Answer: The correct option is (D) 36.
Step-by-step explanation: We are given to find the value of 'y' that would make OP parallel to LN.
MO = 28 units, OL= 14 units, Pl = 18 units and MP = y = ?
From the figure, we have
if OP ║ LN, then we must have
∠MOP = ∠MLN
and
∠MPO = ∠MNL.
Since ∠M is common to both the triangles MOP and MLN, so by AAA postulate, we get
ΔMOP similar to ΔMLN.
We know that the corresponding sides of two similar triangles are proportional, so
[tex]\dfrac{MO}{ML}=\dfrac{MP}{MN}\\\\\\\Rightarrow \dfrac{MO}{MO+OL}=\dfrac{MP}{MP+PN}\\\\\\\Rightarrow \dfrac{28}{28+14}=\dfrac{y}{y+18}\\\\\\\Rightarrow \dfrac{28}{42}=\dfrac{y}{y+18}\\\\\\\Rightarrow \dfrac{2}{3}=\dfrac{y}{y+18}\\\\\\\Rightarrow 2y+36=3y\\\\\Rightarrow y=36.[/tex]
Thus, the required value of 'y' is 36.
(D) is the correct option.
Answer:
Option D. y= 36
Step-by-step explanation:
In the given figure there are two triangles ΔMPO and ΔMNL
If two sides OP and LN are parallel and lines MN, ML are transverse respectively.
Then ∠MPO =∠MNL [ corresponding angles ]
and ∠MOP = ∠MLN [ corresponding angles ]
and ∠M is common to both the triangles.
Now by the property of AAA, ΔMPO & ΔMNL are similar
Now we know in similar triangles corresponding sides are in same ratio.
[tex]\frac{MP}{MN}=\frac{MO}{ML}[/tex]
[tex]\frac{y}{y+18}=\frac{28}{28+14}=\frac{28}{42}=\frac{2}{3}[/tex]
By cross multiplication
3y = 2(y + 18)
3y = 2y + 36
y = 36
Option D. y = 36 is the answer.
