Question

If $D$ is the midpoint of $BC$ of a right angled triangle $ABC$,$(\angle BAC={90}^{\circ })$ such that triangle $ADC$ is an equilateral $△$, then $a^2:b^2:c^2$ is

• A. $1:3:4$
• B. $3:1:4$
• C. $4:3:1$
• D. $4:1:3$

Answer 1

The correct option is D $4:1:3$

As shown in figure, $b=\frac{a}{2}$ as $\Delta ADC$ is equilateral
$sinC=sin{60}^{\circ }=\frac{\sqrt{3}}{2}=\frac{c}{a}$
$\therefore c=a\frac{\sqrt{3}}{2}$
Thus $a^2:b^2:c^2=a^2:\frac{a^2}{4}:\frac{3a^2}{4}=4:1:3$
Hence D is correct