AP EAMCET · Maths · Vector Algebra
If \(A B C D\) is a cyclic quadrilateral with \(R\) as the radius of the circumcircle and \((\mathrm{AB})^2+(C D)^2=4 R^2\) then
- A \(\bar{b} \cdot \bar{c}-\bar{a} \cdot \bar{d}=0\)
- B \(\bar{a} \cdot \bar{c}-\bar{b} \cdot \bar{d}=0\)
- C \(\bar{a} \cdot \bar{b}+\bar{c} \cdot \bar{d}=0\)
- D \(\bar{a} \cdot \bar{c}+\bar{b} \cdot \bar{d}=0\)
Answer & Solution
Correct Answer
(C) \(\bar{a} \cdot \bar{b}+\bar{c} \cdot \bar{d}=0\)
Step-by-step Solution
Detailed explanation
We know that if \(A B C D\) is a cyclic quadrilateral with \(\mathrm{R}\) as the radius of the circumcircle and \[ \mathrm{AB}^2+\mathrm{CD}^2=4 \mathrm{R}^2 \] \(\therefore \mathrm{ABCD}\) is a rhombus.…
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