Chapter 9.2, Problem 9WE

(a)

To determine

To Find : The type of the given quadrilateral.

Answer to Problem 9WE

The figure is Square.

Explanation of Solution

Given:

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

The tangent meet at Z.

  

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)

Then addition of Adjutants angle is 180° .

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

Similarly prove that $\angle Z=90°$ .

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

Therefore, this quadrilateral is Rectangle.

Opposite side of rectangle are equal.

Thus,

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

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

Then, it can be concluded that,

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

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

(b)

To determine

To Find : The value of OZ.

Answer to Problem 9WE

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

Explanation of Solution

Given:

OX=5

  

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

Use Pythagoras Theorem,

In triangle OXZ,

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

Simplify;

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

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