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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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GRE triangle in a circle.jpg [ 147.8 KiB | Viewed 617 times ]
A triangle is inscribed in the circle with center O. If the radius is 5, find the area of the shaded region.
Show: ::
\( \approx 13.5\)
A triangle is inscribed in the circle with center O. I
[#permalink]
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
A triangle is inscribed in the circle with center O. I
[#permalink]
05 May 2023, 07:37
05 May 2023, 07:37
The area of triangle seems to be 25. Am I wrong? By 45 45 90 ratios, the side corresponding to 90 is 10 so other two are 10/sqrt 2.
A triangle is inscribed in the circle with center O. I
[#permalink]
05 May 2023, 07:43
05 May 2023, 07:43
Expert Reply
Attachment:
GRE triangle in a circle.jpg [ 78.76 KiB | Viewed 568 times ]
I made a mistake above sorry. I did not double the original value. work and reply here at the same time is not that easy
The area of the triangle is \(\frac{b*h}{2}=\frac{10*5}{2}=25\)
Half area of the circle is 39.5 -area triangle 25=14.5
14.5 is the sum of the two areas
The shaded is half that = 7.25
The triplete is \(x:x:x \sqrt{2}\) but we do not need in this case
_________________
A triangle is inscribed in the circle with center O. I
[#permalink]
05 May 2023, 07:52
05 May 2023, 07:52
Expert Reply
So to recap the process for these questions and similar
Find the area of the circle.
Halving that we do have half of the circle area.
Find the area of the triangle.
Subtract the area of the triangle from the half area of the circle so we have the sum of the two left portions.
The shaded portion is half that.
regards
_________________
Find the area of the circle.
Halving that we do have half of the circle area.
Find the area of the triangle.
Subtract the area of the triangle from the half area of the circle so we have the sum of the two left portions.
The shaded portion is half that.
regards
_________________
gmatclubot
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