1. ## circle

In the figure , AB and CD are chords of a circle with centre O . They intersect at right angle at E . THe perpendiculars from teh centre O meet AB at H and CD at K . Given that EA=27 cm , EB=13 cm , and ED=9cm , prove that CD=30 cm . Hence calculate OH and the radius of the circle .

2. Hello thereddevils
Originally Posted by thereddevils
In the figure , AB and CD are chords of a circle with centre O . They intersect at right angle at E . THe perpendiculars from teh centre O meet AB at H and CD at K . Given that EA=27 cm , EB=13 cm , and ED=9cm , prove that CD=30 cm . Hence calculate OH and the radius of the circle .
There's a theorem - see for instance here - that tells you that in the diagram you have here:
$\displaystyle EB \times EA = ED \times EC$
(It derives from the fact that the triangles $\displaystyle ECB$ and $\displaystyle EAD$ are similar. Can you see why? Hint: use angles in the same segment.)

So, if we plug in the numbers given:
$\displaystyle 13\times27=9\times EC$

$\displaystyle \Rightarrow EC = 3\times 13 = 39$

$\displaystyle \Rightarrow CD = 30$ cm
Now (although it doesn't look like it in your diagram) $\displaystyle K$ is the mid-point of $\displaystyle CD$ (and $\displaystyle H$ the mid-point of $\displaystyle AB$). So $\displaystyle CK = \tfrac12 CD = 15$ cm. Also $\displaystyle OHED$ is a rectangle. So $\displaystyle OH = KE = 39-15=24$ cm.

$\displaystyle AH = \tfrac12 AB = 7$

$\displaystyle \Rightarrow OA^2 = 7^2 + 24^2 = 25^2$

$\displaystyle \Rightarrow$ radius of circle $\displaystyle = 25$ cm.

Hello thereddevilsThere's a theorem - see for instance here - that tells you that in the diagram you have here:
$\displaystyle EB \times EA = ED \times EC$
(It derives from the fact that the triangles $\displaystyle ECB$ and $\displaystyle EAD$ are similar. Can you see why? Hint: use angles in the same segment.)

So, if we plug in the numbers given:
$\displaystyle 13\times27=9\times EC$

$\displaystyle \Rightarrow EC = 3\times 13 = 39$

$\displaystyle \Rightarrow CD = 30$ cm
Now (although it doesn't look like it in your diagram) $\displaystyle K$ is the mid-point of $\displaystyle CD$ (and $\displaystyle H$ the mid-point of $\displaystyle AB$). So $\displaystyle CK = \tfrac12 CD = 15$ cm. Also $\displaystyle OHED$ is a rectangle. So $\displaystyle OH = KE = 39-15=24$ cm.

$\displaystyle AH = \tfrac12 AB = 7$

$\displaystyle \Rightarrow OA^2 = 7^2 + 24^2 = 25^2$

$\displaystyle \Rightarrow$ radius of circle $\displaystyle = 25$ cm.