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Question:
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Let $A B C$ be an acute-angled triangle with area $R$. Then,
$$
\sqrt{a^2 b^2-4 R^2}+\sqrt{b^2 c^2-4 R^2}+\sqrt{c^2 a^2-4 R^2}=
$$
Options:
$$
\sqrt{a^2 b^2-4 R^2}+\sqrt{b^2 c^2-4 R^2}+\sqrt{c^2 a^2-4 R^2}=
$$
Solution:
2458 Upvotes
Verified Answer
The correct answer is:
$\frac{a^2+b^2+c^2}{2}$
$$
\begin{aligned}
& \sqrt{a^2 b^2-4 R^2}=\sqrt{a^2 b^2-(a b \sin c)^2} . \\
& =a b \cos C
\end{aligned}
$$
Similarly, $\sqrt{b^2 c^2-4 R^2}=b c \cos A$
$$
\sqrt{c^2 a^2-4 R^2}=c a \cos B
$$
$\therefore$ Given, expression
$$
\begin{aligned}
& =a b \cos C+b c \cos A+c a \cos B \\
& =\frac{a^2+b^2-c^2}{2}+\frac{b^2+c^2-a^2}{2}+\frac{c^2+a^2-b^2}{2} \\
& =\frac{a^2+b^2+c^2}{2}
\end{aligned}
$$
\begin{aligned}
& \sqrt{a^2 b^2-4 R^2}=\sqrt{a^2 b^2-(a b \sin c)^2} . \\
& =a b \cos C
\end{aligned}
$$
Similarly, $\sqrt{b^2 c^2-4 R^2}=b c \cos A$
$$
\sqrt{c^2 a^2-4 R^2}=c a \cos B
$$
$\therefore$ Given, expression
$$
\begin{aligned}
& =a b \cos C+b c \cos A+c a \cos B \\
& =\frac{a^2+b^2-c^2}{2}+\frac{b^2+c^2-a^2}{2}+\frac{c^2+a^2-b^2}{2} \\
& =\frac{a^2+b^2+c^2}{2}
\end{aligned}
$$
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