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22 Aug 2012, 14:40
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Its simple actually. If one angle is 120, the adjacent angle has to be 60 (parallelogram rule).
Draw a parallelogram and draw diagonals. Apply 30-60-90 angle rules and you will get the answer. Catch is, you are asked ratio of the diagonal lengths, not the actual lengths.
Draw a parallelogram and draw diagonals. Apply 30-60-90 angle rules and you will get the answer. Catch is, you are asked ratio of the diagonal lengths, not the actual lengths.
22 Aug 2012, 21:32
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What is the approximate area of a circle that has an equilateral triangle with an area of 4√3 inscribed in it?
Image attached
O is the center of the circle and triangle ABC is equilateral.
AD is the angle bisector and is perpendicular bisector of the other side...
So, AO:OD = 2:1
(Radius = 2x)
In Triangle ADB, applying Pythagorean theorem we have
a^2 = (3x)^2 + (a/2)^2
=> x = a/(2*(sqrt 3))
given that area of triangle = 4 sqrt3 = (sqrt 3)*a^2/4
so, a = 4
so, x = a/(2*(sqrt 3)) = 2/(sqrt 3)
radius = 2x = 4/ (sqrt3)
Area of circle = pie r^2 = (22/7) * 16/3 = 352/21
Hope it helps!
Image attached
O is the center of the circle and triangle ABC is equilateral.
AD is the angle bisector and is perpendicular bisector of the other side...
So, AO:OD = 2:1
(Radius = 2x)
In Triangle ADB, applying Pythagorean theorem we have
a^2 = (3x)^2 + (a/2)^2
=> x = a/(2*(sqrt 3))
given that area of triangle = 4 sqrt3 = (sqrt 3)*a^2/4
so, a = 4
so, x = a/(2*(sqrt 3)) = 2/(sqrt 3)
radius = 2x = 4/ (sqrt3)
Area of circle = pie r^2 = (22/7) * 16/3 = 352/21
Hope it helps!
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File comment: Figure
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23 Aug 2012, 20:45
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NYC5648
The distance covered by the first man is S and the one run by the second man is 1000-S (meters). We need to find out X=1000-2S.
Their time is equal hence
s/15 = (1000-s)/25
5s=3000-3s
s=375
1000-2S=1000-750=250
