Tuffest question

By- Vikash Rahii
Q.
The diagram involves a rectangle with dimensions of 10 and 20. Inside the rectangle is one diagonal. There are also two circles of equal size that are tangent to each other and the rectangle.what is the area of red spot?

 
step-1
 The area of the red regions can be solved by subtraction. The red regions are equal to the area of the triangle (formed by the rectangle and its diagonal) minus the area of the portions of circles.


The area of the triangle is half that of the rectangle, which is (20)(10)/2 = 100.
The two regions of the circle together combine to form the area of one circle. This is seen by symmetry: the diagonal divides each circle into two regions. The region below the diagonal in one half is exactly equal to the region above the diagonal in the other half. The circle has a radius that is half the rectangle’s height of 10, so its radius is 5. The circle has an area of π(5²) = 25π.
The area of the red regions is therefore 100 – 25π ≈ 21.460.
Here is the picture for the second problem.









The area of the red regions is now 100 – 25π – (lower left piece). So we focus on solving for the piece in the lower left.
Consider a single circle inside of a square with length 10, and draw a line from its diagonal to halfway on one side. We can solve for the area of the region in the lower left by subtraction as follows.








Now we need to solve for the area of the right triangle, the area of the lower right piece, and the area of the circular piece.
The area of the right triangle is (10)(5)/2 = 25 because its base is the square side and its height is half the square’s height.
The area of the lower right piece is the area of 1/4 of the large square (so its side length is 5) minus the area of 1/4 of the circle (radius 5).


The area of the lower right piece is therefore 25 – 25π/4 = 25(1 – π/4).
The area of the circular region is a bit harder. We again solve it by subtraction. It is the area of a sector of the circle minus the area of an isosceles triangle.










The area of a circular sector with angle α and radius r is r²α/2.
The area of a triangle with adjacent sides a and b and an angle between them β is (ab sin β)/2.
The circle has a radius of 5, so the isosceles triangle also has its two equal sides equal to 5.
Consider the right triangle with sides 5 and 10. Let θ be the angle opposite the side of 5. The circle’s radius is parallel to the triangle’s side of 10, meaning the triangle in the circle has an angle equal to θ. Because the triangle is isosceles, its other angle is θ. The circle’s central angle is therefore π – 2θ because all of the triangle’s angles must add up to π radians.
We can substitute these values and simplify using the identity sin(π – x) = sin x.











We can solve for the angle θ as well. In a right triangle where its opposite side is 5 and its adjacent side is 10, the angle θ is therefore the inverse tangent of 5/10 = 1/2. Thus θ = tan^-1(1/2).
The right triangle with legs of 5 and 10 has a hypotenuse of √(5² + 10²) = √125. This means cos θ = 10/√125 and sin θ = 5/√125.
Now we substitute these values and simplify the formulas as necessary, using the identity sin(2y) = 2(sin y)(cos y).



We can put all this together to get the area of the lower left region is 10 – 25π/4 + 25 tan^-1(1/2) ≈ 1.956.




Recalling the area of the red regions is 100 – 25π – (lower left piece), we get to the the answer of 90 – 18.75π – 25 tan^-1(1/2) ≈ 19.504.
 this is right answer..