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DS area of square to area of circle
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Being defeated is often only a temporary condition. Giving up is what makes it permanent.
Hi!
IMO D
The easiest way to solve this problem is to remember formulas.
(A) c = π*d = π*XY, π = pi = constant and answer is incorrect
(B) look at diagonals in a square, they have the same width
If the diagonal is given we can easily calculate the length of a side( Pythagorean theorem)
a^2 = (0.5d)^2 + (0.5d)^2
a = 0.5 root(2)d = 0.5root(2)XY
A = a^2
A = 0.5(XY)^2 answer is incorrect
(C) p = 2a + b, given the fact that the triangle is a isosceles right itriangle with a leg XY we can calculate the lenght of hypotenuse( Pythagorean theorem)
a = XY
b^2 = a^2 + a^2 = 2(XY)^2
b = 0.5root(2)XY
p = XY( 2 + root(2)) answer is incorrect
(D) we can imagine different circles with chords that are the same length
it is correct answer
(E) A = (a^2)root(3)/4 = (XY^2)root(3)/4 incorrect
hope it helps!
Best
IMO D
The easiest way to solve this problem is to remember formulas.
(A) c = π*d = π*XY, π = pi = constant and answer is incorrect
(B) look at diagonals in a square, they have the same width
If the diagonal is given we can easily calculate the length of a side( Pythagorean theorem)
a^2 = (0.5d)^2 + (0.5d)^2
a = 0.5 root(2)d = 0.5root(2)XY
A = a^2
A = 0.5(XY)^2 answer is incorrect
(C) p = 2a + b, given the fact that the triangle is a isosceles right itriangle with a leg XY we can calculate the lenght of hypotenuse( Pythagorean theorem)
a = XY
b^2 = a^2 + a^2 = 2(XY)^2
b = 0.5root(2)XY
p = XY( 2 + root(2)) answer is incorrect
(D) we can imagine different circles with chords that are the same length
it is correct answer
(E) A = (a^2)root(3)/4 = (XY^2)root(3)/4 incorrect
hope it helps!
Best
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if you have any questions, send me a private message!
should you find this post useful, please click on "thanks" button
