91Ó°ÊÓ

Two congruent circles intersect at $C$ and $D$.

$\begin{array}{l}CQ=QR\\ QR=CR\\ CR=RD\end{array}$

The sides of the quadrilateral $QDRC$ are equal and all angles of the diagonals bisect each other at right angle, it is a rhombus.

a.

As the sides of the quadrilateral $QDRC$are equal and all angles of the diagonals bisect each other at right angle, it is a rhombus.

b.

In localid="1648893720620" $\mathrm{Δ}\text{COQ}$and $\mathrm{Δ}\text{COR}$,

$\begin{array}{c}∠CQO=∠CRO\left(\text{given}\right),\text{and}\\ CO=CO\text{ }\left(\text{common}\right).\end{array}$

So, by SAS rule,

$\begin{array}{c}\mathrm{Δ}COQâ‰\dots \mathrm{Δ}COR.\\ \end{array}$

By congruent property,

$\begin{array}{c}∠COQ=∠COR,\\ ∠QCO=∠OCR,\\ ∠COQ+∠COR=180°,\\ 2∠COQ=180°,\\ ∠COQ=90°.\end{array}$

Therefore, $CD$is a perpendicular bisector of $QR$.

c.

$CD$is a perpendicular bisector of $QR$.

$\begin{array}{l}QC=17cm\text{ }\\ QR=30cm\end{array}$

$\begin{array}{c}\text{±õ²ÔΔ²Ï°¿°ä:}\\ C{Q}^2=O{Q}^2+O{C}^2,\\ {17}^2={15}^2+O{C}^2,\\ O{C}^2={17}^2−{15}^2,\\ O{C}^2=64,\\ OC=8\text{cm.}\\ \text{Now,since:}\\ CD=OC+OD\text{and}\\ OC=OD,\\ \text{wecanwrite:}\\ CD=2OC,\\ CD=16\text{cm.}\end{array}$

Therefore, $CD=16\text{cm}$.