Question

Draw $\odot O$with perpendicular radii $\overline{OX}$and $\overline{OY}$. Draw tangents to the circle at X and $Y$.

$\left(a\right).$ If the tangents meet at $Z,$ what kind of figure is $OXYZ?$ Explain.

$\left(b\right).$ If $OX=5,$ find $OZ$.

Short answer

a. The figure is, Square.

b. The value of $OZ$ is, $5\sqrt{2}$.

Step-by-step solution

Part a. Step 1. Given information.

Given:

The circle with a center $O$with perpendicular segment $OX$and segment $OY$.

The tangent meet at $Z$.

Now, figure drawn by the information is,

In quadrilateral $OXYZ,$$OX\perp XZ\text{and}OY\perp YZ.$

$OX$ and $OY$ is tangent to a circle.

That is;

$\begin{array}{l}OX\left|\right|YZ\\ OY\left|\right|XZ\end{array}$ .......(1)

Step 2. Concept used.

Then addition of Adjutants angle is $180°$.

$\begin{array}{l}\angle X+\angle Y=180°\\ \begin{array}{l}90°+\angle Y=180°\\ \angle Y=90°\end{array}\end{array}$

Similarly prove that, $\angle Z=90°$.

Then all angles of quadrilateral are $90°$ and opposite side are parallel.

Therefore, this quadrilateral is Rectangle.

Step 3. Opposite side of rectangle are equal.

Thus,

$\begin{array}{l}OX=YZ\\ OY=XZ\mathrm{......}\left(2\right)\end{array}$

But $YX=XY$ (Tangent to the circle to same point)

Then, it can be concluded that,

$\begin{array}{l}OX=OY\\ =YZ\\ =XY\end{array}$

Then, all side is equal and all angles are $90°$.

Part b. Step 1. Given information.

The given value is,

$OX=5$

In the above figure this is a square then all side is same.

Step 2. Use Pythagoras Theorem.

In triangle $OXZ,$

$\begin{array}{l}\begin{array}{l}OZ²=OX²+XZ²\\ OZ²=5²+5²\end{array}\\ OZ²=50\end{array}$

Step 3. Simplify.

$\begin{array}{l}\sqrt{O{Z}^2}=\sqrt{50}\\ OZ=5\sqrt{2}\end{array}$

Therefore, the value of $OZ$is, $OZ=5\sqrt{2}$.