Bet You Can't Solve this SAT Question

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Brain Station

Күн бұрын

Пікірлер: 47

@brain_station_videos 11 күн бұрын

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@samdaman2510 Ай бұрын

There is no way this was on the sat

@heenakhandelwal8608 Ай бұрын

This was the toughest and most satisfying question on this channel. What a beautiful presentation! I was shocked!

@gidonbezborodko4324 Ай бұрын

Nah, it’s EZ math

@VedanshBhatia Ай бұрын

@@gidonbezborodko4324Sure, then factorise 16x square minus 9y square minus 6y minus 1.

@caramelldansen8485 18 күн бұрын

@@gidonbezborodko4324 did you actually solve it tho? (It's usually people like this who couldn't even get anywhere near solving it)

@gidonbezborodko4324 18 күн бұрын

@@caramelldansen8485 yes I did

@Pyreshade Ай бұрын

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.

@marioalb9726 Ай бұрын

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

@fgdgtryhdfgrsgrtsr1749 Ай бұрын

"Why is this so hard?" Looks at answer "Oh, the answer is an approximate value"

@monroeclewis1973 Ай бұрын

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.

@brain_station_videos Ай бұрын

Nice solution!

@tejaamuthuraam2458 Ай бұрын

I couldn't understand clearly, why do you link the law of sines, from where is the central angle located, I am very blank, didn't understand your way of approach, how would it be possible to make a fraction out of 360,from where do you took the reference???

@altoclef4249 Ай бұрын

Or you can use an integral and solve it in a minute

@ArkticGamer Ай бұрын

Can you explain how , i am weak in math

@Megusta508 Ай бұрын

@@ArkticGamerwrite x as a function of y then integrate

@Naman-m9k Ай бұрын

@@Megusta508I did this way

@HobinderSinghA Ай бұрын

i did using coordinate geometery took wayy to long, also it was approx [1.23..]

@Descrypto Ай бұрын

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

@Maverick-115C Ай бұрын

That doesn't give you the are of the small part

@BruceKoerner 18 күн бұрын

You wasted your time. If you only solve the lower left quadrant of the problem, it is much simpler.

@ShineHtetAung-wv5so 26 күн бұрын

This is so easy to solve, i literally got the equation just by looking at the thumbnail, and my equation does not contain sine, cosine, let alone integretation. So here's the equation: Blue part = (The rectangle's area - The semi-circle's area) / 4 The reason why is divide by 4 at last is the blue part is just 1/4 of left area, Don't get it? Forget that there's a line making the triangle, then the picture becomes a semi-circle in a rectangle, add the line again, the left side of the left part is divided into half. Imagine the right had a line as the left also, now the left part is equally divided into 4. Let's get back into the equation, the equation to find a semi circle's area is 1/2 * pi * radious^2 Blue part = ((4*8) - (1/2 * 3.14159 * 4 * 4)) / 4 Blue part = (32 - 25.13272) / 4 Blue part = 6.86728 / 4 Blue part = 1.71682 My answer may not be accurate exactly, cuz i use PI and pi is infinite, i did use it as 3.14159, tho :D

@ShineHtetAung-wv5so 26 күн бұрын

The area of the blue part is 1.71982unit^2, i forgot the unit squared thing, sorry

@brain_station_videos 26 күн бұрын

But the answer is 1.25 sq. units

@RealQinnMalloryu4 Ай бұрын

(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).

@srinivaskaruturi982 Ай бұрын

Nah I figured the answer before this guy explained we had this topic in mensuration

@Релайм 28 күн бұрын

i nearly done it but i think i wrongly calculated sector because i through the angle feta was 45 degree

@zecaaabrao3634 Ай бұрын

Just integrate

@adityagupta8082 Ай бұрын

I did by another property of circle by joing top left corner with the intersection of diagnol of squear and circle making 90° and then same taking angle...

@chrisd.9319 Ай бұрын

This was great! How is 180 degrees equal pi - 2 theta at 2:18?

@shreesayajha Ай бұрын

i think he means 180 degrees is the same as pi not 180 = 2pi - 2theta

@Electro_Zap Ай бұрын

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).

@_Unknown420_ Ай бұрын

@@Electro_Zap What are radians 😭 Is it radii, sectors or segments of the same length??

@_Unknown420_ Ай бұрын

Nvm I just remembered they’re like radius units on the circumference/peripheral of the circle

@_Unknown420_ Ай бұрын

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)