Question

In the figure $\angle PQR={60}^0$ and ray $QT$ bisects $\angle PQR$. $SegBA\parallel rayQPandsegBC\parallel rayQR$. If $BC=8$, find the perimeter of $\square ABCQ$

• A. $16(1+\sqrt{7})$
• B. $16(1+\sqrt{5})$
• C. $16(1+\sqrt{2})$
• D. $16(1+\sqrt{3})$

Answer 1

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

Given: PQ bisects $\angle PQR$, $AB\perp BQ$, $AC\perp QR$
In $△BCQ$
$\frac{BC}{QC}=tan\left(\angle BQC\right)$
$\frac{BC}{QC}=tan30$ (BQ bisects $\angle BQC$)
$QC=\frac{BC}{tan30}$
$QC=8\sqrt{3}$
Now,In $△QAB$ and $△QBC$,
$BQ=BQ$ (Common)
$\angle BQA=\angle BQC$ (BQ bisects $\angle Q$)
$\angle BAQ=\angle BCQ$ (Each ${90}^{\circ }$)
Thus, $△QAB\cong △QCB$ (ASA rule)
hence, $AB=BC=8$ cm (Corresponding sides)
$AQ=CQ=8\sqrt{3}$ cm (Corresponding sides)
Thus, Perimeter of $\square ABCQ$ = $AB+BC+CQ+AQ$
Perimeter of $\square ABCQ$ = $8+8+8\sqrt{3}+8\sqrt{3}$
Perimeter of $\square ABCQ$ = $16(1+\sqrt{3})$ cm