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A triangle is inscribed in the circle with center O. I
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Updated on: 05 May 2023, 07:04
Updated on: 05 May 2023, 07:04
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A triangle is inscribed in the circle with center O. If the radius is 5, find the area of the shaded region.
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\( \approx 13.5\)
A triangle is inscribed in the circle with center O. I
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05 May 2023, 07:03
05 May 2023, 07:03
Expert Reply
In this question is just a matter of working outside in.
we know the radius that is 5.
Calculate the area of the circle \(C= \pi r^2=\pi·5^2≈78.53982\) Therefore we assume for easier calculation that the area is 79
The entire area where we do have the triangle is half that so is \(39.5\)
The radius 5 is also the side of the triangle and the height is also 5 because is also the radius 5. So the area of the triangle is \(\frac{b*h}{2}=\frac{5*5}{2}=12.5\)
The two areas outside the triangle BUT inside the half of the circle we are considering is \(39.5\) (area half of the circle) minus the area of the triangle \(12.5=39.5-12.5= 27\)
we need the shaded region which is half the sum of the two areas \(\frac{27}{2}=13.5\)
The answer is \(\approx 13.5\)
we know the radius that is 5.
Calculate the area of the circle \(C= \pi r^2=\pi·5^2≈78.53982\) Therefore we assume for easier calculation that the area is 79
The entire area where we do have the triangle is half that so is \(39.5\)
The radius 5 is also the side of the triangle and the height is also 5 because is also the radius 5. So the area of the triangle is \(\frac{b*h}{2}=\frac{5*5}{2}=12.5\)
The two areas outside the triangle BUT inside the half of the circle we are considering is \(39.5\) (area half of the circle) minus the area of the triangle \(12.5=39.5-12.5= 27\)
we need the shaded region which is half the sum of the two areas \(\frac{27}{2}=13.5\)
The answer is \(\approx 13.5\)
A triangle is inscribed in the circle with center O. I
[#permalink]
05 May 2023, 07:23
05 May 2023, 07:23
Carcass wrote:
In this question is just a matter of working outside in.
we know the radius that is 5.
Calculate the area of the circle \(C= \pi r^2=\pi·5^2≈78.53982\) Therefore we assume for easier calculation that the area is 79
The entire area where we do have the triangle is half that so is \(39.5\)
The radius 5 is also the side of the triangle and the height is also 5 because is also the radius 5. So the area of the triangle is \(\frac{b*h}{2}=\frac{5*5}{2}=12.5\)
The two areas outside the triangle BUT inside the half of the circle we are considering is \(39.5\) (area half of the circle) minus the area of the triangle \(12.5=39.5-12.5= 27\)
we need the shaded region which is half the sum of the two areas \(\frac{27}{2}=13.5\)
The answer is \(\approx 13.5\)
we know the radius that is 5.
Calculate the area of the circle \(C= \pi r^2=\pi·5^2≈78.53982\) Therefore we assume for easier calculation that the area is 79
The entire area where we do have the triangle is half that so is \(39.5\)
The radius 5 is also the side of the triangle and the height is also 5 because is also the radius 5. So the area of the triangle is \(\frac{b*h}{2}=\frac{5*5}{2}=12.5\)
The two areas outside the triangle BUT inside the half of the circle we are considering is \(39.5\) (area half of the circle) minus the area of the triangle \(12.5=39.5-12.5= 27\)
we need the shaded region which is half the sum of the two areas \(\frac{27}{2}=13.5\)
The answer is \(\approx 13.5\)
That's an isosceles right angled triangle. Diameter is 10 and so sides of the triangle are 5 rt 2 not 5
