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:
$EB \times EA = ED \times EC$
(It derives from the fact that the triangles $ECB$ and $EAD$ are similar. Can you see why? Hint: use angles in the same segment.)

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

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

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

$AH = \tfrac12 AB = 7$

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

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

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

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

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

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

$AH = \tfrac12 AB = 7$

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

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