Spectre927
New member
- Joined
- Feb 22, 2006
- Messages
- 12
I have a circle inscribed in a triangle. Triangle ABC. Area of triangle is 12in^2 and perimeter is 20 inches. Is that enough info to find the area of the inscribed circle?
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Inscribed Circle's area
- Thread starter Spectre927
- Start date
Hello, Spectre927!
We're expected to know a certain theorem.
Given a triangle with perimeter \(\displaystyle P\) and the radius of the inscribed circle is \(\displaystyle r\)
\(\displaystyle \;\;\)the area of the triangle is: \(\displaystyle \,A\;=\;\frac{1}{2}Pr\;\) **
So we have: \(\displaystyle \,12\;=\;\frac{1}{2}(20)r\;\;\Rightarrow\;\;r = 1.2\)
Therefore, the area of the circle is: \(\displaystyle \,\pi r^2\;=\;\pi(1.2^2) \;=\;1.44\pi\) in<sup>2</sup>
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
**\(\displaystyle \;\;\) Proof
\(\displaystyle O\) is the center of the inscribed circle with radius \(\displaystyle r\).
The sides of the triangle are tangent to the circle,
\(\displaystyle \;\;\)hence, the radii \(\displaystyle r\) are perpendicular to the sides.
We have three smaller triangles: \(\displaystyle \,\Delta AOB,\;\Delta BOC,\;\Delta COA\)
\(\displaystyle \;\;\)Area of \(\displaystyle \Delta AOB\;=\;\frac{1}{2}\cdot c\cdot r\)
\(\displaystyle \;\;\)Area of \(\displaystyle \Delta BOC\;=\;\frac{1}{2}\cdot a\cdot r\)
\(\displaystyle \;\;\)Area of \(\displaystyle \Delta COA\;=\;\frac{1}{2}\cdot b\cdot r\)
The area of \(\displaystyle \Delta ABC\:=\:\frac{1}{2}\cdot c\cdot r\,+\,\frac{1}{2}\cdot a\cdot r\,+\,\frac{1}{2}\cdot b\cdot r\;=\;\frac{1}{2}(a\,+\,b\,+\,c)r\)
Since \(\displaystyle a\,+\,b\,+\,c\;=\;P\), the perimeter,
\(\displaystyle \;\;\) we have: \(\displaystyle \,A\;=\;\frac{1}{2}Pr\;\;\) . . . ta-DAA!
Yes, it is . . .I have a circle inscribed in triangle ABC.
Area of triangle is 12 in<sup>2</sup> and perimeter is 20 inches.
Is that enough info to find the area of the inscribed circle?
We're expected to know a certain theorem.
Given a triangle with perimeter \(\displaystyle P\) and the radius of the inscribed circle is \(\displaystyle r\)
\(\displaystyle \;\;\)the area of the triangle is: \(\displaystyle \,A\;=\;\frac{1}{2}Pr\;\) **
So we have: \(\displaystyle \,12\;=\;\frac{1}{2}(20)r\;\;\Rightarrow\;\;r = 1.2\)
Therefore, the area of the circle is: \(\displaystyle \,\pi r^2\;=\;\pi(1.2^2) \;=\;1.44\pi\) in<sup>2</sup>
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
**\(\displaystyle \;\;\) Proof
Code:
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/ * *O * \
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B-*- - - - - -* * * - - - - - * C
aThe sides of the triangle are tangent to the circle,
\(\displaystyle \;\;\)hence, the radii \(\displaystyle r\) are perpendicular to the sides.
We have three smaller triangles: \(\displaystyle \,\Delta AOB,\;\Delta BOC,\;\Delta COA\)
\(\displaystyle \;\;\)Area of \(\displaystyle \Delta AOB\;=\;\frac{1}{2}\cdot c\cdot r\)
\(\displaystyle \;\;\)Area of \(\displaystyle \Delta BOC\;=\;\frac{1}{2}\cdot a\cdot r\)
\(\displaystyle \;\;\)Area of \(\displaystyle \Delta COA\;=\;\frac{1}{2}\cdot b\cdot r\)
The area of \(\displaystyle \Delta ABC\:=\:\frac{1}{2}\cdot c\cdot r\,+\,\frac{1}{2}\cdot a\cdot r\,+\,\frac{1}{2}\cdot b\cdot r\;=\;\frac{1}{2}(a\,+\,b\,+\,c)r\)
Since \(\displaystyle a\,+\,b\,+\,c\;=\;P\), the perimeter,
\(\displaystyle \;\;\) we have: \(\displaystyle \,A\;=\;\frac{1}{2}Pr\;\;\) . . . ta-DAA!
Spectre927
New member
- Joined
- Feb 22, 2006
- Messages
- 12
Oooook. I see. Thank you. That was for my little sis and I dont remember much geometry. I wasnt sure if she'd witten everything down. Thanks again.