Question

Match $\text{List I}$ with $\text{List II}$ and select the correct answer using the code given below the lists :

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& \text{In a}△ABC,\text{if}{a}^2+b^2+c^2=ab+bc+ca,\text{then}& \text{(P)}& △ABC\text{is an equilateral triangle}\\ \text{(B)}& \text{In a}△ABC,\text{if}{a}^2+2b^2+c^2=2bc+2ab,\text{then}& \text{(Q)}& △ABC\text{is a right angled triangle}\\ \text{(C)}& \text{In a}△ABC,\text{if}{a}^2+b^2+c^2=\sqrt{2}a(b+c),\text{then}& \text{(R)}& △ABC\text{is a scalene triangle}\\ \text{(D)}& \text{In a}△ABC,\text{if}{a}^2+b^2+c^2=ca+\sqrt{3}ab,\text{then}& \text{(S)}& A={90}^{\circ },B={45}^{\circ },C={45}^{\circ }\end{array}$

Which of the following is the only CORRECT combination?

• A. $\text{(C)}\to \text{(S)},\text{(D)}\to \text{(R)}$
• B. $\text{(C)}\to \text{(Q)},\text{(D)}\to \text{(P)}$
• C. $\text{(C)}\to \text{(Q)},\text{(D)}\to \text{(S)}$
• D. $\text{(C)}\to \text{(P)},\text{(D)}\to \text{(R)}$

Answer 1

The correct option is A $\text{(C)}\to \text{(S)},\text{(D)}\to \text{(R)}$

$\left(C\right)$
$a^2+b^2+c^2-\sqrt{2}ab-\sqrt{2}ac=0$
$\Rightarrow 2a^2+2b^2+2c^2-2\sqrt{2}ab-2\sqrt{2}ac=0$
$\Rightarrow (a^2-2\sqrt{2}ab+2b^2)+(a^2-2\sqrt{2ac}+2c^2)=0$
$\Rightarrow (a-\sqrt{2}b{)}^2+(a-\sqrt{2}c{)}^2=0$
$\Rightarrow a=\sqrt{2}b=\sqrt{2}c$
Now, $\cos A=\frac{b^2+c^2-a^2}{2bc}=0$
$\Rightarrow A={90}^{\circ }$
and $B=C\Rightarrow B=C={45}^{\circ }$

$\left(D\right)$
$a^2+b^2+c^2-ca-\sqrt{3}ab=0$
$\Rightarrow 4a^2+4b^2+4c^2-4ca-4\sqrt{3}ab=0$
$\Rightarrow (3a^2-4\sqrt{3}ab+4b^2)+(a^2-4ca+4c^2)=0$
$\Rightarrow (\sqrt{3}a-2b{)}^2+(a-2c{)}^2=0$
$\Rightarrow a=\frac{2b}{\sqrt{3}}=2c$
$\Rightarrow b=\frac{\sqrt{3}}{2}a$ and $c=\frac{a}{2}$
Now, $\cos A=\frac{b^2+c^2-a^2}{2bc}=0$
$\Rightarrow A={90}^{\circ }$
Then $C={30}^{\circ }$ and $B={60}^{\circ }$