GPREP PS #12

[hardikrs]

Figure attached.

Please explain your answer.

Thanks,

[touche]

Join o and Q.
angle formed by two triangles at apex = 180-35-35=110..............1.
Intercepted arc O R =9pi.....................2.
110=(9pi+ arc PQ)/2
220-9pi=arc PQ
Are the options ok?

[eravi11]

it's A..
to find the length of minor arc PQ, we need to find angle subtented at the center by segment PQ.

the angle is 40.
thus length of arc PQ = (40/360) * 2*pi*9= 2pi.

[touche]

Originally Posted by eravi11

it's A..
to find the length of minor arc PQ, we need to find angle subtented at the center by segment PQ.

the angle is 40.
thus length of arc PQ = (40/360) * 2*pi*9= 2pi.

Couldnt get this, could you explain I would be most indebted!

[eravi11]

sure touche..here it is..

let O be center of circle..
since OP=OR=9 and angle ORP=35, thus angle OPR=35
we already know angle QPR=35, since PQ!!OR.

Hence angle OPQ= angleOPR+angleQPR =35+35=70.

Now OP=OQ=9, thus angle OQP=70

now in traingle POQ, angle POQ= 180-(70+70)=40.

now u know the next steps to calculate the length of arc PQ...

hth

[touche]

That was simple and efficient!
Thanks! would remain indebted!

[clock60]

agree with A
the center of the circle will be C radius of the circle will be 18/2=9
angle ORP=RPQ=35 (as cross angles with paralell lines OR and QP)
if we join center of the circle C with point P we have triangle RCP isosceles
as CP=critical reasoning=radius
angles CRP=CPR=35
and now join point C with Q we have isosceles triangle PCQ as PC=QC=radius
angle CPQ=CQR=70 (35+35)
so anfle PCQ=180-(70+70)=40
and finally arc PQ
40/360*2*pi*R=40/360*2*pi*9=2*pi