A circle of radius $R$ is circumscribed about a right triangle $ABC$. If $r$ is the radius of circle inscribed in triangle, then what is the area of the triangle (in terms of $r$ and $R$)?
I have no idea how to start.
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Quanto
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Raghav Dixit
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Related: https://math.stackexchange.com/questions/1582061/finding-the-radius-of-a-circle-inside-of-a-triangle – Matti P. Nov 01 '19 at 13:08
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I suggest you look for similar problems, I'm sure you'll find ... – Matti P. Nov 01 '19 at 13:08
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@Matti P. We have to find area not Radius. – Raghav Dixit Nov 01 '19 at 13:10
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2Welcome to Math.SE! The community here prefers/expects questions to include something of what the asker knows. (What have you tried? Where did you get stuck? etc) This information helps answers tailor their responses to best serve you, without duplicating your effort or having to guess what techniques you've used. (It also helps convince people that you aren't simply trying to get them to do your homework for you. Note that "I have no idea" claims tend to be interpreted this way.) Please edit your question to add details. – Blue Nov 01 '19 at 13:18
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As is the case with a lot or problems like this, you can start by drawing a diagram and trying to identify every piece of information you can deduce from the diagram. What happened then? (You cannot post your diagram until you have more reputation, but you can link to it.) – David K Nov 01 '19 at 18:30
3 Answers
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Let $a$ and $b$ be the sides. The area is,
$$A= \frac12 r(a+b+2R)$$
Substitute $$(a+b)^2=a^2+b^2+ 2ab=4R^2+4A$$
to obtain,
$$\frac{A^2}{r^2}-(\frac {2R}{r}+1)A=0$$
Then, solve for the area,
$$A=r\left(2R+r\right)$$
Quanto
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How did you get the step $\frac{(ab)^2}{r^2}-\left(\frac{4R}{r}+2\right)ab=0$ from equation $(1)$? – vidyarthi Nov 01 '19 at 14:02
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My question is still not clear after your editing, could you elaborate a bit? – vidyarthi Nov 01 '19 at 14:12
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Now perfectly clear. The fact that you used $A$ and $\frac{ab}{2}$ interchangibly confused me – vidyarthi Nov 01 '19 at 14:23
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$A$ and $\frac{ab}{2}$ are both the area of the triangle. This prof goes a long the lines I hinted in my answer. – Alberto Saracco Nov 01 '19 at 21:23
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The area of the triangle is then $r^2 + r(a-r) + r(b-r) = r^2 + r((a - r) + (b - r))$ where we note that $(a-r) + (b-r) = c = 2R$. Therefore, the area is $r^2 + 2rR$.
Michael Biro
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Hint. Let the sides of the triangle be $a,b,c$. You know:
$c^2=a^2+b^2$ (Pythagoras)
$c=2R$ (circumscribed circle)
Double of area $2A=ab=r(a+b+c)$ (inscribed circle)
Can you proceed from here? Try to determine $a,b$.
Alberto Saracco
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