Question

In the figure, $\angle AQR={60}^o$ and ray $QT$ bisects $\angle AQR$. Seg $BA$ $\perp$ ray $QA$ and seg $BC$ $\perp$ ray $QR$. If $BC=8$, find the perimeter of $\square$ $ABCQ$.

• A. $8(1+\sqrt{3})$ cm
• B. $16(1+\sqrt{3})$ cm
• C. $12(1+\sqrt{3})$ cm
• D. $9(1+\sqrt{3})$ cm

Answer 1

The correct option is B $16(1+\sqrt{3})$ cm

A point which lies on the bisector of an angle is equidistant from its sides.

$\therefore BC=AB=8$ cm.

In $△BQC$,

$\angle C={90}^o$, $\angle BQC={30}^o$

$\therefore \angle QBC={60}^o$

$\therefore △BQC$ is a ${30}^o-{60}^o-{90}^o$ triangle

$\therefore BC=\frac{1}{2}BQ$ and $QC=\frac{\sqrt{3}}{2}BQ$

$\therefore BQ=2\times BC$ and $QC=\frac{\sqrt{3}}{2}\times 2BC$

$\therefore BQ=16$ units and $QC=8\sqrt{3}$ units

$\therefore QC=AB=8\sqrt{3}$ units

Perimeter $\square ABCQ=AB+BC+QC+AQ$

$=8\sqrt{3}+8+8\sqrt{3}+8$

$=16+16\sqrt{3}$

$=16(1+\sqrt{3})$ cm

So, option B is correct.