Question

$$\begin{gathered} \mathrm{In}\mathrm{the}\mathrm{figure},\square \mathrm{PQRV}\mathrm{is}\mathrm{a}\mathrm{trapezium}\mathrm{in}\mathrm{which}\mathrm{seg}\mathrm{PQ}\parallel \mathrm{seg}\mathrm{VR}. \\ \mathrm{SR}=4\mathrm{and}\mathrm{PQ}=6.\mathrm{Find}\mathrm{VR}. \end{gathered}$$

Answer 1

From the figure,
VR = VT + TS + SR
$\Rightarrow$ VR = VT + 6 + 4 [TS = PQ = 6 and SR = 4]
$\Rightarrow$ VR = VT + 10 …(1)
Now, we need to find VT.

In Δ SQR, $\angle$QSR = 90° and $\angle$SQR = 45°. Then,
$\angle$QRS = 180° – ( $\angle$QSR + $\angle$SQR) = 180° – (90° + 45°) = 180° – 135° = 45°
Since $\angle$SQR = $\angle$QRS = 45°, QS = SR = 4
Quadrilateral PQRV is a trapezium, PT and QS are the heights of the trapezium. Thus, PT = QS = 4.

Now, in ΔVPT, $\angle$VTP = 90° and $\angle$VPT = 60°. Then,
$\angle$PVT = 180° – ( $\angle$VTP + $\angle$VPT) = 180° – (90° + 60°) = 180° – 150° = 30°
Thus, Δ VPT is a 30°-60°-90° triangle.
By the 30°-60°-90° triangle theorem, we get:
PT = $\frac{1}{2}$ × VP
$\Rightarrow$ VP = 2 × PT = 2 × 4 = 8
And
VT = $\frac{\sqrt{3}}{2}$ × VP = $\frac{\sqrt{3}}{2}$ × 8 = $4\sqrt{3}$

Now, on substituting VT = $4\sqrt{3}$ in equation (1), we get:
VR = VT + 10 = ($4\sqrt{3}$ + 10) units.