In the figure (attached) here, lines l and m are parallel. I

m/l = PQ/PA (Similar triangles)

=> m/l = 3/2

AB = 3pi

AB subends a right triangle at center

so 90/360 = 3pi/2pir (r is radius of circle)

=> 1/4 = 3/2r

=> r = 6

Noe there is an isoceles triangle formed by AB at center

and l = AB (hypotenus) = root(2) * 6

So m = 3/2 * root(2) * 6 = 9root(2)

Answer - E

First of all, the figure should make you think of similar triangles. Evaluate whether the triangles are similar - they have a common angle. Also, the two sides, l and m are parallel so corresponding angles of the transversal will be equal. So triangle PAB is similar to triangle PQR (say the unnamed lower right vertex of the triangle is R). So by similar triangles, PA /PQ = 2/3 implies l/m = 2/3 too.

Now, AB subtends 45 degree angle at the circle so it will subtend a 90 degree angle at the centre of the circle which means that arc AB is 1/4 of the total circumference of the circle. Since length of arc AB is \(3\pi\), circumference of the circle is \(4*3\pi = 12\pi\). So radius is 6.

Now consider the triangle formed by l and the central angle subtended by AB. This will be an isosceles right triangle so 45-45-90 triangle giving ratio of sides as \(1:1:\sqrt{2}\). The length of the two legs will be 6 each so l \(= 6*\sqrt{2}\).

Since \(l/m = 2/3 = 6*\sqrt{2}/m\)

\(m = 9*\sqrt{2}\)

Answer (E)

Guessing:
If nothing comes to mind or time is very short, improve your chances of correct answer by eliminating some options easily. Note that length of l will be something less than arc AB so less than \(3\pi\).
So m will be \(3/2 * 3\pi = 4.5*3 = 13.5\) (approximately)
So m is a bit less than 13.5
Options (A), (C) and (D) are too large so they are out. Option (E) is the closest.

Angle at the center for arc AB = 90deg
=> (90/360)*2 pi R = arc AB
=> (90/360)*2 pi R = 3 pi
=> radius = 6

triangle OAB is right isosceles, right angled at O
OA=OB=radius=6
AB = 6root2 (pythagoras)

PA/PQ=2/3=l/m
m=(3/2) 6root2
=9root2

1. First of all redraw figure
2. Central angle of inscribed angle of 45 is 90. Therefore since radius makes this an isosceles triangle we have an isosceles right triangle.
3. We need to find the measurements, so we need the radius, we are given that the arc represents 1/4 of the circumference (Since central angle is 90) therefore, 2 pi * r / 4 = 3 pi --> r = 6
4. The hypothenuse of our isosceles right triangle will be 6 sqrt (2), as per the relation of measure in right isosceles triangles 45-45-90.
5. Since the ratio of the small triangle to the large triangle is 2:3 and since both sides are parallel, then 3/2 * 6sqrt (2) = 9sqrt (2)

Hence answer is E

Hope its clear
Cheers
J

Solution: E

AB forms 45 at circumference, it will form 90 at the centre of the circle

Let the radius of the circle be r

Arc length AB = 3π = (90/360) 2π * r = π * r /4
r = 6

Since arc AB forms 90 at centre

\(l = AB = r \sqrt{2} = 6\sqrt{2}\)

m/l = 3/2

\(m = l * 3/2 = 6\sqrt{2} *3/2 = 9\sqrt{2}\)

IMO E

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