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Q9.

Expert-verifiedFound in: Page 336

Book edition
Student Edition

Author(s)
Ray C. Jurgensen, Richard G. Brown, John W. Jurgensen

Pages
227 pages

ISBN
9780395977279

**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 $OZ$.**

**a.** The figure is, **Square**.

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

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

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

That is;

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

Then addition of Adjutants angle is $180\xb0$.

$\begin{array}{l}\angle \text{}X\text{}+\text{}\angle Y\text{}=180\xb0\\ \begin{array}{l}90\xb0\text{}+\text{}\angle Y=\text{}180\xb0\\ \angle Y\text{}=90\xb0\end{array}\end{array}$

Similarly prove that, $\angle Z\text{}=90\xb0$.

Then all angles of quadrilateral are $90\xb0$ and opposite side are parallel.

Therefore, this quadrilateral is Rectangle.

Thus,

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

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

Then, it can be concluded that,

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

Then, all side is equal and all angles are $90\xb0$.

The given value is,

$OX=5$

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

In triangle $OXZ,$

$\begin{array}{l}\begin{array}{l}OZ\xb2\text{}=OX\xb2\text{}+\text{}XZ\xb2\\ OZ\xb2\text{}=5\xb2\text{}+\text{}5\xb2\end{array}\\ OZ\xb2\text{}=50\end{array}$$\begin{array}{l}\sqrt{O{Z}^{2}}=\sqrt{50}\\ OZ=5\sqrt{2}\end{array}$

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

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