Hi
Please help solve the attached.
Thanks,
J
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Geometry
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Remember this:
"To maximize the area of a triangle given a base, make it a 90 degree isoceles, 90-45-45" -MGMAT book
Therefore with radius 1 (base =1), the height equals 1 and thus the area equalls 1/2.
"To maximize the area of a triangle given a base, make it a 90 degree isoceles, 90-45-45" -MGMAT book
Therefore with radius 1 (base =1), the height equals 1 and thus the area equalls 1/2.
HI
In this problem it is given that one vertex of the triangle is at the center of a unit circle.
From this we conclude that if you join this vertex to other two vertex somewhere at the circle, then length of these two sides of triangle will be equal to radius of circle.
Let these sides be a, b , therefore a = b =radius of circle = 1
Let angle between these two sides is C.
Then area of the triangle =1/2*a*b* sin C
=1/2*1*1* sin C = 1/2* sin C
here we see area of the triangle solely depends on angle C.
Therefor, area of the triangle will be maximum where sin C is maximum.
Maximu value that sin function takes is 1, cosequently (sin C)max =1.
Hence, required area of the triangle = 1/2*1 =1/2.
Therefore, option B is correct.
In this problem it is given that one vertex of the triangle is at the center of a unit circle.
From this we conclude that if you join this vertex to other two vertex somewhere at the circle, then length of these two sides of triangle will be equal to radius of circle.
Let these sides be a, b , therefore a = b =radius of circle = 1
Let angle between these two sides is C.
Then area of the triangle =1/2*a*b* sin C
=1/2*1*1* sin C = 1/2* sin C
here we see area of the triangle solely depends on angle C.
Therefor, area of the triangle will be maximum where sin C is maximum.
Maximu value that sin function takes is 1, cosequently (sin C)max =1.
Hence, required area of the triangle = 1/2*1 =1/2.
Therefore, option B is correct.