2
$\begingroup$

enter image description here

All the lengths in red had been given in the problem statement. The task is to find $BC$. The green ones were calculated from the information in red.

Here's two ways I tried to solve it (and I would like feedback):


  1. The green line is drawn parallel to $QC$. Suppose, the other unlabelled end of the green line be point $R$. So the line can be called $PR$.

    In $\triangle BPR$,

    $PC=QC=9, \space\space PC=3\times AP\\PB=6, \space\space PB=3 \times AQ\\\therefore \triangle BPR \sim \triangle APQ\\BD=4\times3, \space\space \boxed{BD=12}$

    In quadrilateral $PQCD$, opposite sides are parallel to each other and it is therefore a parallelogram. So $DC=4$.

    $BC=BD+DC\\BC=12+4\\\boxed{BC=16}$


  1. In $\triangle ABC$, $AB=2+6=2(1+3)=\boxed{2\times 4}\\AC=3+9=3(1+3)=\boxed{3\times 4}\\BC=?$

    In $\triangle APQ$, $AP=\boxed{2}\\AC=\boxed{3}\\PQ=4$

    $\because \triangle ABC\sim \triangle APQ,\\BC=PQ\times4=4\times4=16,\\\boxed{BC=16}$

$\endgroup$
  • $\begingroup$ The red $9$ above the green line should be written in green, right? $\endgroup$ – 5xum Jan 24 '18 at 9:41
  • $\begingroup$ Also, when you wrote $PC=QC$, I think you meant $PR=QR$. $\endgroup$ – 5xum Jan 24 '18 at 9:42
  • $\begingroup$ @5xum Problem fixed. Thanks for noticing! $\endgroup$ – Soha Farhin Pine Jan 24 '18 at 9:45
  • $\begingroup$ I still see $PC=QC$ which is not true. $\endgroup$ – 5xum Jan 24 '18 at 9:48
  • $\begingroup$ I don't see the point D anywhere $\endgroup$ – Darkrai Jan 24 '18 at 9:53
2
$\begingroup$

There are so many typos that it's kind of hard to read your proof.

  1. You write $PC=QC$, but you probably mean $PR=QR$.
  2. You write $PC=3\cdot AP$, which is not true.
  3. You write $PB = 3\cdot AQ$ which is not true (since $6\neq 3\cdot 3$)
  4. You write $\Delta BPR\sim\Delta APQ$ which is almost ok, but not entirely - be careful to write the corners in the correct order!
  5. In the middle of your proof, you start talking about a point called "$D$" you did not previously define (i.e. "in the quadrilateral $PQCD$", there is no previous mention of $D$ - you probably meant $PQCR$.

Your solution is extremely sloppy.


Don't take this the wrong way. It's very obvious to me that you correctly solved the problem. You used the correct similar triangles to calculate all the values. Your answer in the end is correct.

But your description showing how you got to the answer is a horrible mess. If I was grading this solution, I would give you maybe half of all possible points.

$\endgroup$
0
$\begingroup$

Your question doesn't mention $PQ$ is parallel to $BC$ but your answer does.

So assuming that line $PQ$ is parallel to $BC$ then by similar triangles

( Considering the triangles $APQ$ and $ABC$) $$\frac {AP}{AB}= \frac {AQ}{AC}=\frac {PQ}{BC}$$ Hence we get $$\frac {3}{12}=\frac {4}{BC}$$ $$\Rightarrow BC=16$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.