Maximum Quadrilateral Area

Geometry Level 5

Two sides of a quadrilateral have length a a , while the other two are of length b b . Let the maximum possible area of the quadrilateral be A A .

Let the area of an ellipse with the major and minor axes as a a and b b , be B B .

Then the value of B A \frac{B}{A} is ...


The answer is 0.7854.

Janardhanan Sivaramakrishnan by Janardhanan Sivaramakris… , 42

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2 solutions

Max area of a quadrilateral is that of a square, and that of an ellipse is a circle. So it is circle in a square. a=b.
π 4 ∗ a 2 a 2 = 0.7854 \dfrac{\frac \pi 4* a^2}{a^2}=0.7854 .

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If two sides are of length a and the other two have length b, then there are two opposite vertices, each adjoining one side of length a and one of length b.

The area of the quadrilateral would thus be 2 × 1 2 a b sin ⁡ ( θ ) 2\times\frac{1}{2}ab\sin(\theta) , where θ \theta is the included angle between the two sides. Thus, the maximum quadrilateral area would be A = a b A=ab .

The area of ellipse with a a and b b as the major and minor axes would be B = π 4 a b B=\frac{\pi}{4}ab

Thus, B A = π 4 = 0.785 \frac{B}{A}=\frac{\pi}{4}=\boxed{0.785}

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isn't the area of ellipse = π a b = \pi a b ?

Sarthak Rath - 6 years, 3 months ago

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It is π a b \pi a b , if a a and b b are semi-major and semi-minor axes. Here a a and b b are the major and minor axes.

Janardhanan Sivaramakrishnan - 6 years, 3 months ago

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