In the figure shown, line segments QS and RT are diameters of the circ

Question

https://gmatclub.com/forum/download/file.php?id=45056 In the figure shown, line segments QS and RT are diameters of the circle. If the distance between Q and R is \frac{8}{\sqrt{2}} , what is the area of the circle? (A) 4 \pi (B) 8 \pi (C) 16 \pi (D) 32 \pi (E) 64 \pi Untitled.png

Bunuel · Accepted Answer

The distance between two points always means the shortest distance between those two points, which is the length of a straight line between them.

Hope it's clear.

Bunuel · Answer

Let the center of the circle be O. Then OQ and OR will be the radii. Now as triangle OQR is right angle then QR will be its hypotenuse hence  QR^2=(\frac{8}{\sqrt{2}})^2=r^2+r^2  -->  2r^2=32  -->  r^2=16  -->  area=\pi{r^2}=16\pi .

Answer: C.

Sarang · Answer

Line joining Q n R makes right isoscelleous(45, 45, 90) triangle where hypotenus = 8/sqrt2

Thus radius = 8/(sqrt2 * sqrt2) = 4

Area = pie*r^2 = 16 pie

Answer:- C

mvictor · Answer

please use the math formulas to remove all ambiguities...
we can draw a line from Q to R, with a length of

\frac{8}{\sqrt{2}}

we know that both legs are radii. applying the 45-45-90 triangle rules, we can identify the radius, but it's easier to write the pythagorean formula:

(\frac{8}{\sqrt{2}})^2 = \frac{64}{2} = 2r^2 
 32 = 2r^2 
 r^2 = 16.

Area of a circle is:  pi*r^2

C

scorpionmba · Answer

Hi Bunuel,

COuld you please tell why do we consider the distance between Q and R not as the length of arc from Q to R ???

herein · Answer

Hi Bunuel,

COuld you please tell why do we consider the distance between Q and R not as the length of arc from Q to R ???

Bunuel

monirjewel

In the figure shown, line segments QS and RT are diameters of the circle. If the distance between Q and R is 8/root2, what is the area of the circle?
(A) 4 pie
(B) 8 Pie
(C) 16 pie
(D) 32 Pie
(E) 64 Pie

Let the center of the circle be O. Then OQ and OR will be the radii. Now as triangle OQR is right angle then QR will be its hypotenuse hence \(QR^2=(\frac{8}{\sqrt{2}})^2=r^2+r^2\) --> \(2r^2=32\) --> \(r^2=16\) --> \(area=\pi{r^2}=16\pi\).

Answer: C.

[/quote]

The distance between two points always means the shortest distance between those two points, which is the length of a straight line between them.

Hope it's clear.[/quote]

Hi Bunuel,

How can I make such assumption on the GMAT, unless otherwise the information that you provided above is mentioned. I am not sure whether I would call this a common sense to assume that the problem is talking about the "shortest distance" and not the arc length. If possible, please share some similar questions or relevant material, in which certain assumptions like the one stated above is considered a fundamental knowledge.

Also, I must say that your contribution to the success of vary many GMAT takers is immeasurable. Thanks a lot for everything you have done, and for everything you are doing.

RussianDude · Answer

hey herein, I almost made the same mistake when attempted this question. As far as I know, the GMAT always specifically tells you whenever the arch length shall be used. There is no reference to arch length in this question, so the shortest distance shall be used, as suggested by Bunuel.

Nathanlambson · Answer

How did you know that you needed to square 8 / \sqrt{2}?

bp2013 · Answer

couldn't we just use the 45/45/90 theorem that the sides of a triangle are x, x, and xsqrt2? then 8/sqrt2 = xsqrt 2 --> 8 = 2x, x = 4 (radius = 4), area = 16 pie

BrentGMATPrepNow · Answer

If QS and RT are diameters of the circle, then QS and RT both pass through the center of the circle. 
Since QS and RT intersect, their point of intersection must be the center of the circle (which we'll denote with  O )
This means OQ and OR are both radii of the circle, which we'll denote with   r

https://i.imgur.com/kX9vueK.png 
At this point, we can see that ∆QRO is a RIGHT TRIANGLE, which means we can apply the  Pythagorean Theorem 
We get:   r  ² +   r  ² = ( 8/√2 )²
Simplify: 2r² = 64/2
Simplify more: 2r² = 32
Divide both sides by 2 to get: r² = 16
Solve:   r = 4

What is the area of the circle?  
Area of circle = π(radius)²
= π(  4  )²
= 16π

Answer: C

Cheers,
Brent

jaysonbeatty12 · Answer

I agree with the math as detailed above. AND - you absolutely should NOT do the math here. You should estimate. The answer choices from a to e here are each 100 percent bigger than the one above. When answer choices are spread out like this you can save time by estimating. Estimate root 2 as 1.5.

The hypotenuse is around 5.5 (8 / root 2 is about 5.5). The radius is therefore about 3.7 - in a 45-45-90 triangle the hypotenuse is 140% of the leg distance. 3.7 squared is close to 16 and far from any other answer choice. C is correct.

Almost every GMAT PS question has shortcuts - this is by design. Finding these shortcuts is the path to GMAT mastery. The GMAT is NOT a test of your ability to calculate.

bpdulog · Answer

Hi,

How do we assume the distance between Q and R is a straight line and not the length of the arc?

Bunuel · Answer

Answered here: https://gmatclub.com/forum/in-the-figur ... l#p1672959

dcummins · Answer

Info that helps solve this question:
- The distance between two points is a straight line.
- since QS and RT are the diamters of the circle, (Q,0) and (0,R) are radii and thus equal

Since we have a 45:45:90 triangle and one side is known, we can apply the side ratios to determine the radii

x: x: x\sqrt{2} 
 x\sqrt{2} =\frac{8}{\sqrt{2}} 
Multiply equation by  \sqrt{2}  to get:
2x = 8
thus x =4 (radii = 4)

Solve using  pi*r^2 = 16Pi

adityasuresh · Answer

The distance between q & r = 4 root 2. Since oq=op=radius, angle opposite 90 is in the ration root2 *x from stem we know pq=4 root 2
This implies op =oq=radius =4 
Area = pie r^2 =16pie

In the figure shown, line segments QS and RT are diameters of the circ

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In the figure shown, line segments QS and RT are diameters of the circ

Prep Toolkit

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