Question

If a, b, c, d are in proportion, then prove that 11a²+9ac/11b²+9bd=a²+3ac/b²+3bd

Answer 1

Step-by-step explanation:

For this we have to prove that

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

i have added a picture in it you will

understand it well

11a²+9ac/11b²+9bd=a²+3ac/ b²+3bd

a(11a+9c)/b(11b+9d)=a(a+3c)/b(b+3d)

11a+9c/a+3c=11b+9d/b+3d

11ab+33ad+9bc+27cd=11ab+33bc+9ad+11ab

ad=bc

a/b=c/d

hence prooved

Answer 2

Step-by-step explanation:

For this we have to prove that

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

b

a

=

d

c

11a²+9ac/11b²+9bd=a²+3ac/ b²+3bd

a(11a+9c)/b(11b+9d)=a(a+3c)/b(b+3d)

11a+9c/a+3c=11b+9d/b+3d

11ab+33ad+9bc+27cd=11ab+33bc+9ad+11ab

ad=bc

a/b=c/d

hence prooved