Geometry Practice Test 4

ABCD is a quadrilateral.
AB || DC
AD || BC
In equilateral ΔABE, perimeter = 6 units
Side = 2 cm
Area of the equilateral triangle = $\frac{\sqrt{3}}{4}{\mathit{a}}^2$
= $\frac{\sqrt{3}\times 2\times 2}{4}$ = $\sqrt{3}{\mathrm{units}}^2$ ... (1) ${\mathrm{units}}^2$
In the right angle triangle ΔADE,
AE = 2 unit
DE = 1 unit
${\left(\mathit{A}\mathit{D}\right)}^2+{\left(\mathit{D}\mathit{E}\right)}^2$ = ${\left(\mathit{A}\mathit{E}\right)}^2$
${\left(\mathit{A}\mathit{D}\right)}^2+{\left(1\right)}^2$ = ${\left(2\right)}^2$
AD = $\sqrt{4-1}$ = $\sqrt{3}$
Area of Δ ADE = $\frac{1}{2}\times 1\times \sqrt{3}$ = $\frac{\sqrt{3}}{2}$ ... (2)
Similarly, area of Δ ECB = $\frac{1}{2}\times 1\times \sqrt{3}$ = $\frac{\sqrt{3}}{2}$ ... (3)
Adding(1), (2) and (3),
Area of quadrilateral ABCD = $\sqrt{3}+\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}$
= $\frac{2\sqrt{3}+\sqrt{3}+\sqrt{3}}{2}$ = $\frac{4\sqrt{3}}{2}$
= $2\sqrt{3}$ square units.