Let ABC be an acute-angled triangle with area | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) We know that the area $R = \frac{1}{2}ab \sin C$,so $2R = ab \sin C$,which implies $4R^2 = a^2b^2 \sin^2 C$. Substituting this into the first term

Vedclass · Accepted Answer

$\frac{a^2+b^2+c^2}{2}$

Vedclass · Answer

(C) We know that the area $R = \frac{1}{2}ab \sin C$,so $2R = ab \sin C$,which implies $4R^2 = a^2b^2 \sin^2 C$. Substituting this into the first term: $\sqrt{a^2b^2 - 4R^2} = \sqrt{a^2b^2 - a^2b^2 \sin^2 C} = \sqrt{a^2b^2(1 - \sin^2 C)} = \sqrt{a^2b^2 \cos^2 C} = ab \cos C$. Similarly,$\sqrt{b^2c^2 - 4R^2} = bc \cos A$ and $\sqrt{c^2a^2 - 4R^2} = ca \cos B$. Using the Law of Cosines,$\cos C = \frac{a^2+b^2-c^2}{2ab}$,so $ab \cos C = \frac{a^2+b^2-c^2}{2}$. Thus,the expression becomes: $\frac{a^2+b^2-c^2}{2} + \frac{b^2+c^2-a^2}{2} + \frac{c^2+a^2-b^2}{2} = \frac{a^2+b^2+c^2}{2}$.

Let $ABC$ 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} = $

Similar Questions

In $\triangle ABC$,if $(r_2-r_1)(r_3-r_1)=2 r_2 r_3$,then $2(r+R)=$

In $\triangle ABC$,if $a=3, b=4, c=6$,then $\frac{\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}}{\cot A+\cot B+\cot C}=$

In any $\triangle ABC$,with usual notations,$c(a \cos B - b \cos A) =$

In $\Delta ABC,$ if $\cot A, \cot B, \cot C$ are in $A.P.,$ then $a^2, b^2, c^2$ are in

The inradius of the triangle whose sides are $3, 5, 6$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module