Geometry Practice Test 2

Let OE = x
OD = y
AB = 6 cm BC = 2 cm
OC = $\sqrt{50}$ = radius.
By Pythagoras theorem,

${\left(\mathit{O}\mathit{A}\right)}^2$ = $\mathit{O}{\mathit{D}}^2+\mathit{A}{\mathit{D}}^2$ and $\mathit{O}{\mathit{C}}^2$ = $\mathit{E}{\mathit{C}}^2+\mathit{E}{\mathit{O}}^2$
${\left(\sqrt{50}\right)}^2$ = ${\mathit{y}}^2+{\left(6-\mathit{x}\right)}^2$
${\left(\sqrt{50}\right)}^2$ = ${\mathit{x}}^2+{\left(\mathit{y}+2\right)}^2$
50 = ${\mathit{y}}^2+36+{\mathit{x}}^2-12\mathit{x}$ and
50 = ${\mathit{x}}^2+{\mathit{y}}^2+4+4\mathit{y}$
Equating both the equations, we get:
- 12x + 36 = 4 + 4y
12x + 4y = 32
3x + y = 8
By hit and trial, there are only two possibilities i.e., x = 1 and y = 5
and x = 2 and y = 2 [but both x and y are not equal, so these values are rejected.]
Hence, x = 1 and y = 5
In ΔBOE,
${\left(\mathit{O}\mathit{B}\right)}^2$ = ${\left(\mathit{O}\mathit{E}\right)}^2+\mathit{E}{\mathit{B}}^2$
(Since EB = OD)
$\mathit{O}{\mathit{B}}^2$ = $\mathit{O}{\mathit{E}}^2+\mathit{O}{\mathit{D}}^2$
$\mathit{O}{\mathit{B}}^2$ = ${\mathit{x}}^2+{\mathit{y}}^2$
$\mathit{O}{\mathit{B}}^2$ = $1^2+5^2$ = 26