Question

If in a $\Delta ABC,{{a}^{2}}+{{b}^{2}}+{{c}^{2}}=8{{R}^{2}},$ where R = circumradius, then the triangle is
(a) Equilateral
(b) Isosceles
(c) Right angled
(d) None of these

Answer 1

Hint:For solving this question, it is given that R is a circumradius. We use the concept of sine rule. Using the trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we simplify our equation. On solving further, we get all the angles of the triangles and with the help of angles we easily check whether it is equilateral, isosceles or right-angle triangle.

Complete step-by-step answer:
Given: ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=8{{R}^{2}}$

Using the sine rule,
$\dfrac{a}{\operatorname{Sin}A}=\dfrac{b}{\operatorname{Sin}B}=\dfrac{c}{\operatorname{Sin}C}=2R$
By this the value of a = 2RsinA, b = 2RsinB and c = 2RsinC.
Putting the value of a, b and c in the equation ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=8{{R}^{2}}$.
$\Rightarrow {{\left( 2R\sin A \right)}^{2}}+{{\left( 2R\sin B \right)}^{2}}+{{\left( 2R\sin C \right)}^{2}}=8{{R}^{2}}$
Tanking $4{{R}^{2}}$ common from the left-hand side, we have:
$\Rightarrow 4{{R}^{2}}\left[ {{\left( \sin A \right)}^{2}}+{{\left( \sin B \right)}^{2}}+{{\left( \sin C \right)}^{2}} \right]=8{{R}^{2}}$
We can write $8{{R}^{2}}\text{ as }4{{R}^{2}}\times 2$,
$\Rightarrow 4{{R}^{2}}\left[ {{\left( \sin A \right)}^{2}}+{{\left( \sin B \right)}^{2}}+{{\left( \sin C \right)}^{2}} \right]=4{{R}^{2}}\times 2$
$4{{R}^{2}}$ cancel out from both sides, we have:
$\Rightarrow {{\left( \sin A \right)}^{2}}+{{\left( \sin B \right)}^{2}}+{{\left( \sin C \right)}^{2}}=2$
We know the identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, from this we get the value of ${{\cos }^{2}}\theta \text{=1-si}{{\text{n}}^{2}}\theta $. And, we know that all the angles in the triangles are $\pi (i.e{{.180}^{\circ }}).$ Using this,
\[\begin{align}
  & A+B+C=\pi \\
 & \therefore A+B=\pi -C \\
 & \Rightarrow {{\sin }^{2}}\left( \pi -\left( A+B \right) \right)=2-{{\sin }^{2}}B-{{\sin }^{2}}A \\
 & \Rightarrow {{\sin }^{2}}\left( A+B \right)=\left( 1-{{\sin }^{2}}A \right)+\left( 1-{{\sin }^{2}}B \right) \\
 & \Rightarrow {{\left( \sin A\cos B+\cos A\sin B \right)}^{2}}={{\cos }^{2}}A+{{\cos }^{2}}B \\
\end{align}\]
Now, by using the expansion ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$, we get
\[\begin{align}
  & \Rightarrow \left( {{\sin }^{2}}A \right)\left( {{\cos }^{2}}B \right)+\left( {{\cos }^{2}}A \right)\left( {{\sin }^{2}}B \right)+2\left( \sin A \right)\left( \cos B \right)\left( \cos A \right)\left( \sin B \right)={{\cos }^{2}}A+{{\cos }^{2}}B \\
 & \Rightarrow \left( 1-{{\cos }^{2}}A \right)\left( {{\cos }^{2}}B \right)+\left( 1-{{\cos }^{2}}B \right)\left( {{\cos }^{2}}A \right)={{\cos }^{2}}A+{{\cos }^{2}}B-2\left( \sin A \right)\left( \cos B \right)\left( \cos A \right)\left( \sin B \right) \\
 & \Rightarrow L.H.S={{\cos }^{2}}B-\left( {{\cos }^{2}}A \right)\left( {{\cos }^{2}}B \right)+{{\cos }^{2}}A-\left( {{\cos }^{2}}B \right)\left( {{\cos }^{2}}A \right) \\
 & \Rightarrow R.H.S={{\cos }^{2}}A+{{\cos }^{2}}B-2\left( \sin A \right)\left( \cos B \right)\left( \cos A \right)\left( \sin B \right) \\
\end{align}\]
Now, cancelling the common terms from both sides and takin the common factors afterwards, we get
\[\begin{align}
  & \Rightarrow -2\left( {{\cos }^{2}}A \right)\left( {{\cos }^{2}}B \right)=-2\left( \sin A \right)\left( \cos B \right)\left( \cos A \right)\left( \sin B \right) \\
 & \Rightarrow \left( -2\cos A\cos B \right)\left( \cos A \right)\left( \cos B \right)=-2\left( \sin A \right)\left( \cos B \right)\left( \cos A \right)\left( \sin B \right) \\
 & \Rightarrow \left( \cos A \right)\left( \cos B \right)=\left( \sin A \right)\left( \sin B \right) \\
 & \Rightarrow \cos A\cos B-\sin A\sin B=0 \\
\end{align}\]
Putting the value of (cos (A + B) = cosAcosB – sinAsinB) in the above expression, we have:
$\begin{align}
  & \Rightarrow \cos \left( A+B \right)=0 \\
 & \because \cos \dfrac{\pi }{2}=0 \\
 & \therefore \left( A+B \right)=\dfrac{\pi }{2} \\
 & \text{Also, }A+B+C=\pi \\
 & \therefore C=\pi -\left( A+B \right) \\
 & C=\pi -\dfrac{\pi }{2} \\
 & C=\dfrac{\pi }{2} \\
\end{align}$
So, the triangle is a right-angle triangle having a right angle at C.
Therefore, option (c) is correct.

Note: The circumradius of a triangle is a radius of the circle inside which the triangle can be inscribed.Students must be careful while converting the angles by using complementary and supplementary angle formulas. The conversion should be in such a way to cancel common terms from both sides.