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Joined: 16 May 2014
Posts: 592
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GRE 1: Q165 V161
Practice Q#5- Coordinate Geometry Circle
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03 Aug 2014, 11:00
03 Aug 2014, 11:00
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Question Stats:
75% (02:26) correct
24% (02:04) wrong based on 33 sessions
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Try this tricky Geometry question
math1.JPG [ 28.8 KiB | Viewed 5264 times ]
Attachment:
math1.JPG [ 28.8 KiB | Viewed 5264 times ]
Re: Practice Q#5
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04 Aug 2014, 17:30
04 Aug 2014, 17:30
3
Lets name the point of intersection of the perpendicular from OQ on the line OR as X.
Thus forming a triangle named OQX. Now, from the figure angle QOR = 180-150=30 degrees, this makes triangle OQX a 30-60-90 triangle with OQ=4. From this, we know that the sides of the triangles are in a ratio of \(\sqrt{3}\):2:1, hence in triangle OQX, OQ = 4,\(OX = 2\sqrt{3}\), QX=2.
As angle QOR = 30 degree, the length of the arc \(QR = (30/360) X 2pi X OQ\) = \(2pi/3\).
As per, the question the perimeter of the shaded region would be,
length of QX + length of XR + length of arc QR = \(2 + (4 - 2\sqrt{3}) + 2pi/3\) = \(2pi/3 - 2\sqrt{3} + 6\)
Hence, the correct answer would be (D)
Thus forming a triangle named OQX. Now, from the figure angle QOR = 180-150=30 degrees, this makes triangle OQX a 30-60-90 triangle with OQ=4. From this, we know that the sides of the triangles are in a ratio of \(\sqrt{3}\):2:1, hence in triangle OQX, OQ = 4,\(OX = 2\sqrt{3}\), QX=2.
As angle QOR = 30 degree, the length of the arc \(QR = (30/360) X 2pi X OQ\) = \(2pi/3\).
As per, the question the perimeter of the shaded region would be,
length of QX + length of XR + length of arc QR = \(2 + (4 - 2\sqrt{3}) + 2pi/3\) = \(2pi/3 - 2\sqrt{3} + 6\)
Hence, the correct answer would be (D)
Practice Q#5- Coordinate Geometry Circle
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04 Jan 2024, 01:48
04 Jan 2024, 01:48
Let us call the point where a line from Q drops on to the line OR at an angle of 90 degrees as P. We have to find the perimeter of the figure PQO.
We now have a triangle PQO, which by the given measures is a 30-60-90 triangle.
Now the sides of a 30-60-90 triangle are in the ratio \(x:\sqrt{3}x:2x\) (short side:long side:hypotenuse)
Now we know that the hypotenuse OQ is nothing but the radius and is thus equal to 4 and is equal to 2x in the above ratio. This means that the value of x is 2. From this the we know that the value of the long side OP is \(\sqrt{3}*2 = 2*\sqrt{3}\). And the length of the short side PQ is x, which is equal to 2.
OQ= 4
OP = \(2\sqrt{3}\)
perimeter of shaded region - 2.4.png [ 20.16 KiB | Viewed 503 times ]
PQ = 2
perimeter of shaded region - 2.1.png [ 19.44 KiB | Viewed 515 times ]
Now we the know value of one side of the shaded region - PQ = 2.
Now to calculate the value of PR.
From the figure, PR = OR - OP
OR = 4 (the radius)
OP = \(2\sqrt{3}\) (the long side of triangle PQO)
PR = OR - OP
PR = \(4 - 2\sqrt{3}\)
perimeter of shaded region - 2.2.png [ 20.44 KiB | Viewed 504 times ]
Now we know the value of one more side - PR - of the shaded region PQR.
Now we need to know the value of the third side of the shaded region - the curve QR.
This can be easily found out since we are given the diameter of the circle PS = 8. This means the radius is 4 and the circumference is \(2*\pi*4 = 8*\pi.\)
Now we want to know the value of the central angle QOR. We know this to be 30 degs since it is a part of 30-60-90 triangle.
Now the length of the Arc QR can be found out from the ratio
\(\dfrac{30}{360} = \dfrac{x}{8*\pi}\)
\(\dfrac{1}{12} = \dfrac{x}{8*\pi}\)
\(x = \dfrac{8*\pi}{12} \)
\(x = \dfrac{2*\pi}{3}\)
\(QR = \dfrac{2*\pi}{3}\)
perimeter of shaded region - 2.3.png [ 20.28 KiB | Viewed 509 times ]
Now we can find the perimeter of the shaded region
PQR = PQ + PR + QR = \(2 + 4 - 2\sqrt{3} + \dfrac{2*\pi}{3} = \dfrac{2*\pi}{3} - 2\sqrt{3} + 6\)
The answer is Choice D - \(\dfrac{2*\pi}{3} - 2\sqrt{3} + 6\)
Attachment:
File comment: Drop point P from Q
perimeter of shaded region - 2.png [ 18.86 KiB | Viewed 508 times ]
perimeter of shaded region - 2.png [ 18.86 KiB | Viewed 508 times ]
We now have a triangle PQO, which by the given measures is a 30-60-90 triangle.
Now the sides of a 30-60-90 triangle are in the ratio \(x:\sqrt{3}x:2x\) (short side:long side:hypotenuse)
Now we know that the hypotenuse OQ is nothing but the radius and is thus equal to 4 and is equal to 2x in the above ratio. This means that the value of x is 2. From this the we know that the value of the long side OP is \(\sqrt{3}*2 = 2*\sqrt{3}\). And the length of the short side PQ is x, which is equal to 2.
OQ= 4
OP = \(2\sqrt{3}\)
Attachment:
perimeter of shaded region - 2.4.png [ 20.16 KiB | Viewed 503 times ]
PQ = 2
Attachment:
perimeter of shaded region - 2.1.png [ 19.44 KiB | Viewed 515 times ]
Now we the know value of one side of the shaded region - PQ = 2.
Now to calculate the value of PR.
From the figure, PR = OR - OP
OR = 4 (the radius)
OP = \(2\sqrt{3}\) (the long side of triangle PQO)
PR = OR - OP
PR = \(4 - 2\sqrt{3}\)
Attachment:
perimeter of shaded region - 2.2.png [ 20.44 KiB | Viewed 504 times ]
Now we know the value of one more side - PR - of the shaded region PQR.
Now we need to know the value of the third side of the shaded region - the curve QR.
This can be easily found out since we are given the diameter of the circle PS = 8. This means the radius is 4 and the circumference is \(2*\pi*4 = 8*\pi.\)
Now we want to know the value of the central angle QOR. We know this to be 30 degs since it is a part of 30-60-90 triangle.
Now the length of the Arc QR can be found out from the ratio
\(\dfrac{30}{360} = \dfrac{x}{8*\pi}\)
\(\dfrac{1}{12} = \dfrac{x}{8*\pi}\)
\(x = \dfrac{8*\pi}{12} \)
\(x = \dfrac{2*\pi}{3}\)
\(QR = \dfrac{2*\pi}{3}\)
Attachment:
perimeter of shaded region - 2.3.png [ 20.28 KiB | Viewed 509 times ]
Now we can find the perimeter of the shaded region
PQR = PQ + PR + QR = \(2 + 4 - 2\sqrt{3} + \dfrac{2*\pi}{3} = \dfrac{2*\pi}{3} - 2\sqrt{3} + 6\)
The answer is Choice D - \(\dfrac{2*\pi}{3} - 2\sqrt{3} + 6\)
