Arc & Semicircle

[ndcruz]

If arc PQR above is a semicircle, what is the length of diameter PR?
(1) a = 4
(2) b = 1

SPOILER: OA: D

[california]

1. a = 4

PQ = Sqrt(20)
QR = Sqrt(b^2+4)
PQR is right angle triangle as arc PQR is semi circle
PR^2 = PQ^2 + QR^2
(4+b)^2 = 20 + b^2 +4
16+b^2+8b = b^2+24
8b = 8
b = 1
Diameter = 5.....Sufficient

2. b=1
a can be determined as way as above ......Sufficient
Ans: D

[Dynamo]

Same logic and same calculation as california

its D i guess

[jimmyjamesdonkey]

Is there a way to solve this using similar triangles??? I thought when you drop a perpendicular height from a 90 degree angle to the opposite side you get 3 similar triangles????

[somu]

I too was trying to solve it using similar triangles but couldn't figure it out

[Ankost]

Similar triangles' approach
1. ratio 2/a is the same as b/2, a=4 2/4=b/2 b=1 D=4+1=5 Suff.
2. ratio b/2 is the same as 2/a, b=1 1/2=2/a a=4 D=4+1=5 Suff.

[gmatcraze]

Originally Posted by Ankost

Similar triangles' approach
1. ratio 2/a is the same as b/2, a=4 2/4=b/2 b=1 D=4+1=5 Suff.
2. ratio b/2 is the same as 2/a, b=1 1/2=2/a a=4 D=4+1=5 Suff.

Can you help to explain why the ratio is not taken as 2/a = 2/b?
Somebody please help to explain this .... not clear with similar triangle concept ...

[gmatcraze]

Originally Posted by Ankost

Similar triangles' approach
1. ratio 2/a is the same as b/2, a=4 2/4=b/2 b=1 D=4+1=5 Suff.
2. ratio b/2 is the same as 2/a, b=1 1/2=2/a a=4 D=4+1=5 Suff.

Can someone help to explain the similar triangles concept on the basis of the above solution .... Just wondering why the ratio is not taken as 2/a = 2/b? Please help to clarify .... Thanks

[Makumajon]

While applying the similar triangle properties rule, be careful to compare a side of one triangle with its appropriate counterpart from the other triangle. Here, consider the two split triangles: QMP & QMR (M is the point in which the perpendicular intersects the base PR.

QMP vs. QMR:
QMP=QMR...(1) (because each angle is 90 degree)
Angle QPM=Angle RQM...(2) (because each angle is complementary to angle PQM)
Angle PQM=Angle QRM...(3) (because each angle is complementary to angle RQM)

Once you have identified the equal angles, you now apply the side ratio relationship. You must choose side opposite to corresponding angles.

(1) gives you PQ/RQ
(2) gives you QR/RM
(3) gives you PM/QR.

So, PQ/RQ=QR/RM=PM/QR. Here you do not need to first ratio.

So, QR/RM=PM/QR
=> PM*RM=QR^2. This relationship in split right triangles is so favorite among GMAT, GRE, SAT, CAT, MBA tentmakers that you can memorize it.

[venkat007]

IMO : D
using pthogerous theorem got ab = 4.
so we can know the values of a and b from both of the given conditions seperately.