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17 Nov 2020, 01:15
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C
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Difficulty:
45%
(medium)
Question Stats:
64% (01:57) correct
36%
(02:13)
wrong
based on 99
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What is the circumference of the semicircle in the figure shown?
A. \(14\pi\sqrt{6}\)
B. \(7\pi\sqrt{6}\)
C. \(\frac{7\pi\sqrt{6}}{2}\)
D. \(\frac{7\pi\sqrt{6}}{3}\)
E. \(\frac{7\pi\sqrt{6}}{4}\)
Attachment:
2020-10-27_14-08-58.png [ 30.13 KiB | Viewed 3486 times ]
17 Nov 2020, 02:19
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Length of side of an isosceles triangle with angles 45 deg = 14 cos30 = 14 \sqrt{3} /2 = 7 \sqrt{3} = the other equal side of an isosceles triangle
The hypotenuse of this isosceles triangle = Diameter of the semi-circle = \sqrt{2} x 7 \sqrt{3} = 7 \sqrt{6}
Radius of semi circle = 7 \sqrt{6} / 2
Circumference of semi-circle = pi x 7 \sqrt{6} / 2 = \(\frac{7\pi\sqrt{6}}{2}\)
ANSWER C IMO
The hypotenuse of this isosceles triangle = Diameter of the semi-circle = \sqrt{2} x 7 \sqrt{3} = 7 \sqrt{6}
Radius of semi circle = 7 \sqrt{6} / 2
Circumference of semi-circle = pi x 7 \sqrt{6} / 2 = \(\frac{7\pi\sqrt{6}}{2}\)
ANSWER C IMO
Bunuel
17 Nov 2020, 03:31
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IMO: C
Top Triangle's is of 30,60,90 deg property hence, Side will be in the ration of 1:Sqrt3:2
therefore, Base = 14 x (sqrt3)/2 = 7 x (sqrt3)
Therefore hypotenuse of below triangle in diagram = 7 x (sqrt6) (note: the lower triangle is isosceles right-angle triangle hence, we can apply Pythagoras theorem)
Therefore Dia of Circle= 7 x (sqrt6)
Hence, circumference of semi cirlce = (pi x Dia)/2 = Pi x 7 x (sqrt6) /2
ANSWER: C
Top Triangle's is of 30,60,90 deg property hence, Side will be in the ration of 1:Sqrt3:2
therefore, Base = 14 x (sqrt3)/2 = 7 x (sqrt3)
Therefore hypotenuse of below triangle in diagram = 7 x (sqrt6) (note: the lower triangle is isosceles right-angle triangle hence, we can apply Pythagoras theorem)
Therefore Dia of Circle= 7 x (sqrt6)
Hence, circumference of semi cirlce = (pi x Dia)/2 = Pi x 7 x (sqrt6) /2
ANSWER: C
17 Nov 2020, 19:39
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Triangle 1:
The triangle on top is 30-60-90 triangle
So, we know the ratio of sides of the triangle is \(x:x\sqrt{3}:2x\)
From the triangle 1 on top, we know the side 2x=14, so x=7
Base of triangle 1 = \(7\sqrt{3}\)
Triangle 2:
The triangle in the bottom is 45-45-90 triangle
So, we know the ratio of sides fof the triangle is x:x:\(x\sqrt{2}\)
The hypotenuse of the triangle with side \(x\sqrt{2}\) = \(7\sqrt{3}\)*\(\sqrt{2}\)
The hypotenuse of the triangle 2 = \(7 \sqrt{6}\)
Circle:
The hypotenuse of the triangle is the diameter of the circle.
Diameter = \(7 \sqrt{6}\)
Circumference = d\(\pi\) = \(7 \sqrt{6}*\pi\)
Circumference of semicircle = \( (7 \sqrt{6}*\pi) / 2\)
Ans: C
The triangle on top is 30-60-90 triangle
So, we know the ratio of sides of the triangle is \(x:x\sqrt{3}:2x\)
From the triangle 1 on top, we know the side 2x=14, so x=7
Base of triangle 1 = \(7\sqrt{3}\)
Triangle 2:
The triangle in the bottom is 45-45-90 triangle
So, we know the ratio of sides fof the triangle is x:x:\(x\sqrt{2}\)
The hypotenuse of the triangle with side \(x\sqrt{2}\) = \(7\sqrt{3}\)*\(\sqrt{2}\)
The hypotenuse of the triangle 2 = \(7 \sqrt{6}\)
Circle:
The hypotenuse of the triangle is the diameter of the circle.
Diameter = \(7 \sqrt{6}\)
Circumference = d\(\pi\) = \(7 \sqrt{6}*\pi\)
Circumference of semicircle = \( (7 \sqrt{6}*\pi) / 2\)
Ans: C
