Let $A B C$ be an acute-angled triangle with area $R$. Then, $$ \sqrt{a^2 b^2-4 R^2}+\sqrt{b^2 c^2-4…
Let $A B C$ be an acute-angled triangle with area $R$. Then,
$$
\sqrt{a^2 b^2-4 R^2}+\sqrt{b^2 c^2-4 R^2}+\sqrt{c^2 a^2-4 R^2}=
$$
- $a+b+c$
- $a^2+b^2+c^2$
- $\frac{a^2+b^2+c^2}{2}$
- $2\left(a^2+b^2+c^2\right)$
Solution
$
\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}
$
Asked in: AP EAMCET 2022 (07 Jul Shift 2)
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