Question

In a $\Delta ABC,ifAB=BC=CA=2a$ and $AD\perp BCthen$

• A. i) $AD=\sqrt[a]{a2}$
  (ii) area $\left(\Delta ABC\right)=\sqrt{5}{a}^2$
• B. i) $AD=\sqrt[a]{a2}$
  (ii) area $\left(\Delta ABC\right)=\sqrt{3}{a}^3$
• C. i) $AD=\sqrt[a]{a3}$
  (ii) area $\left(\Delta ABC\right)=\sqrt{3}{a}^2$ is proved
• D. i) $AD=\sqrt[a]{a5}$
  (ii) area $\left(\Delta ABC\right)=\sqrt{5}{a}^2$

Answer 1

The correct option is C i) $AD=\sqrt[a]{a3}$

(ii) area $\left(\Delta ABC\right)=\sqrt{3}{a}^2$ is proved
(i) Here, $AD\perp BC$.
Clearly, $\Delta ABC$ is an equilateral triangle.
Thus, in $\Delta ABD$ and $\Delta ACD$
$AD=AD$ [Common]
$\angle ADB=\angle ADC$ [${90}^o$ each]
and $AB=AC$
$\therefore$ By RHS congruency condition
$\Delta ABD\cong \Delta ACD$
$\Rightarrow BD=DC=a$
Now, $\Delta ABD$ is a right angled triangle
$\therefore AD=\sqrt{{AB}^2-{AD}^2}$ [Using Pythagoras Theorem]
$AD=\sqrt{{4a}^2-a^2}=\sqrt{3}aor\sqrt[a]{a3}$
(ii) Area$\left(\Delta ABC\right)=\frac{1}{2}\times BC\times AD$
$=\frac{1}{2}\times 2a\times \sqrt[a]{a3}$
$=a^2\sqrt{3}$.