Question

if lm = 6, what is the perimeter of △pkq?

Explanation:

Step1: Use the property of similar triangles

Since \(PQ\parallel JL\), \(\triangle KPQ\sim\triangle KJL\).
The ratio of corresponding sides is equal. \(\frac{KP}{KJ}=\frac{KQ}{KL}\).
We know \(KJ=(x - 6)+3=x - 3\), \(KL=x + 5\), \(KP=x - 6\), \(KQ=x\).
Also, because of the parallelogram \(PJMQ\) (since \(PQ\parallel JM\) and \(PJ\parallel QM\)), \(PQ = LM = 6\).
By the similarity of \(\triangle KPQ\) and \(\triangle KJL\), \(\frac{KP}{KJ}=\frac{PQ}{JL}\).
Another way: Using the basic proportionality theorem (Thales' theorem).
Since \(PQ\parallel JL\), \(\frac{KP}{KJ}=\frac{KQ}{KL}\). Substituting \(KJ=x - 3\), \(KL=x + 5\), \(KP=x - 6\), \(KQ=x\) gives \(\frac{x-6}{x - 3}=\frac{x}{x + 5}\).
Cross - multiply: \((x - 6)(x + 5)=x(x - 3)\).
Expand: \(x^{2}+5x-6x - 30=x^{2}-3x\).
Simplify: \(x^{2}-x - 30=x^{2}-3x\).
Subtract \(x^{2}\) from both sides: \(-x - 30=-3x\).
Add \(3x\) to both sides: \(2x-30 = 0\). Add \(30\) to both sides: \(2x=30\), so \(x = 15\).
Then \(KP=x - 6=15 - 6 = 9\), \(KQ=x = 15\), \(PQ = 6\).

Step2: Calculate the perimeter of \(\triangle PKQ\)

The perimeter \(P\) of \(\triangle PKQ\) is \(P=KP+KQ+PQ\).
Substitute \(KP = 9\), \(KQ = 15\), \(PQ = 6\) into the formula: \(P=9 + 15+6\).

Answer:

\(30\)