Question : In a $\triangle$LMN, OP is a line segment drawn parallel to the side MN. OP intersects the sides LM and LN at O and P, respectively. If LM = 15 cm, OM = 4 cm, and PN = 5 cm, then what is the length (in cm) of the side LN?
Option 1: 16.25
Option 2: 18.75
Option 3: 20.25
Option 4: 22.75
Correct Answer: 18.75
Solution :
OP || MN
LM = 15 cm
OM = 4 cm
PN = 5 cm
LO = LM – OM = 15 – 4 = 11 cm
Let LN = $x$ cm
LP = LN – PN = $(x-5)$ cm
In $\triangle LOP$ and $\triangle LMN$,
$\angle LOP = \angle LMN$ (corresponding angles)
$\angle LPO = \angle LNM$ (corresponding angles)
$\angle OLP = \angle MLN$ (common angle)
By AAA similarity, $\triangle LOP$ ~ $\triangle LMN$
$\frac{LO}{LM} = \frac{LP}{LN}$
⇒ $\frac{11}{15} = \frac{(x-5)}{x}$
⇒ $11x = 15x - 75$
⇒ $4x = 75$
⇒ $x = \frac{75}{4} = 18.75$
Hence, the correct answer is 18.75.
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