Question

lf one side of a triangle is $\frac{1}{\sqrt{2}}$ that of an other and the angles opposite to these sides are equal, then the triangle is

• A. isosceles
• B. right angled
• C. equilateral
• D. right angled isosceles

Answer 1

The correct option is D right angled isosceles

Let the sides be $a,b,c$ and angles be $A,B,C$
Now $b=\frac{1}{\sqrt{2}}c$ ...(i) and
$A=B$
Hence by sine Rule
$a=b$
Now assuming it to be a right angled triangle with $c^{\prime }$ as the hypotenuse we apply Pythagorean theorem,
$\sqrt{a^2+b^2}$
$=\sqrt{2b^2}$
$=b\sqrt{2}$
$=c^{\prime }$
$\therefore \frac{c^{\prime }}{\sqrt{2}}=b$
$\therefore \frac{c^{\prime }}{\sqrt{2}}=\frac{c}{\sqrt{2}}$ from ...(i)
Hence, $c^{\prime }=c$
Hence, the above triangle is an isosceles right angled triangle right angled at C.