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

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Re: In the figure shown, line segments QS and RT are diameters of the circ [#permalink]
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

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Re: In the figure shown, line segments QS and RT are diameters of the circ [#permalink]
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
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Re: In the figure shown, line segments QS and RT are diameters of the circ [#permalink]
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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 wrote:
monirjewel wrote:
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 $$QR^2=(\frac{8}{\sqrt{2}})^2=r^2+r^2$$ --> $$2r^2=32$$ --> $$r^2=16$$ --> $$area=\pi{r^2}=16\pi$$.

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Re: In the figure shown, line segments QS and RT are diameters of the circ [#permalink]
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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 wrote:
monirjewel wrote:
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 $$QR^2=(\frac{8}{\sqrt{2}})^2=r^2+r^2$$ --> $$2r^2=32$$ --> $$r^2=16$$ --> $$area=\pi{r^2}=16\pi$$.

[/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.
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Re: In the figure shown, line segments QS and RT are diameters of the circ [#permalink]
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herein wrote:
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.

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.
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Re: In the figure shown, line segments QS and RT are diameters of the circ [#permalink]
mvictor wrote:
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

How did you know that you needed to square 8 / \sqrt{2}?
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Re: In the figure shown, line segments QS and RT are diameters of the circ [#permalink]
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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
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Re: In the figure shown, line segments QS and RT are diameters of the circ [#permalink]
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monirjewel wrote:
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) $$64 \pi$$

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

At this point, we can see that ∆QRO is a RIGHT TRIANGLE, which means we can apply the Pythagorean Theorem
We get: r² + r² = (8/√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?
= π(4
= 16π

Cheers,
Brent
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Re: In the figure shown, line segments QS and RT are diameters of the circ [#permalink]
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Kudos
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.
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Re: In the figure shown, line segments QS and RT are diameters of the circ [#permalink]
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Hi,

How do we assume the distance between Q and R is a straight line and not the length of the arc?
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Re: In the figure shown, line segments QS and RT are diameters of the circ [#permalink]
bpdulog wrote:
Hi,

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

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Re: In the figure shown, line segments QS and RT are diameters of the circ [#permalink]
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 $$pi*r^2 = 16Pi$$
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Re: In the figure shown, line segments QS and RT are diameters of the circ [#permalink]
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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