Question

In $\Delta ABC,\overline{PQ}\parallel \overline{AB}$, P,Q are on BC and CA respectively if CQ: QA=1:3 are CP=4 then BC=......

Answer 1

We have,

$In\Delta ABC$

P and Q are mid point of side BC and CA respectively.

Given that,

$CQ:QA=1:3$ and $CP=4$

Then, BC=?

So,

We know that,

Similar triangle theorem,

$In\Delta ABC$

$\frac{CQ}{QA}=\frac{CP}{PB}$

$\frac{1}{3}=\frac{4}{PB}$

$PB=12$

Now,

$BC=BP+PB$

$BC=12+4$

$BC=16$

Hence, this is the answer.