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14 May 2019, 23:45
Kudos
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
45% (02:29) correct
55%
(02:57)
wrong
based on 40
sessions
History
Date
Time
Result
Not Attempted Yet
The figure below shows 13 circles of radius 1 within a larger circle. All the intersections occur at points of tangency. What is the area of the region, shaded in the figure, inside the larger circle but outside all the circles of radius 1 ?
A. \(4\pi\sqrt{3}\)
B. \(7\pi\)
C. \(\pi(3\sqrt{3} + 2)\)
D. \(10\pi(\sqrt{3} - 1)\)
E. \(\pi(\sqrt{3} + 6)\)
fc485e1337895925d85375cce1a9d3908ca8b7e3.png [ 42.31 KiB | Viewed 6002 times ]
A. \(4\pi\sqrt{3}\)
B. \(7\pi\)
C. \(\pi(3\sqrt{3} + 2)\)
D. \(10\pi(\sqrt{3} - 1)\)
E. \(\pi(\sqrt{3} + 6)\)
Attachment:
fc485e1337895925d85375cce1a9d3908ca8b7e3.png [ 42.31 KiB | Viewed 6002 times ]
16 May 2019, 00:21
Kudos
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For the radius of the bigger circle:
Centre smaller circle with two adjacent circles and 4th circle touching the bigger circle's circumference is taken..
Now since all are tangent.. we will have a right angled triangle of hypotenuse 2, and base of 1.. so altitude will be. 3^(1/2).The radius of the bigger circle will be 2* (3^(1/2)) +1
Now shaded area = area of big circle - area of smaller circles
=option A
Posted from my mobile device
Centre smaller circle with two adjacent circles and 4th circle touching the bigger circle's circumference is taken..
Now since all are tangent.. we will have a right angled triangle of hypotenuse 2, and base of 1.. so altitude will be. 3^(1/2).The radius of the bigger circle will be 2* (3^(1/2)) +1
Now shaded area = area of big circle - area of smaller circles
=option A
Posted from my mobile device
11 Jan 2022, 22:24
Kudos
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The four circles can be viewed as the sides of an equilateral as shown in the figure below.
The center of the bigger circle, is the centroid of the equilateral triangle.
Hence, the length of AO is \(\frac{\sqrt 3 * S}{ 3}\)
S - Side of the equilateral triangle
The side of the equilateral triangle is equal to the sum of diameter of two circles and radius of the other two circles where the vertex of the triangle lies.
Thus S = 2 + 2 + 1 + 1 = 6
AO = \(\frac{\sqrt 3 * 6}{ 3}\)
AO = \( 2 * \sqrt 3 \)
The radius if the bigger circle is OZ
OZ = AO + AZ
OZ = \( 2 * \sqrt 3 \) + 1
Area of the shaded region = Area of the bigger circle - ( 13 * Area of the smaller circles)
Area of the shaded region = \( \pi (2 * \sqrt 3 + 1) ^ 2 \) - (13 * \( \pi (1) ^ 2 \))
Area of the shaded region = \( 12\pi + 1\pi + 4\sqrt 3 \pi - 13 \pi \)
Area of the shaded region = \( 4\sqrt 3 \pi \)
The center of the bigger circle, is the centroid of the equilateral triangle.
Hence, the length of AO is \(\frac{\sqrt 3 * S}{ 3}\)
S - Side of the equilateral triangle
The side of the equilateral triangle is equal to the sum of diameter of two circles and radius of the other two circles where the vertex of the triangle lies.
Thus S = 2 + 2 + 1 + 1 = 6
AO = \(\frac{\sqrt 3 * 6}{ 3}\)
AO = \( 2 * \sqrt 3 \)
The radius if the bigger circle is OZ
OZ = AO + AZ
OZ = \( 2 * \sqrt 3 \) + 1
Area of the shaded region = Area of the bigger circle - ( 13 * Area of the smaller circles)
Area of the shaded region = \( \pi (2 * \sqrt 3 + 1) ^ 2 \) - (13 * \( \pi (1) ^ 2 \))
Area of the shaded region = \( 12\pi + 1\pi + 4\sqrt 3 \pi - 13 \pi \)
Area of the shaded region = \( 4\sqrt 3 \pi \)
Attachments
fc485e1337895925d85375cce1a9d3908ca8b7e3.png [ 45.64 KiB | Viewed 4521 times ]
