For trigonomitry values use Sin ( 0, 30 or pi/6, 45 or pi/4, 60 or pi/3, 90 or pi/2) = 0 , 1/2, racine(2)/2, racine(3)/2, 1 Cos the opposite Tan use tang= sin/ cos
We're told that PQR is an isosceles triangle with at least one side length of 7. This gives us two possible cases. If PQR has two sides with length seven, then the length of the third side can be between 0 and 14. This means the perimeter can be anything between 14 and 28. If PQR has one side with length seven then the other two sides must be the same. The smallest length they could be is a little over 3.5 each, but there is no maximum length. It's possible to have a triangle with side lengths of 7, 100, and 100. It would be a very small and skinny triangle, but it's a viable triangle. This means the minimum the perimeter can be is 14, but there is theoretically no maximum. Therefore, we can pick answer choices (C), (D), (E), and (F) as answers to this question. I hope that helps!
For the purposes of the GRE, this is a standard result you can use without proving it. Proving that the side lengths of a 30:60:90 triangle are in the ratio x:x*sqrt(3):2x involves using trigonometry, which is beyond the scope of the GRE, so I'm not going to do it here. If you want a full explanation of why this works, check out the linked video below. I hope that helps! ruclips.net/video/SFL4stapeUs/видео.html
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For trigonomitry values use
Sin ( 0, 30 or pi/6, 45 or pi/4, 60 or pi/3, 90 or pi/2) = 0 , 1/2, racine(2)/2, racine(3)/2, 1
Cos the opposite
Tan use tang= sin/ cos
Sum of angles of polygan= 180 ×(n-2) regular or not
Angle of each regular polygon = 180 (n-2)/n
Isoscele troanfle with two equl sides
The length of one size of a triangle < the sum of the two others
In a right triangle: each side is less than the hypotenuse
In comparative question, you dont have to calculate something, you can use inequloties formulas
Similar triangle theorem is just thales propotional theorem using 2 height od traingle as parallel also
how to attempt such questions within GRE's time limit
in the 3rd question why didnt we follow the constraint that perimeter has to be less than 28 ?
We're told that PQR is an isosceles triangle with at least one side length of 7. This gives us two possible cases.
If PQR has two sides with length seven, then the length of the third side can be between 0 and 14. This means the perimeter can be anything between 14 and 28.
If PQR has one side with length seven then the other two sides must be the same. The smallest length they could be is a little over 3.5 each, but there is no maximum length. It's possible to have a triangle with side lengths of 7, 100, and 100. It would be a very small and skinny triangle, but it's a viable triangle.
This means the minimum the perimeter can be is 14, but there is theoretically no maximum. Therefore, we can pick answer choices (C), (D), (E), and (F) as answers to this question.
I hope that helps!
in Q1 why it is x:xroot3:2x
For the purposes of the GRE, this is a standard result you can use without proving it. Proving that the side lengths of a 30:60:90 triangle are in the ratio x:x*sqrt(3):2x involves using trigonometry, which is beyond the scope of the GRE, so I'm not going to do it here. If you want a full explanation of why this works, check out the linked video below.
I hope that helps!
ruclips.net/video/SFL4stapeUs/видео.html