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The area of the circle above, with center C, is
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12 Oct 2020, 07:33
12 Oct 2020, 07:33
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The area of the circle above, with center C, is \(36\pi\) , and \(x > 90\)
Quantity A |
Quantity B |
The area of the triangle |
18 |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Kudos for the right answer and explanation
Re: The area of the circle above, with center C, is
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Updated on: 15 Oct 2020, 03:42
Updated on: 15 Oct 2020, 03:42
3
The area of any triangle with a x° angle in the domain of 90° < x < 180°, is less than the area of a triangle with x=90°. Having said that, we can calculate the area of a 90 degress triangle:
Area of the circle
\(Á_{circle} = \pi*r = 36\pi\)
\(r = 6\)
but, we know that the area of a triangle is given by:
\(Á_{triangle} = \frac{b*h}{2}\)
and, because we have a isoseles, the area is:
\(Á_{triangle} = \frac{r*r}{2} = \frac{6*6}{2}\)
\(Á_{triangle} = 18\)
but, x is MORE THAN 90°, therefore, we are completely sure that the area of any triangle in that domain must be less than 18.
Option B
Area of the circle
\(Á_{circle} = \pi*r = 36\pi\)
\(r = 6\)
but, we know that the area of a triangle is given by:
\(Á_{triangle} = \frac{b*h}{2}\)
and, because we have a isoseles, the area is:
\(Á_{triangle} = \frac{r*r}{2} = \frac{6*6}{2}\)
\(Á_{triangle} = 18\)
but, x is MORE THAN 90°, therefore, we are completely sure that the area of any triangle in that domain must be less than 18.
Option B
Re: The area of the circle above, with center C, is
[#permalink]
15 Oct 2020, 02:28
15 Oct 2020, 02:28
1
cote wrote:
The area of any triangle with a x° angle in the domain of 90° < x < 180°, is less than the area of a triangle with x=90°. Having said that, we can calculate the area of a 90 degress triangle:
Area of the circle
\(Á_{circle} = \pi*r = 36\pi\)
\(r = 6\)
but, we know that the area of a triangle is given by:
\(Á_{triangle} = \frac{b*h}{2}\)
and, because we have a isoseles, the area is:
\(Á_{triangle} = \frac{r*r}{2} = \frac{6*6}{2}\)
\(Á_{triangle} = 18\)
but, x is LESS THAN 90°, therefore, we are completely sure that the area of any triangle in that domain must be less than 18.
Option B
Area of the circle
\(Á_{circle} = \pi*r = 36\pi\)
\(r = 6\)
but, we know that the area of a triangle is given by:
\(Á_{triangle} = \frac{b*h}{2}\)
and, because we have a isoseles, the area is:
\(Á_{triangle} = \frac{r*r}{2} = \frac{6*6}{2}\)
\(Á_{triangle} = 18\)
but, x is LESS THAN 90°, therefore, we are completely sure that the area of any triangle in that domain must be less than 18.
Option B
it is mentioned in question that
x is > 90.,
therefore ans should be A
