MHT CET · Maths · Properties of Triangles
In \(\triangle A B C\), with usual notations, \(a \cos \mathrm{~B}=\mathrm{b} \cos \mathrm{A}, a \cos \mathrm{C} \neq \mathrm{c} \cos \mathrm{A}\) then \(\mathrm{A}(\triangle \mathrm{ABC})\) _____ sq. units.
- A \(\frac{c}{2} \sqrt{4 a^2-b^2}\)
- B \(\frac{c}{4} \sqrt{4 a^2-c^2}\)
- C \(\frac{b}{2} \sqrt{4 b^2-c^2}\)
- D \(\frac{b}{4} \sqrt{4 b^2-c^2}\)
Answer & Solution
Correct Answer
(B) \(\frac{c}{4} \sqrt{4 a^2-c^2}\)
Step-by-step Solution
Detailed explanation
\(a \frac{a^2+c^2-b^2}{2ac} = b \frac{b^2+c^2-a^2}{2bc} \implies a^2+c^2-b^2 = b^2+c^2-a^2 \implies a=b\) \(\mathrm{A}(\triangle \mathrm{ABC}) = \frac{1}{2} c \cdot h_c\)
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