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 a triangle $ABC$,if $\tan \frac{A}{2} : \tan \frac{B}{2} : \tan \frac{C}{2} = 15 : 10 : 6$,then $\frac{a}{b-c} =$

With usual notations in $\Delta ABC$,$a=3$,$c=2$ and $\sin C=\frac{2}{3}$,then $\angle A=$

With usual notations,in $\triangle ABC$,the lengths of two sides are $10 \text{ cm}$ and $9 \text{ cm}$ respectively. If angles $A, B, C$ are in $A$.$P$.,then the perimeter of $\triangle ABC$ is:

In $\triangle ABC$,if $a=5$ and $\tan \frac{A-B}{2}=\frac{1}{4} \tan \frac{A+B}{2}$,then $\sqrt{a^2-b^2}=$

In $\triangle ABC$,if $8R^2 = a^2 + b^2 + c^2$,then the triangle is a/an

Vedclass Test Series

Exam Paper Generator

Online Exam Module