∴ RHS = \(\displaystyle \frac{\enclose{downdiagonalstrike}{c}k(k - 2)}{\enclose{downdiagonalstrike}{c}(k^2 - 4)}\)

∴ RHS = \(\displaystyle \frac{k\enclose{downdiagonalstrike}{(k -2)}}{(k + 2)\enclose{downdiagonalstrike}{(k -2)}}\)

∴ RHS = \(\displaystyle \frac{k}{k + 2}\) ... (ii)

From (i) and (ii),

 LHS = RHS

∴ \(\displaystyle \frac{a}{a + 2b}\) = \(\displaystyle \frac{a - 2b}{a - 4c}\)

a, b, c are in continued proportion. ... (Given)

∴ \(\displaystyle \frac{a}{b}\) = \(\displaystyle \frac{b}{c} = k \text{ (say)}\)

∴ \(\displaystyle b = ck\) and
\(\displaystyle a = bk = ck \times k = k^2\)

Now,

 LHS = \(\displaystyle \frac{b}{b + c}\)

∴ LHS = \(\displaystyle \frac{ck}{ck + c}\)

∴ LHS = \(\displaystyle \frac{\enclose{downdiagonalstrike}{c}k}{\enclose{downdiagonalstrike}{c}(k + 1)}\)

∴ LHS = \(\displaystyle \frac{k}{k + 1}\) ... (i)

Also,

  RHS = \(\displaystyle \frac{a - b}{a - c}\)

∴ RHS = \(\displaystyle \frac{ck^2 - ck}{ck^2 - c}\)

∴ RHS = \(\displaystyle \frac{\enclose{downdiagonalstrike}{c}k(k - 1)}{\enclose{downdiagonalstrike}{c}(k^2 - 1)}\)

∴ RHS = \(\displaystyle \frac{k\enclose{downdiagonalstrike}{(k - 1)}}{(k + 1)\enclose{downdiagonalstrike}{(k - 1)}}\)

∴ RHS = \(\displaystyle \frac{k}{k + 1}\) ... (ii)

From (i) and (ii),

 LHS = RHS

∴ \(\displaystyle \frac{b}{b + c}\) = \(\displaystyle \frac{a - b}{a - c}\)

 \(\displaystyle \frac{12x^2 + 18x + 42}{18x^2 + 12x + 58}\) = \(\displaystyle \frac{2x + 3}{3x + 2}\)

Multiplying the numerator and denominator of the second ratio by − 6x,

 \(\displaystyle \frac{12x^2 + 18x + 42}{18x^2 + 12x + 58}\) = \(\displaystyle \frac{2x + 3}{3x + 2} \times \frac{- 6x}{- 6x}\)

∴\(\displaystyle \frac{12x^2 + 18x + 42}{18x^2 + 12x + 58}\) = \(\displaystyle \frac{- 12x^2 - 18x}{- 18x^2 - 12x}\)

Using Theorem on Equal Ratios,

∴ Each Ratio = \(\displaystyle \frac{\enclose{downdiagonalstrike}{12x^2} + \enclose{updiagonalstrike}{18x} + 42 - \enclose{downdiagonalstrike}{12x^2} - \enclose{updiagonalstrike}{18x}}{\enclose{downdiagonalstrike}{18x^2} + \enclose{updiagonalstrike}{12x} + 58 - \enclose{downdiagonalstrike}{18x^2} - \enclose{updiagonalstrike}{12x}}\)

∴ Each Ratio = \(\displaystyle \frac{42}{58}\)

∴ Each Ratio = \(\displaystyle \frac{21}{29}\)

∴ \(\displaystyle \frac{12x^2 + 18x + 42}{18x^2 + 12x + 58}\) = \(\displaystyle \frac{2x + 3}{3x + 2}\) = \(\displaystyle \frac{21}{29}\)

Now, let’s use the 2^nd and the 3^rd ratio.

 \(\displaystyle \frac{2x + 3}{3x + 2}\) = \(\displaystyle \frac{21}{29}\)

By cross multiplication,

  \(\displaystyle 29(2x + 3)\) = \(\displaystyle 21(3x + 2)\)

∴ \(\displaystyle 29(2x + 3)\) = \(\displaystyle 21(3x + 2)\)

∴ \(\displaystyle 58x + 87 = 63x + 42\)

∴ \(\displaystyle 58x - 63x = 42 - 87\)

∴ \(\displaystyle -5x = -45\)

i.e. \(\displaystyle 5x = 45\)

∴ \(\displaystyle x = \frac{45}{5}\)

∴ \(\displaystyle x = 9\)

 \(\displaystyle \frac{2x - 3y}{3z + y}\) = \(\displaystyle \frac{z - y}{z - x}\) = \(\displaystyle \frac{x + 3z}{2y - 3x}\) ... (Given)

Multiplying the numerator and denominator of the 2^nd

 \(\displaystyle \frac{2x - 3y}{3z + y}\) = \(\displaystyle \frac{- 3(z - y)}{- 3(z - x)}\) = \(\displaystyle \frac{x + 3z}{2y - 3x}\)

∴ \(\displaystyle \frac{2x - 3y}{3z + y}\) = \(\displaystyle \frac{- 3z + 3y}{- 3z + 3x}\) = \(\displaystyle \frac{x + 3z}{2y - 3x}\)

Using Theorem on Equal Ratios,

  Each Ratio = \(\displaystyle \frac{2x - \enclose{downdiagonalstrike}{3y} - \enclose{downdiagonalstrike}{3z} + \enclose{downdiagonalstrike}{3y} + x + \enclose{downdiagonalstrike}{3z}}{\enclose{updiagonalstrike}{3z} + y - \enclose{updiagonalstrike}{3z} + \enclose{updiagonalstrike}{3x} + 2y - \enclose{updiagonalstrike}{3x}}\)

∴ Each Ratio = \(\displaystyle \frac{\enclose{updiagonalstrike}{3}x}{\enclose{updiagonalstrike}{3}y}\)

∴ Each Ratio = \(\displaystyle \frac{x}{y}\)

 \(\displaystyle \frac{by + cz}{b^2 + c^2}\) = \(\displaystyle \frac{cz + ax}{c^2 + a^2}\) = \(\displaystyle \frac{ax + by}{a^2 + b^2}\) ... (Given)

Using Theorem on Equal Ratios,

 Each Ratio = \(\displaystyle \frac{by + cz + cz + ax + ax + by}{b^2 + c^2 + c^2 + a^2 + a^2 + b^2}\)

∴ Each Ratio = \(\displaystyle \frac{2ax + 2by + 2cz}{2a^2 + 2b^2 + 2c^2}\)

∴ Each Ratio = \(\displaystyle \frac{\enclose{downdiagonalstrike}{2}(ax + by + cz)}{\enclose{downdiagonalstrike}{2}(a^2 + b^2 + c^2)}\)

∴ Each Ratio = \(\displaystyle \frac{ax + by + cz}{a^2 + b^2 + c^2}\) ... (i)

∴ \(\displaystyle \frac{ax + by + cz}{a^2 + b^2 + c^2}\) = \(\displaystyle \frac{by + cz}{b^2 + c^2}\) = \(\displaystyle \frac{cz + ax}{c^2 + a^2}\) = \(\displaystyle \frac{ax + by}{a^2 + b^2}\) ... (ii)

Again, using Theorem on Equal Ratios,

 Each Ratio = \(\displaystyle \frac{(ax + by + cz) - (by + cz)}{(a^2 + b^2 + c^2) - (b^2 + c^2)}\)

∴ Each Ratio = \(\displaystyle \frac{ax + \enclose{downdiagonalstrike}{by} + \enclose{downdiagonalstrike}{cz} - \enclose{downdiagonalstrike}{by} - \enclose{downdiagonalstrike}{cz}}{a^2 + \enclose{downdiagonalstrike}{b^2} + \enclose{downdiagonalstrike}{c^2} - \enclose{downdiagonalstrike}{b^2} - \enclose{downdiagonalstrike}{c^2}}\)

∴ Each Ratio = \(\displaystyle \frac{ax}{a^2}\)

∴ Each Ratio = \(\displaystyle \frac{x}{a}\) ... (iii)

Again, using Theorem on Equal Ratios,

 Each Ratio = \(\displaystyle \frac{(ax + by + cz) - (cz + ax)}{(a^2 + b^2 + c^2) - (c^2 + a^2)}\)

∴ Each Ratio = \(\displaystyle \frac{\enclose{downdiagonalstrike}{ax} + by - \enclose{downdiagonalstrike}{cz} - \enclose{downdiagonalstrike}{cz} - \enclose{downdiagonalstrike}{ax}}{\enclose{downdiagonalstrike}{a^2} + b^2 + \enclose{downdiagonalstrike}{c^2} - \enclose{downdiagonalstrike}{c^2} - \enclose{downdiagonalstrike}{a^2}}\)

∴ Each Ratio = \(\displaystyle \frac{by}{b^2}\)

∴ Each Ratio = \(\displaystyle \frac{y}{b}\) ... (iv)

Again, using Theorem on Equal Ratios,

 Each Ratio = \(\displaystyle \frac{(ax + by + cz) - (ax + by)}{(a^2 + b^2 + c^2) - (a^2 + b^2)}\)

∴ Each Ratio = \(\displaystyle \frac{\enclose{downdiagonalstrike}{ax} + \enclose{downdiagonalstrike}{by} + cz - \enclose{downdiagonalstrike}{ax} - \enclose{downdiagonalstrike}{by}}{\enclose{downdiagonalstrike}{a^2} + \enclose{downdiagonalstrike}{b^2} + c^2 - \enclose{downdiagonalstrike}{a^2} - \enclose{downdiagonalstrike}{b^2}}\)

∴ Each Ratio = \(\displaystyle \frac{cz}{c^2}\)

∴ Each Ratio = \(\displaystyle \frac{z}{c}\) ... (v)

From (iii), (iv) and (v), we have:

 \(\displaystyle \frac{x}{a} = \frac{y}{b} = \frac{z}{c}\) ... (vi)