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$A B C$ is an equilateral triangle inscribed in a circle of centre $O$ and radius $5 \mathrm{~cm}$. Let the diameter through $C$ meet the circle again at $\mathrm{D}$.
Assertion (A): $A D \cdot B D < O B \cdot O C$
Reason (R): $2\left(A D^{2}+B D^{2}\right)=C D^{2}=100$ sq cm
Options:
Assertion (A): $A D \cdot B D < O B \cdot O C$
Reason (R): $2\left(A D^{2}+B D^{2}\right)=C D^{2}=100$ sq cm
Solution:
1628 Upvotes
Verified Answer
The correct answer is:
A is false but $\mathrm{R}$ is true
(A) : Consider $A D \cdot B D < O B \cdot O C$
Now, $\mathrm{OA}=5 \mathrm{~cm}$ (Radius), $\mathrm{OB}=5 \mathrm{~cm}$ (Radius) and $\mathrm{OC}=5 \mathrm{~cm}($ Radius $)$
Since, $\Delta O A D$ and $\triangle O B D$
are congruent by SAS therefore
$$
A D=O A=5 \mathrm{~cm}
$$
and $B D=O B=5 \mathrm{~cm}$
Thus, $A D \cdot B D=5 \times 5=25$
and $\quad O B \cdot O C=5 \times 5=25$
Thus, we have
$A D \cdot B D=25=O B \cdot O C$
$$
\begin{aligned}
\operatorname{Now}(\mathrm{R}): & 2\left(A D^{2}+B D^{2}\right) \\
&=2[25+25]=100 \\
\text { and } \quad & C D^{2}=(10)^{2}=100
\end{aligned}
$$
Thus, $2\left(A D^{2}+B D^{2}\right)=C D^{2}=100$ sq. $\mathrm{cm}$.
Hence, (A) is false and (R) is true.
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