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Geometry Puzzle: Find the Area of the Rectangle

A square with four quarters of circles, one in each of its vertices. A red rectangle is at the center/centre of the square such that each of its vertices belongs to a quarter of circle. A vertical double arrow with the length of the square's side is labeled/labelled ten.

The sides of a rectangle with perimeter 12 are parallel to the sides of the square with side length 10.

What is the area of the rectangle?

Solution

Let xx and yy be the length and the width of the rectangle, respectively.

The radii of the quarters of circle is half the side length of the square, i.e., 5.

Let θ\theta be the marked acute angle in the right(-angled) triangle which hypotenuse is the radius of the quarter of circle.

A square with four quarters of circles, one in each of its vertices. A red rectangle is at the center/centre of the square such that each of its vertices belongs to a quarter of circle. Its length is labeled/labelled x and its width is labeled/labelled y. A right(-angled) triangle is at the bottom left of the square, where its hypotenuse is the radius of the quarter of circle and it is labeled/labelled five. The angle made by the hypotenuse with its base is labeled/labelled teta.

As its perimeter is 12, we have:

2x+2y=122x+2y=12
x+y=6(1)x+y=6\;\;\;\left(1\right)

Now, consider the right(-angled) triangle. The side length of the square is

5cos⁡θ+x+5cos⁡θ=105\cos \theta + x + 5\cos \theta = 10
10cos⁡θ+x=1010\cos \theta + x = 10
x=10−10cos⁡θ(2)x = 10 – 10\cos \theta\;\;\;\left(2\right)

and also

5sin⁡θ+y+5sin⁡θ=105\sin \theta + y + 5\sin \theta = 10
10sin⁡θ+y=1010\sin \theta + y = 10
y=10−10sin⁡θ(3)y = 10 – 10\sin \theta\;\;\; \left(3\right)

Substituting (2) and (3) into (1) we get:

10−10cos⁡θ+10−10sin⁡θ=610 – 10\cos \theta + 10 – 10\sin \theta = 6
20−10cos⁡θ−10sin⁡θ=620 – 10\cos \theta – 10\sin \theta = 6
10(cos⁡θ+sin⁡θ)=1410\left(\cos \theta + \sin \theta \right) = 14
cos⁡θ+sin⁡θ=1410\cos \theta + \sin \theta = \frac{14}{10}
cos⁡θ+sin⁡θ=75(4)\cos \theta + \sin \theta = \frac{7}{5}\;\;\;\left(4\right)

Squaring (4) we have:

(cos⁡θ+sin⁡θ)2=4925\left( \cos \theta + \sin \theta \right)^2= \frac{49}{25}
cos2⁡θ+2cos⁡θsin⁡θ+sin2⁡θ=4925\cos^2 \theta + 2\cos \theta \sin \theta + \sin^2 \theta = \frac{49}{25}

As sin2⁡θ+cos2⁡θ=1\sin^2 \theta + \cos^2 \theta = 1, we get:

2cos⁡θsin⁡θ+1=49252\cos \theta \sin \theta + 1 = \frac{49}{25}
2cos⁡θsin⁡θ=24252\cos \theta \sin \theta = \frac{24}{25}
cos⁡θsin⁡θ=1225(5)\cos \theta \sin \theta = \frac{12}{25}\;\;\;\left(5\right)

The area of the rectangle is xyxy, i.e.,

(10−10cos⁡θ)(10−10sin⁡θ)\left(10 – 10\cos \theta \right) \left(10 – 10\sin \theta \right)
100−100sin⁡θ−100cos⁡θ+100cos⁡θsin⁡θ100 – 100\sin \theta – 100\cos \theta + 100\cos \theta \sin \theta
100−100(sin⁡θ+cos⁡θ)+100cos⁡θsin⁡θ(6)100 – 100 \left(\sin \theta + \cos \theta \right) + 100\cos \theta \sin \theta\;\;\;\left(6\right)

Substituting (4) and (5) into (6) we get the area of the rectangle:

100−100×75+100×1225100 – 100 \times \frac{7}{5} + 100 \times \frac{12}{25}
100−140+48100 – 140 + 48
88

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