Cat for me is the best GRE instructor..I like she progressively builds the skills of her students with a clear plan what she will do.and achieve as objectives..only teachers can see how she is professional..
More so when the programs we are applying for aren’t in any way related to what the test consists of. Very irrelevant. Neither math nor rare complicated English words are required for my program at all
ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-05Jsne8RUkk.html Since there is a right angle at the top tip of the big triangle, we can put a "height" from the center downward creating a square on the left side. divide 10 in half where we get 5. since squares have 90 degree corners (split into 45 degree angles) the sides will be equal to each other, so 5 would be the answer.
Confused on your steps for the last questions regarding two circles and square... how would the answer be (C) Pi *R^2 when that would be finding the area of the whole circle when you are only asked to find the area of the shaded region?
Great question! The key here is that Cat defines r as the radius of the small circle (so the area of the small circle is indeed 𝜋r²). From here, use 45-45-90 logic to see that the radius of the large circle is r√2-so its area is 𝜋(r√2)² = 2𝜋r². Finally, notice that the shaded area = large circle area - small circle area = 2𝜋r² - 𝜋r² = 𝜋r².
Careful! In the smaller triangle, the sides are indeed 5 and the hypotenuse is 5√2. But in the larger triangle, the sides are 5√2, so the hypotenuse is 5√2√2 = 5*2 = 10.
@@LilGirth Thanks for the knowledge. I had the same doubt, too and I was struck at this point for a longtime and finally an internet hero to my rescue!
Yes! Both rectangles and squares are defined as having four sides that meet at right angles, but squares also need to have all four sides the same length.
@@manhattanprepgre7390 Thanks for your response, I really appreciate. So just because the information length is not being given , therefore it's tantamount to being a rectangle and not square?
The hypotenuse of an isosceles right triangle (i.e. a right triangle where the other two angles are 45 degrees) is always the length of the side times root 2. So for the triangle on the left, the hypotenuse is x rt 2. That hypotenuse then becomes the side of the triangle on the right, which is also 45 - 45 - 90. As a result, x * rt 2 * rt 2 = 10.
One way is knowing that with an isosceles triangle, the hypotenuse is always the length of the two shorter sides times the square root of 2. The other is the way you did it, I think you may have just made an arithmetic mistake. Working out that equation yields (8r)^2 and if you take the square root of both sides to solve for the diameter you get 2 root 2*r. Perhaps what you missed was the next step because you are solving in terms of radius, so in this case you would divide that result by 2 and you would end up with her result
@@kingwein89 for question 51 why did she use x*root 2*root2 =10. Isn't the 10 the hypothenus? If yes shouldn't the question be (x*root2 plus x*root2) all squared?
