Question

In \(\triangle A B C\), with usual notations, \(a \cos B=b \cos A, a \cos C \neq c \cos A\) then \(\mathrm{A}(\triangle \mathrm{ABC})\) \(\qquad\) sq. units.

• A. \(\frac{c}{2}\sqrt{4a^{2}-b^{2}}\)
• B. \(\frac{c}{4}\sqrt{4a^{2}-c^{2}}\)
• C. \(\frac{b}{2}\sqrt{4b^{2}-c^{2}}\)
• D. \(\frac{b}{4}\sqrt{4b^{2}-c^{2}}\)

Hint:

The condition a cos B = b cos A forces a = b, so the triangle is isosceles; then find its area using the base and height.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
\(a\cos B=b\cos A\) with the sine rule gives \(\sin A\cos B=\sin B\cos A\), so \(\sin(A-B)=0\) and \(A=B\). Hence \(a=b\).

Step 2: Use the Second Condition
If also \(a=c\), the triangle is equilateral and \(a\cos C=c\cos A\). The condition \(a\cos C \ne c\cos A\) rules this out, so the triangle is isosceles with equal sides \(a\) and base \(c\).

Step 3: Height
The height on base \(c\) is \(\sqrt{a^{2}-\dfrac{c^{2}}{4}}=\dfrac{1}{2}\sqrt{4a^{2}-c^{2}}\).

Step 4: Area
Area \(=\dfrac{1}{2}\cdot c\cdot\dfrac{1}{2}\sqrt{4a^{2}-c^{2}}=\dfrac{c}{4}\sqrt{4a^{2}-c^{2}}\), option (B).

Final Answer:
The area is \(\dfrac{c}{4}\sqrt{4a^{2}-c^{2}}\) square units. \[ \boxed{\text{(B) }\dfrac{c}{4}\sqrt{4a^{2}-c^{2}}} \]