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GmatPrep Geometry Qs
Source: Beat The GMAT — Problem Solving |
I got 2 pie as the answer.
Whats the OA?
Here is my method.
1. PQ || OR. Therefore, angle(ORP)=35 ..(alternate angle theorem)
2. Diameter OR subtends a right angle at P. Therefore, angle(OPR)=90
angle(OPR) + angle (PRO) + angle (POR)= 180
i.e. 90+35+ angle(POR)=180
angle(POR)=55
3. minor arc(OP) = 2* angle (PRO) ...(inscribed angle theorem)
minor arc (OP) = 2*35= 70.
4. arc(POR) = minor arc(OP) + arc (OR)
arc(OR) = 180 ..(since semi-circle)
Therefore, arc(POR) = 70 +180 = 250
Again, by incribed angle theorem,
angle(PQR) = (1/2) * [arc(POR)] = (1/2)*250 = 125
5. In triangle (PQR),
angle(PQR) + angle (PRQ) + angle (QPR) = 180
125 + angle (PRQ) + 35 = 180
angle(PRQ)=20
6. minor arc (PQ) = 2 * angle (PRQ)
= 2 * 20
= 40
7. length of arc = (40/360) * 2 * pie * radius
= 2pie
Well, not sure if this method is right. There could be an easier method.
Please post the original answer.
Whats the OA?
Here is my method.
1. PQ || OR. Therefore, angle(ORP)=35 ..(alternate angle theorem)
2. Diameter OR subtends a right angle at P. Therefore, angle(OPR)=90
angle(OPR) + angle (PRO) + angle (POR)= 180
i.e. 90+35+ angle(POR)=180
angle(POR)=55
3. minor arc(OP) = 2* angle (PRO) ...(inscribed angle theorem)
minor arc (OP) = 2*35= 70.
4. arc(POR) = minor arc(OP) + arc (OR)
arc(OR) = 180 ..(since semi-circle)
Therefore, arc(POR) = 70 +180 = 250
Again, by incribed angle theorem,
angle(PQR) = (1/2) * [arc(POR)] = (1/2)*250 = 125
5. In triangle (PQR),
angle(PQR) + angle (PRQ) + angle (QPR) = 180
125 + angle (PRQ) + 35 = 180
angle(PRQ)=20
6. minor arc (PQ) = 2 * angle (PRQ)
= 2 * 20
= 40
7. length of arc = (40/360) * 2 * pie * radius
= 2pie
Well, not sure if this method is right. There could be an easier method.
Please post the original answer.
ZMy answer is also A. 2 pi
PQ is parallel to diameter OR
1) PRO = QPR = 35
2) Arc PRO = Arc QPR = 70 (35x2)
Now, a circle = 360 degree
Arc PQ = 360- Arc OR - Arc PO - Arc QR = 350-180-70-70=40
length of the arc = (2*pi*r*theta)/360
= (2*pi*9*40)/360
= 2 pi
PQ is parallel to diameter OR
1) PRO = QPR = 35
2) Arc PRO = Arc QPR = 70 (35x2)
Now, a circle = 360 degree
Arc PQ = 360- Arc OR - Arc PO - Arc QR = 350-180-70-70=40
length of the arc = (2*pi*r*theta)/360
= (2*pi*9*40)/360
= 2 pi
Angle ORP = Angle RPQ = 35deg
Angle OPR = 90 deg (OR is the dia)
Angle OPQ + Angle QRO = 180 deg (cyclic quad)
So, angle PRQ = 20deg
Therefore, Angle subtended by minor arc PQ = 2 x 20 = 40deg (inscribed angle)
Length of arc = (2 x pi x 9 x 40)/360 = 2pi
Angle OPR = 90 deg (OR is the dia)
Angle OPQ + Angle QRO = 180 deg (cyclic quad)
So, angle PRQ = 20deg
Therefore, Angle subtended by minor arc PQ = 2 x 20 = 40deg (inscribed angle)
Length of arc = (2 x pi x 9 x 40)/360 = 2pi
