Question

In a right angled triangle $ABC$, $AB=$ $10\sqrt{3}$ cm and $BC=20$ cm, $\angle A={90}^{\circ }$. An equilateral triangle $ABD$ is constructed with base $AB$ and with vertex $D$ at a maximum possible distance from $C$. Find the length of $CD$

• A. $10\sqrt{7}$ cm
• B. $10\sqrt{11}$ cm
• C. $10\sqrt{12}$ cm
• D. $10\sqrt{14}$ cm

Answer 1

Answer: (A)

$10\sqrt{7}$ cm

Two equilateral triangles can be termed with the base AB.
But vertex D should be at maximum distance from C.
Therefore, triangle ABD will be preferred.
Now,$\Delta ABD$ is equilateral triangle.
$\therefore AB=BD=AD=10\sqrt{3}$cm
In right $\Delta ABC$
${\left(BC\right)}^2={\left(AB\right)}^2+{\left(AC\right)}^2$
${\left(20\right)}^2={\left(10\sqrt{3}\right)}^2+{\left(AC\right)}^2$
${\left(AC\right)}^2=400-300$
$AC=10$
In $\Delta ACD,\angle CAD={90}^0+{60}^0={150}^0$

$\cos \frac{5\pi }{6}=\frac{{10\sqrt{3}}^2+{10}^2-{CD}^2}{2\left(10\sqrt{3}\right)\left(10\right)}$
$-\frac{\sqrt{3}}{2}=\frac{300+100-{\left(CD\right)}^2}{200\sqrt{3}}$
$-300=400-{\left(CD\right)}^2$
${\left(CD\right)}^2=700$
$CD=10\sqrt{7}$cm