1) What is the area of the square circumscribed by the circle with a radius of 10*sqrt(2) units?

2) If the smaller angles in an obtuse triangle are x and 2x, what is a possible value for x?

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- Mar 30th 2008, 03:08 AM #1

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- Mar 30th 2008, 03:30 AM #2

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1)

Well since the square contains the circle of radius 10 sqrt(2) hence the length of one side of the square is 2 times the radius of the circle (diameter) which is 20sqrt(2), from this you can find the area of the square:

Area of square = (length of one side)^2

= [20 sqrt(2)]^2

= 400(2)

= 800

2)

for an obtuse triangle, the obtuse angle must be larger than 90 degree and smaller than 180 degree.

From an internal angle of a triangle which sums up to 180 degree,

90 < obtuse angle < 180

-90 < -obtuse angle < -180

180 < -obtuse angle <90

180 - 180 <180 - obtuse angle <180 - 90

0 < 180 - obtuse angle < 90

and we know that:

obtuse angle + x + 2x = 180

3x = 180 - obtuse angle

therefore;

0 < 3x < 90

where 3x can take any value in the range such as 80

which gives us x=26.6667 degree ,2x=53.3333 degree and the obtuse angle = 100