SAT Practice - Hard Geometry Problems

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Can you solve these HARD geometry problems? If so you are on your way to getting a perfect score!

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Such a great video with great explanations, you don't get enough credit!

solar

Really liked the solution to the first problem :) I did a system of equations and it was a bit long, definitely clever to see it as an area.

TheCalcSeries

Can you solve more this type of hard questions??❤❤🎉🎉🎉😊😊😊

GAMEOVER-izlm

Amazing video!! You are so underrated and I'm going to share your videos to my friends.

I have a question regarding the second to last problem. How did you get your final answer (in purple) from the equation (written in red) right above of it? Why did you divide the equation by 2 if the fraction is 1/4?

Please let me know! Thanks!

nstnx

Hey! Why in the first question will the area of the big triangle be equal to the small one?

adityabhardwaj

The first one is not challenging. Use the area of triangle to solve it. For this triangle, there are two ways to solve the area of the right triangle, which is the usual half of altitude*height and the half of leg*leg. 8*6=x*10 (x is the length BD)

randaya

Hi, isn't the answer at question about the perimeter of shaded region must be: 9/2pi-9?

sol'n:
the diameter of the square is equal to the diameter of the circle: biscecting the square, we get a right triangle, using pythagorean theorem, we can arrive that the hypothenus is equal to 6sqrt(2).
c=sqrt(6^2+6^2) => c=6sqrt(2).
radius: 6sqrt(2)/2 => r=3sqrt(2).

area of circle is pi(r)^2
* A=pi(3sqrt(2))^2
* A=18pi
we can set-up an equation for the shaded region:
let x equal to the area of shaded region:

*4x + 36 = 18pi
because there are four identical shape cut by the circle.
solving for x:

4x = 18pi - 36
x= (18pi - 36)/2

simplifying further;
[ x= 9/2pi - 9 ]

jevonsmartirez