In the figure above, the circumference of the circle is 20π. Which of

Among all the rectangles inscribed in a circle , the square has the maximum area.
Lets assume the side of the square is a , Hence the are a is \(a^2\)
According to the Q-stem for the circle : 2πr=20π ==>r=10 ==> 2r=20 = diameter of the circle = diagonal of the square

As diagonal of the square is 20, hence , \(20^2\) = \(a^2\)+\(a^2\) ==> 2\(a^2\)=400==>\(a^2\)=200

Hence I would go for Option B.
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Dear u1983.
Why are you considering it as square?
In question, rectangle is given. Could you explain this?

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Solution

Given:
• The circumference of the circle given is 20π

To find:
• The maximum possible are of the rectangle inscribed in the circle

Approach and Working:
• If the radius of the circle is r, then 2πr = 20π

o Or, radius r = 10

• Hence, the diameter of the circle = 2r = 20 = the diagonal of the rectangle

Now, for a rectangle inscribed within a circle, the area will be maximum when all the sides are equal (basically it is a square)
• Therefore, the diagonal length of the square with side a = a√2 = 20

o Hence, a = 10√2

• So, the area = 10√2 * 10√2 = 200

Hence, the correct answer is option B.

Answer: B
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Let us consider a rectangle with perimeter 24 units.
Now if perimeter is 24, the length and breadth of the rectangle can have possible values as (11, 1), (10, 2), (9, 3), (8, 4), (7, 5), (6, 6)
If we calculate the area in every individual case, we can see the following:



You can see, keeping the perimeter constant, the area of a rectangle will be maximum when the side lengths are equal, or it denotes a square
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Think about it - what is the constraint on a rectangle inside a fixed circle?

It can have very very little area (say the blue rectangle). Then we can start expanding it till it becomes the yellow rectangle (which is actually a square). If we try to expand it any more, its area will start decreasing again till it becomes very very small again (the green rectangle).



Hence, there is no minimum area of the rectangle but there is a maximum possible area (when it is a square) dependent on the value of the radius.

If the radius of the circle is 10 and diameter is 20, the area of the rectangle will be maximum when it is a square and each side of square will be \(\frac{20}{\sqrt{2}}\) (since diagonal of square = diameter of circle) and hence the area of the square will be 200.

Answer (B)
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I solved this using the Pythagoras theorem and got the area to be 192 sq. units. I considered diagonal, 20, to be the hypotenuse and derived the other two lengths through the Pythagorean triplet pairs of 3-4-5 (i.e, 12, 16 and 20 in this case). I approximated based on the answer above that since the answer is closest to 200, option B is correct.

Can anyone help me understand if my approach is wrong, why so?

Every right triangle with hypotenuse 20 will not be 12-16-20 triangle. Pythagorean triplets only give us the triangles with integer sides. But actually the other two sides could take infinite other values. e.g.
1, sqrt(399), 20
2. sqrt(396), 20
etc.

You are assuming that the rectangle with the greatest area is the one in which the two triangles are 3-4-5 triangles. That is not correct. The rectangle with the greatest area will be the square inside the circle.

Check this video explaining acute, obtuse and right triangles: https://www.youtube.com/watch?v=y05XIAWgAT0
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Thank you KarishmaB, that helps explain it better!