Question

$$\begin{gathered} \mathrm{In}\mathrm{the}\mathrm{figure},\angle \mathrm{PQR}=60°\mathrm{and}\mathrm{ray}\mathrm{QT}\mathrm{bisects}\angle \mathrm{PQR}.\text{S}\mathrm{eg}\mathrm{BA}\perp \mathrm{ray}\mathrm{QR}. \\ \mathrm{If}\mathrm{BC}=8,\mathrm{find}\mathrm{the}\mathrm{perimeter}\mathrm{of}\square \mathrm{ABCQ}. \end{gathered}$$

Answer 1

In ΔBQC, $\angle$BCQ = 90° and $\angle$BQC = $\frac{1}{2}$ × $\angle$PQR = $\frac{1}{2}$ × 60° = 30°. Then,
$\angle$CBQ = 180° – ($\angle$BCQ + $\angle$BQC) = 180° – (90° + 30°) = 180° – 120° = 60°

Thus, Δ BQC is a 30°-60°-90° triangle.
By the 30°-60°-90° triangle theorem, we get:
BC = $\frac{1}{2}$× BQ $\Rightarrow$ BQ = 2 × BC = 2 × 8 = 16
QC = $\frac{\sqrt{3}}{2}$× BQ = $\frac{\sqrt{3}}{2}$ × 16 = $8\sqrt{3}$

Now, in ΔBQA and ΔBQC,
$\angle$BAQ = $\angle$BCQ [Each is equal to 90°]
$\angle$BQA = $\angle$BQC [Each is half of $\angle$PQR]
BQ = BQ [Common]
Therefore, by the AAS congruency property,
ΔBQA $\cong$ ΔBQC

Then,
AB = BC = 8 and QA = QC = $8\sqrt{3}$
Therefore, the perimeter of quadrilateral ABCQ is
AB + BC + CQ + QA = 8 + 8 + $8\sqrt{3}$ + $8\sqrt{3}$ = 16 +16$\sqrt{3}$ = 16(1 +$\sqrt{3}$ ) units.