Question

Construct a quadrilateral PQRS with PQ = 4 cm, QR = 6 cm, PR = 7 cm, PS = 5 cm and $\angle$ PQS = ${40}^o$ and find its area.

Answer 1

Given:
PQ = 4 cm, QR = 6 cm, PR= 7 cm, PS= 5 cm and $\angle$ PQS = 40.

Steps for construction
Step 1 : Draw a rough diagram and mark the given measurements.
Step 2 : Draw a line segment PQ = 4 cm.
Step 3 : With P and Q as centres draw arcs of radii 7 cm and 6 cm respectively and let them cut at R.
Step 4 : Join $\overline{PR}$ and $\overline{QR}$ .
Step 5 : At Q on $\overline{PQ}$ make $\angle$ PQT whose measure is ${40}^o$ .
Step 6 : With P as centre and radius 5 cm draw an arc. This cuts $\overline{QT}$ at S.
Step 7 : Join $\overline{PS}$. PQRS is the required quadrilateral.
Step 8 : From Q draw $\overline{QX}\perp \overline{PR}$ and from S draw $\overline{SY}\perp \overline{PR}$ . Measure the lengths QX and SY. QX = $h_1$ = 3.1 cm, SY = $h_2$ = 3.9 cm. PR = d = 7 cm.

Calculation of area:
In the quadrilateral ABCD, d = 7 cm, $h_1$ = 3.1 cm and $h_2$ = 3.9 cm.

Area of the quadrilateral ABCD = $\frac{1}{2}d(h_1+h_2)$

$=\frac{1}{2}\left(7\right)(3.1+3.9)$

$=\frac{1}{2}\times 7\times 7$

$=24.5c{m}^2$