In the above figure, ABCD is a rectangle with a circle inscribed insid

Question

https://gmatclub.com/forum/download/file.php?id=49479 In the above figure, ABCD is a rectangle with a circle inscribed inside it. There is another rectangle with an inscribed circle inside it. The colored regions are marked in the figure above. if a dart is fired towards the block, and it hits the block in colored region, what is the probability that the dart hits the region which is red-colored? A) 1/2 B) 1/3 C) 1/4 D) 1/5 E) 1/6 GMATbuster's Collection WhatsApp Image 2019-07-28 at 20.11.32.jpeg

IanStewart · Accepted Answer

The GMAT won't test you on technicalities around the definition of "inscribed" -- unless you know precisely how that word is defined, you wouldn't know if it's possible to inscribe a non-square rectangle in a circle (touching at just two points). You can't actually do that, because of the definition of "inscribed", so the two rectangles here must be squares, but if this were a real GMAT question, it would just tell you both of the quadrilaterals are squares.

It's a pure ratio problem, so we can invent a number - say the edge of the big square is 4. Then the area of that square is 16, and the area of the big circle, which has a radius of 2, is 4π. Subtracting the circle from the square, we get the total area of the four corners (one of which is red), so that total area is 16 - 4π, and the area of one corner is 1/4 of that, or 4 - π. So that's the red-shaded area.

The diameter 4 of the big circle is the diagonal of the small square. Since the diagonal of a square is √2 times an edge of that square, each edge of the small square is 4/√2 = 2√2. So the area of the small square is (2√2)^2 = 8. The small circle has a radius half the length of a side of the small square, so a radius of √2 .Its area is thus 2π. We get the total yellow area by subtracting the area of the small circle from the area of the small square, so the yellow area is 8 - 2π.

Notice now that the yellow area is exactly twice as big as the red area, so 1/3 of the coloured area is red, so that's the probability, if we pick a random point from the coloured regions, that the point is in the red region.

GMATBusters · Answer

The rectangle here indeed is a Square, but it is not mentioned explicitly to increase the level of question a bit.

IanStewart · Answer

I understood that, but that's not how they increase the difficulty level of real GMAT questions.

Rajat1998 · Answer

Lets assume first of all that this is a square bcuz the circle inscribed touches every side. 
Let side be 4

Area of bigger square = 4*4 = 12

Area of bigger circle = 22/7*2*2 = 88/7
Radius is half of the side of sq

Area of the red portion = (12-88/7)/4 = 24/28

Side of inner sq = 4/√2

Area of inner sq = 8

Area of inner circle = 22/7*2/√2*2/√2 = 44/7

Area of yellow region = 8-44/7 = 12/7

Probability = favourable outcome / all outcomes

= Area of red region / area of all colored region
= (24/28)/(12/7+24/28) = 1/3

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Sayon · Answer

All we need to find is Area  \frac{(Colored Red)}{Area (Colored Red) + Area (Colored Yellow)}  ->  (a) 
We need to only find the ratios so we can assume any size for the given figure.
Since a circle is inscribed in the rectangle, the rectangle has to be a square.

Consider Figure A:
 Let, Side of smaller Square RQ = 2

Therefore, 
Radius of smaller circle OT = 1
Radius of bigger circle OS =  \sqrt{2} 
Side of bigger square BD =  2\sqrt{2}

Finding Area of Yellow Regions: 
Consider OHSJ:
Area of Yellow Portion in OSHJ = Area (OSHJ) – Area of Sector (OHJ) = ( 1 * 1) – (\frac{1}{4} * \pi * 1^2) = \frac{4 – \pi}{4} 
Now there are four such yellow regions. So total area of Yellow shaded region =  \frac{4 – \pi}{4} * 4 = 4 – \pi ->  (b)
 
 Finding Area of Red Region: 
Consider OIAK:
Area of Red Portion on OIAK = Area (OIAK) – Area of Sector (OIK) =  (\sqrt{2} * \sqrt{2}) – (\frac{1}{4} * \pi * (\sqrt{2})^2) = \frac{4 – \pi}{2}  ->  (c)

Hence From  (a) , the required probability =  \frac{(c)}{(b) + (c)} = \frac{4 – \pi}{\frac{4 – \pi}{2} + 4 – \pi} = \frac{1}{3}

Answer B

baraa900 · Answer

what is this formula called?

baraa900 · Answer

did you mean that the area of yellow is 6 whereas the area of red is 3? so we have yellow is twice big than red?? as a ratio 1/3 will be greater than 1/6?

Fdambro294 · Answer

Great Combination of Geometry and clever Algebra/Probability question.

(1st) Since a circle is inscribed inside a 4 sided rectangle and that rectangle is inscribed inside another a circle in which all 4 vertices “touch” the perimeter of the circumcircle———> and a circle is a set of points equidistant from a common shared center point ———> all 4 shapes will share 1 common center and the 2 rectangles must be Squares

Given that the player has already hit the colored area, what is the Probability that the dart lands in the Red Area?

Geometric Conditional Probability: probably of a success hit in Red Area given either the Yellow or Red Area must have been hit ———>

(Red Area) / (Yellow Area + Red Area) = ?

(1st) Find area of Yellow Region

Assume the inner Square has Side length = 1 unit

A circle inscribed in a Square will have its Diameter = Side of Square

D = 1 and r = (1/2)

Area of Yellow Shaded Region outside the inscribed circle but inside the square = (area of square) - (area of circle)

= (1) - (1/2)^2 * (pi)

= (4 -(pi)) / 4 = Yellow Area (I)

(2nd) Find the Red Region Area

Next, when a square is inscribed inside a circle, the Diagonal of the inscribed square = Diameter of circumscribed Circle

Diagonal of Inscribed Square = (1) * sqrt(2)

Radius of Circumscribed Circle = sqrt(2) / 2

Finally, this Circumscribed Circle becomes an INSCRIBED Circle for the outermost Square. As stated above, any inscribed circle’s Diameter = Side of Outermost Square = (1) * sqrt(2)

To get the Area of the Red Region, we need to take 1/4th of the Value of ———> (outermost square’s area) - (inscribed circle’s area) =

(1/4)   =

(1/2) - (pi)/(8) =

(4 - (pi)) / 8 = Red Area (II)

Probability = (Red Area) / (Yellow Area + Red Area) = (II) / (I + II) =

____________
  +

Pull out the factor of (1/2) from the Red Area Value in both the NUM and DEN and

Let ——>   = X

This gives us:

(1/2)X
________
(1)X + (1/2)X

Cancel the X in each term, combine the numbers in the DEN

(1/2)
____
(3/2)

= (2) / (2 * 3)

= (1/3)

Answer is = 1/3

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vatsal323 · Answer

It took me 4 odd minutes to solve. Are we sure that we are likely to see such a question on the actual GMAT?

GMATinsight · Answer

Hi  vatsal323

It's a lengthy question for sure. While the probability for such a question to appear in GMAT is low, it is desirable to practice such question and also make mindset in a way that average 2 min per question does NOT mean that every question should eb solved in two minutes. Some question while deserve only 40 seconds, some question might need your 3-4 minutes also for solving. So understand that average does not mean every question should be allocated the same amount of the time.

This question however can be done easily if you start assuming some number for either the biggest square or smallest circle.

Solution

Specify the question first

required Probability =  Area of Red Shaded Region  /  Area of Red Shaded Region + Area of Yellow shaded region

Red shaded area = (Big square - Big circle)/4

Yellow region = Area of smaller square - Area of smaller Circle

e.g., let, Smallest circle has radius = 1
i.e. Side of smaller square = Diameter of smaller circle = 2*1 = 2
Diameter of Bigger circle = Diagonal of Small square = Side√2 = 2√2 (i.e Radius = √2)
Side of bigger square = Diameter of Bigger circle = 2√2

Now,  Red shaded area  = \frac{(2√2)^2 - π(√2)^2}{4} = 2 - \frac{π}{2}

Yellow region  = 2^2 - π*1^2 = 4-π

Required Probability  = \frac{(2 - 0.5π) }{ (2 - 0.5π)+(4-π)} = \frac{1}{3}

Answer: Option B

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