Option B
Apply Pythagoras Theorem [H^2 = B^2 + P^2]
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Geometry - Triangle
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neelgandham
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We know the value of y(Square Root(2)) from the question. Let the angle between the side(L), which is parallel to the surface of the laptop keypad when the screen is exactly perpendicular to the keypad, and the hypotenuse be X, then
Sin X = (1/(Square Root(2)))/(Square Root(2)) = 1/2. So, from the result, we know that X = 30 degrees.
If X = 30 Degrees, Tan X = 1/(Square Root(3)) = (1/(Square Root(2)))/L, i.e. L = (Square Root(3))/(Square Root(2)).
The Area of the triangle = 0.5*L*Other side = 0.5 * [(Square Root(3))/(Square Root(2))]*(1/(Square Root(2))) = (Square Root(3))/4
Answer :B
Sin X = (1/(Square Root(2)))/(Square Root(2)) = 1/2. So, from the result, we know that X = 30 degrees.
If X = 30 Degrees, Tan X = 1/(Square Root(3)) = (1/(Square Root(2)))/L, i.e. L = (Square Root(3))/(Square Root(2)).
The Area of the triangle = 0.5*L*Other side = 0.5 * [(Square Root(3))/(Square Root(2))]*(1/(Square Root(2))) = (Square Root(3))/4
Answer :B
Anil Gandham
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Let the base be 'B", now
B^2 + (y/2)^2 = y^2
B^2 = 3Y^2/4
B = Sqrt(3) * y / 2
Area = 1/2 bh = 1/2 * Sqrt(3) * y / 2 * y/2 = Sqrt(3) * y^2 / 8
= Sqrt(3) * 2/8 = Sqrt(3)/4 B IMO
B^2 + (y/2)^2 = y^2
B^2 = 3Y^2/4
B = Sqrt(3) * y / 2
Area = 1/2 bh = 1/2 * Sqrt(3) * y / 2 * y/2 = Sqrt(3) * y^2 / 8
= Sqrt(3) * 2/8 = Sqrt(3)/4 B IMO
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Anurag@Gurome
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Let the base = xgmatblood wrote:Please help
By Pythagoras Theorem, we get y² = (y/2)² + x²
y² = (y²/4) + x²
(3/4)y² = x²
x = y√3/2
Then area of the triangular field in sq miles = (1/2) * y√3/2 * (y/2) = y²√3/8
Since y² = 2 is given in the question, so area = 2√3/8 = [spoiler]√3/4[/spoiler]
The correct answer is B.
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