I got approximately 1.252, or 4*(-pi+2*sin(2*arctan(1/2))+4*arctan(1/2)). This problem was really fun to work through! I solved this by declaring the shaded area as x. The area encompassed by (rect-semicircle)/2 is broken down into two segments x and y. When the equation is rearranged, (rect-semicircle)/2-y=x. I got y by breaking the upper triangle into two sections, y and the divided semicircle. I let radius equal the semicircle radius. The divided semicircle can be broken down into two segments: an isosceles triangle of equal side lengths radius and height of radius*sin(theta) and base of 2*radius*cos(theta), and sector of angle 2*theta. From here, I derived the equation: y=rect/2-isosceles-sector. To create the final equation, I substituted y and got: (rect-semicircle)/2-(rect/2-isosceles-sector)=x. Now, to substitute values: (radius*2*radius-(pi*radius^2/2)/2)/2-(radius*2*radius/2-radius*sin(theta)*radius*2*cos(theta)/2)-pi*radius^2*2*theta/(2*pi))=x. To simplify the equation: radius^2-pi*radius^2/4-radius^2+radius^2*2*sin(theta)*cos(theta)/2+theta*radius^2=x. radius^2*(1-pi/4-1+2*sin(theta)*cos(theta)/2+theta)=x. I got: radius^2*(-pi/4+sin(2*theta)/2+theta)=x. Now to obtain the final answer: theta=arctan(½) radius=4 Substituting: 16*(-pi/4+sin(2*arctan(½))/2+arctan(½))=x. To remove the fractions: 4*(-pi+2*sin(2*arctan(½))+4*arctan(½))=x. This means that x equals roughly 1.252 square units.
I dropped a line down from the center of the semi circle intersecting the long diagonal at a right angle. This created a small right triangle similar to the large right triangle formed by the diagonal of the rectangle and the other two sides. So 4:8 : : x:2x. Then used the Pythagorean theorem, 4^2=x^2+(2x)^2. X= (4 x sq root 5)/5. That gives you all you need to find the length of the chord and the area of the triangle which you must subtract from the area of the sector to find the area of the segment. Knowing the three sides of the triangle, 4, 4, and 16 x sq root 5/5 (I.e., 2x), I used the Law of Sines to find the central angle with the data I had developed above. Then using the fraction of that angle over 360, I found the area of the sector. The rest was just subtracting various areas from the area of the rectangle (32) to find the residual blue area. I came up with 1.26, good enough for government work. The key to the problem was creating similar triangles. The rest was drudgery.
Much simpler if you realize that the radius is 4, and you have the measurements to calculate the area of the rectangle already. Just subtract half of the circle are from the area of the rectangle already
180 degrees = pi in radians, it's an isosceles triangle implying that two of the angles are the same and a triangle must have a total interior angel of 180 degrees (otherwise written as pi in radians).
I was understanding everything until sin(theta) cos(theta) and tan(theta) showed up (I’m a ninth grader and I have no idea what these functions do or what they are)
I got approximately 1.252, or 4*(-pi+2*sin(2*arctan(1/2))+4*arctan(1/2)).
This problem was really fun to work through!
I solved this by declaring the shaded area as x.
The area encompassed by (rect-semicircle)/2 is broken down into two segments x and y.
When the equation is rearranged, (rect-semicircle)/2-y=x.
I got y by breaking the upper triangle into two sections, y and the divided semicircle.
I let radius equal the semicircle radius.
The divided semicircle can be broken down into two segments: an isosceles triangle of equal side lengths radius and height of radius*sin(theta) and base of 2*radius*cos(theta), and sector of angle 2*theta.
From here, I derived the equation: y=rect/2-isosceles-sector.
To create the final equation, I substituted y and got: (rect-semicircle)/2-(rect/2-isosceles-sector)=x.
Now, to substitute values:
(radius*2*radius-(pi*radius^2/2)/2)/2-(radius*2*radius/2-radius*sin(theta)*radius*2*cos(theta)/2)-pi*radius^2*2*theta/(2*pi))=x.
To simplify the equation:
radius^2-pi*radius^2/4-radius^2+radius^2*2*sin(theta)*cos(theta)/2+theta*radius^2=x.
radius^2*(1-pi/4-1+2*sin(theta)*cos(theta)/2+theta)=x.
I got:
radius^2*(-pi/4+sin(2*theta)/2+theta)=x.
Now to obtain the final answer:
theta=arctan(½)
radius=4
Substituting:
16*(-pi/4+sin(2*arctan(½))/2+arctan(½))=x.
To remove the fractions:
4*(-pi+2*sin(2*arctan(½))+4*arctan(½))=x.
This means that x equals roughly 1.252 square units.
tan α = 4/8 = 1/2 --> α=26,565°
β =180°-2α = 129,87°
A₁= ½b.h = ½8*4 = 16 cm²
A₂= ½R²(β-sinβ)=½4²(β-4/5)=11,314cm²
A₃= R²-¼πR²= 4²(1-π/4)= 3,4339cm²
A = A₁-A₂-A₃ = 1,252 cm²
This was the toughest and most satisfying question on this channel. What a beautiful presentation! I was shocked!
Nah, it’s EZ math
There is no way this was on the sat
(4)^2(8)^2={16+64}=80 360°ABCD/80=40.40 2^20.2^20 2^10.2^10 1^5^5.1^5^5 12^3^2^3.2^3^2^3 1^1^1^1.2^1^1^3 23(ABCD ➖ 3ABCD+2).
I dropped a line down from the center of the semi circle intersecting the long diagonal at a right angle. This created a small right triangle
similar to the large right triangle formed by the diagonal of the rectangle and the other two sides. So 4:8 : : x:2x. Then used the Pythagorean theorem, 4^2=x^2+(2x)^2. X= (4 x sq root 5)/5. That gives you all you need to find the length of the chord and the area of the triangle which you must subtract from the area of the sector to find the area of the segment. Knowing the three sides of the triangle, 4, 4, and 16 x sq root 5/5 (I.e., 2x), I used the Law of Sines to find the central angle with the data I had developed above. Then using the fraction of that angle over 360, I found the area of the sector. The rest was just subtracting various areas from the area of the rectangle (32) to find the residual blue area. I came up with 1.26, good enough for government work. The key to the problem was creating similar triangles. The rest was drudgery.
Nice solution!
i did using coordinate geometery took wayy to long, also it was approx [1.23..]
"Why is this so hard?"
Looks at answer
"Oh, the answer is an approximate value"
Or you can use an integral and solve it in a minute
Can you explain how , i am weak in math
@@ArkticGamerwrite x as a function of y then integrate
@@Megusta508I did this way
How do you make video?? Which software do you use for making video please ❤❤❤❤
My answer is approaching 1.234
Just integrate
Just use calculus
Much simpler if you realize that the radius is 4, and you have the measurements to calculate the area of the rectangle already. Just subtract half of the circle are from the area of the rectangle already
That doesn't give you the are of the small part
Whatd a 2 theta
It's like your 'x' in algebra, it's a variable.
EZ math
This was great! How is 180 degrees equal pi - 2 theta at 2:18?
i think he means 180 degrees is the same as pi not 180 = 2pi - 2theta
180 degrees = pi in radians, it's an isosceles triangle implying that two of the angles are the same and a triangle must have a total interior angel of 180 degrees (otherwise written as pi in radians).
@@Electro_Zap
What are radians 😭
Is it radii, sectors or segments of the same length??
Nvm I just remembered they’re like radius units on the circumference/peripheral of the circle
I was understanding everything until sin(theta) cos(theta) and tan(theta) showed up (I’m a ninth grader and I have no idea what these functions do or what they are)
Easu
So Gooood 😼👍