Question

In $△ABC,P$ divides $AB$ in the ratio $3:5$, and $PQ\parallel BC.$ The length of $PQ$ (in inches) is (All lengths are in inches)

Answer 1

So a $△ABC$ here is given


AB has a length of 8 inches, BC has a length of 10 inches and AC has a length of 7 inches.

Now a line $PQ$ is drawn so that it is parallel to $BC$. And it was also told that it divides $AB$ in the ratio $3:5$ meaning that the top 3 to the bottom 5.
So now we have to find length $PQ$.
Figure out $3:5$ part breaking down. The total length AB is divided into 2 pieces. One is 3 one is 5. When add up it becomes 8.

We know that the line $PQ$ which goes on forever is parallel to BC.



That's the case then we can think about these 2 parallel lines being crossed by 2 transversals. One is $AB$ and one is $AC.$ And corresponding angles will be congruent.



For transversal $AB:\angle ABC=\angle APQ$
For transversal $AC:\angle ACB=\angle AQP$

$△ABC$ is similar to $APQ$
So we can see those two triangles are similar. So that means that the corresponding parts will always be in same proportion. So the ratio of corresponding parts will correspond to the ratio of other corresponding parts.

So to find the length of $PQ$
$\frac{10}{PQ}=\frac{8}{3}$
$\Rightarrow 8\cdot PQ=30$
$\Rightarrow PQ=\frac{30}{8}$
$\Rightarrow PQ=\frac{15}{4}$
$PQ=3.75$ inches
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