Geometry Practice Test 1

In ΔABC,
AB = BC = CA = 3 cm

BD = $\frac{1}{2}$ CD
⇒ BD = 1 cm CD = 2 cm
In the equilateral triangle, median AE is the altitude of ΔABC ⇒ BE = EC = $\frac{3}{2}$
AE = $\frac{\sqrt{3}\mathit{a}}{2}$
AE = $\frac{3\sqrt{3}}{2}$
In ΔAED,
${\left(\mathit{A}\mathit{D}\right)}^2$ = ${\left(\mathit{A}\mathit{E}\right)}^2+{\left(\mathit{D}\mathit{E}\right)}^2$
DE = BE − BD
= $\frac{3}{2}-1$ = $\frac{1}{2}$ cm
$\mathit{A}{\mathit{D}}^2$ = ${\left(\frac{3\sqrt{3}}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2$
AD = $\sqrt{\frac{27}{4}+\frac{1}{4}}$
= $\sqrt{\frac{27+1}{4}}$ = $\sqrt{\frac{28}{4}}$ = $\sqrt{7}$
AD = $\sqrt{7}$ cm