Question

Construct a quadrilateral ABCD with AB = 6 cm, AD = 6 cm, $\angle$ABD = ${45}^o$, $\angle$ BDC = ${40}^o$ and EDBC = ${40}^o$. Find also its area.

Answer 1

Given:
AB = 6 cm, AD = 6 cm, $\angle$ ABD = ${45}^o$, $\angle$ BDC = ${40}^o$ and $\angle$ DBC = ${40}^o$.
Steps for construction
Step 1 : Draw a rough diagram and mark the given measurements.
Step 2 : Draw a line segment AB = 6 cm.
Step 3 : At B on $\overline{AB}$ make $\angle$ ABX whose measure is ${45}^o$ .
Step 4 : With A as centre and 6 cm as radius draw an arc. Let it cut $\overline{BX}$ at D.
Step 5 : Join $\overline{AD}$ .
Step 6 : At B on $\overline{BD}$ make $\angle$ DBY whose measure is ${40}^o$ .
Step 7 : At D on $\overline{BD}$ make $\angle$ BDZ whose measure is ${40}^o$ .
Step 8 : Let $\overline{BY}$ and $\overline{DZ}$ intersect at C. ABCD is the required quadrilateral.
Step 9 : From A draw $\overline{AE}\perp \overline{BD}$ and from C draw $\overline{CF}\perp \overline{BD}$. Then measure the lengths of AE and CF. AE = $h-$1 = 4.2 cm, CF = $h_2$ = 3.8 cm and BD = d = 8.5 cm.
Calculation of area:
In the quadrilateral ABCD, d = 8.5 cm, $h_1$ = 4.2 cm and $h_2$ = 3.8 cm.
Area of the quadrilateral ABCD = $\frac{1}{2}d(h_1+h_2)$
=$\frac{1}{2}\left(8.5\right)(4.2+3.8)$
=$\frac{1}{2}\times 8.5\times 8$
=34 $c{m}^2$