In the diagram shown, $\angle A B C=90^{\circ}, C B \| E D, A B=D F$, $A D=24, A E=25$ and $O$ is the centre of the circle.

Determine the perimeter of $C B D F$.
(A) 39
(D) 43
(B) 40
(C) 42

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-263.jpg?height=393&width=431&top_left_y=243&top_left_x=1432)

## Solution

We start by showing that $\triangle A B C \cong \triangle D F E$.

Since $E D \| C B$ this implies that $\angle D E F=\angle B C A$ because of corresponding angles. Also, $\angle D F A=90^{\circ}$ because it is an angle in a semicircle which also means that $\angle D F E$ is $90^{\circ}$. Thus the two triangles are equiangular. Since $A B=D F, \triangle A B C \cong \triangle D F E$ (ASA). Therefore, $E F=C B$ and $D F=B A$. Using Pythagorus in $\triangle A D E, D E^{2}=25^{2}-24^{2} \Rightarrow D E=7=C A$.

Thus $C E=25-7=18$.

The required perimeter is, $\quad C B+B D+D F+F C=E F+(B D+D F)+F C=(E F+F C)+(B D+D F)$

$$
\begin{aligned}
& =(E F+F C)+(B D+B A), \text { since } D F=B A \\
& =C E+A D \\
& =18+24=42
\end{aligned}
$$

ANSWER: (C)
In the diagram shown, $\angle A B C=90^{\circ}, C B \| E D$, $A B=D F, A D=24, A E=25$ and $O$ is the centre of the circle. Determine the perimeter of $C B D F$.
(A) 39
(D) 43
(B) 40
(E) 44
(C) 42

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-141.jpg?height=303&width=357&top_left_y=201&top_left_x=1342)