Question

1. in (\triangle klm) if (angle kcongangle l), (kl = 9x - 40), (lm = 7x - 37), and (km = 3x + 23), find the value of (x) and the measure of each side.

(x=)
(kl=)
(lm=)
(km=)

Explanation:

Step1: Set up the equation

Since \(\angle K\cong\angle L\), by the Isosceles Triangle Theorem, \(KL = LM\).
So, \(9x - 40=7x - 37\).

Step2: Solve for \(x\)

Subtract \(7x\) from both sides:
\(9x-7x - 40=7x - 7x- 37\)
\(2x-40=- 37\)
Add \(40\) to both sides:
\(2x-40 + 40=-37 + 40\)
\(2x=3\)
Divide both sides by \(2\):
\(x=\frac{3}{2}=1.5\)

Step3: Find \(KL\)

Substitute \(x = 1.5\) into \(KL=9x - 40\)
\(KL=9\times1.5-40=13.5 - 40=-26.5\) (This is wrong, we made a mistake above. Let's correct: Since \(\angle K\cong\angle L\), \(KL = KM\). So \(9x - 40=3x + 23\))

Step4: Solve the correct equation

Subtract \(3x\) from both sides:
\(9x-3x-40=3x - 3x+23\)
\(6x-40 = 23\)
Add \(40\) to both sides:
\(6x-40+40=23 + 40\)
\(6x=63\)
Divide by \(6\):
\(x = 10.5\)

Step5: Find \(KL\)

\(KL=9x-40=9\times10.5-40=94.5-40 = 54.5\)

Step6: Find \(LM\)

\(LM=7x-37=7\times10.5-37=73.5-37 = 36.5\) (Wrong again. Since \(KL = KM\), and in an isosceles triangle \(KL = KM\), then \(LM\) should be equal to them if it's equilateral? No, wait, original problem: \(\angle K\cong\angle L\), so \(KL = KM\).
\(KL=9x-40\), \(KM = 3x + 23\)
\(9x-40=3x + 23\)
\(9x-3x=23 + 40\)
\(6x=63\)
\(x = 10.5\)
\(KL=9\times10.5-40=94.5-40=54.5\)
\(LM=7x-37=7\times10.5-37=73.5 - 37=36.5\) (No, wait, another correction: Since \(\angle K\cong\angle L\), \(\triangle KLM\) is isosceles with \(KL = KM\)
\(KL=9x-40\), \(KM=3x + 23\)
\(9x-40=3x+23\)
\(9x-3x=23 + 40\)
\(6x=63\)
\(x = 10.5\)
\(KL=9\times10.5-40=54.5\)
\(KM=3\times10.5+23=31.5+23 = 54.5\)
\(LM=7\times10.5-37=73.5-37=36.5\) (This is wrong. Wait, no, if \(\angle K\cong\angle L\), then \(KM=LM\). So \(3x + 23=7x-37\)

Step7: Solve the correct - correct equation

\(3x+23=7x - 37\)
Subtract \(3x\) from both sides:
\(23=7x-3x-37\)
\(23 + 37=4x\)
\(60=4x\)
\(x = 15\)

Step8: Find \(KL\)

\(KL=9x-40=9\times15-40=135-40 = 95\)

Step9: Find \(LM\)

\(LM=7x-37=7\times15-37=105 - 37=68\) (No! Wait, \(\angle K\cong\angle L\), so \(KM = LM\). \(KM=3x + 23\), \(LM=7x-37\)
\(3x+23=7x-37\)
\(23 + 37=7x-3x\)
\(60 = 4x\)
\(x = 15\)
\(KL=9x-40=9\times15-40=135-40=95\)
\(KM=3x + 23=3\times15+23=45+23=68\)
\(LM=7x-37=7\times15-37=105-37=68\)

Answer:

\(x = 15\), \(KL=95\), \(LM = 68\), \(KM=68\)