Geometry Practice Test 1

BC = 38 cm, CD = 25 cm, QB = 27 cm
MC = CR = 11 cm.
TD = DM = 14 cm.
In ΔTDM,
∠D = 90°
TD = DM = 14 cm.
By using Pythagoras theorem.

${\left(\mathit{T}\mathit{M}\right)}^2$ = ${\left(\mathit{D}\mathit{T}\right)}^2+{\left(\mathit{D}\mathit{M}\right)}^2$
$\mathit{T}{\mathit{M}}^2$ = ${\left(14\right)}^2+{\left(14\right)}^2$
$\mathit{T}{\mathit{M}}^2$ = 196 + 196
TM = $14\sqrt{2}$ cm
In Δ OTM,
OT = OM = radius = x
$\mathit{O}{\mathit{T}}^2+\mathit{O}{\mathit{M}}^2$ = ${\left(\mathit{T}\mathit{M}\right)}^2$
${\mathit{x}}^2+{\mathit{x}}^2$ = ${\left(14\sqrt{2}\right)}^2$
$2{\mathit{x}}^2$ = 196.2
${\mathit{x}}^2$ = 196
Radius = x = 14 cm