Eight Standard >> Ratio and Proportion | Proportion

If 4 quantities are so related that the ratio of the first quantity to the second quantity is equal to the ratio of the 3rd quantity to the 4th, then these four quantities are said to be in proportion.

Thus, \(a:b=c:d\) then \(a,\ b,\ c,\) and \(d\) are in propotion.
                         \(a:b=c:d\) 
      \( \Rightarrow\ a:b::c:d\)
       \(\Rightarrow\  \frac{a}{b}=\frac{c}{d}\)
      \(\Rightarrow\  ad=bc\)

Three quantities are said to be in continued proportion if a:b=b:c. Here b is called mean proportional between a and c.

   Here 'c' is called third proportional
         \(\frac{a}{b}=\frac{b}{c})
       \(\Rightarrow b^{2}=ac\)
       \(\Rightarrow b=\sqrt{ac}\)

 b is the mean proprtional to a & c

Illustration:
    Find the third proportional between \(\frac{a}{b}+\frac{b}{a}\) and \(\sqrt{a^{2}+b^{2}}\)

Solution: Let the third roportion be x,
    \(\left(\frac{a}{b}+\frac{b}{a}\right):\sqrt{a^{2}+b^{2}}::\sqrt{a^{2}+b^{2}}:x\)
\( \Rightarrow\ x\left(\frac{a}{b}+\frac{b}{a}\right)=(\sqrt{a^{2}+b^{2}})(\sqrt{a^{2}+b^{2}})\)
   \(\Rightarrow\ x\left(\frac{a^{2}+b^{2}}{ab}\right)=a^{2}+b^{2}\)
   \(\Rightarrow\ x=\frac{a^{2}+b^{2}}{1} \times x=\frac{ab}{a^{2}+b^{2}}\)
            =\(ab\)
Third proportional is ab.

Illustration 2:
 Finf mean proportional between \(a+b\) and \(b^{3}+ab^{2}\)

Solution: 
  Let m be the mean proportional 
\( \therefore\ (a+b):x=x:(b^{3}+ab^{2})\)
\(\Rightarrow\ x^{2}=(a+b)(b^{3}+ab^{2})\)
\(\Rightarrow\ x^{2}=(a+b)(b+a)b^{2}\)
\(\Rightarrow\ x^{2}=(a+b)^{2}b^{2}\)
\(\therefore\ =(a+b)b=ab+b^{2}\)

Illustration 3:
If the mean proportional between two numbers is 6 and third proportional to them is 48, find the numbers

Solution:
  Let the numbers be x and y
   x:6=6:y
\(\Rightarrow\ xy=6^{2}\)
\(\Rightarrow\ xy=36\)...(1)

   Also x:y=y:48
  \(\Rightarrow\ y^{2}=48x\)
\(\Rightarrow\ x=\frac{y^{2}}{48}\)...(2)

\(\therefore\) from (1)
    \( \frac{y^{2}}{48}.y=36
  \(\Rightarrow\ y^{3}=36 \times 48\)
  \(\therefore\ y^{3}=6 \times 6 \times 2 \times 2 \times 2\)
                              =\((6 \times 2)^{2}\)
   \( \therefore\ y=6 \times 2\)=12

 \(\therefore\ x=\frac{y^{2}}{48}\)
           =\(\frac{12 \times 12}{48}\)=3

\(\therefore\) Two numers are 3 and 12.