Question

In the given figure, AB || DC and ∠ADC = 90º. If AD = 3 cm and AB = 4 cm, then the length of AC is

• A. $2\sqrt{13+6\sqrt{3}}cm$
• B. $4(13+6\sqrt{3})cm$
• C. $4\sqrt{13+6\sqrt{3}}cm$
• D. $2(13+6\sqrt{3})cm$

Answer 1

The correct option is A $2\sqrt{13+6\sqrt{3}}cm$

Given that AB ║ DC and ∠ADC = 90º
Construct BE ║ AD.
∴ BE = AD = 3 cm and DE = AB = 4 cm
∴ In ΔBEC,
$\tan {30}^{\circ }=\frac{BE}{EC}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{3}{EC}$
$\Rightarrow EC=3\sqrt{3}cm$
In ΔADC, by Pythagoras theorem,
$A{C}^2=A{D}^2+C{D}^2$
$\Rightarrow A{C}^2=(3{)}^2+(4+3\sqrt{3}{)}^2$
$(\because CD=DE+EC)$
$\Rightarrow A{C}^2=9+16+27+24\sqrt{3}$
$\Rightarrow A{C}^2=52+24\sqrt{3}$
$\Rightarrow A{C}^2=4(13+6\sqrt{3})$
$\Rightarrow AC=2\sqrt{13+6\sqrt{3}}cm$
Hence, the correct answer is option (a).