# need help for circles properties question

• May 1st 2010, 12:15 AM
tempq1
need help for circles properties question
In the diagram,the chord AB is perpendcular to the diameter CD,AB=70cm and CE=25cm.Find
b)the area of triangle AOB,where O is the centre of the circle.

http://i275.photobucket.com/albums/j...ngming/1-1.jpg

ans is:
a)37cm
b)420cm^2

i nid help on how to get the answer.
• May 1st 2010, 02:34 AM
Debsta
Quote:

Originally Posted by tempq1
In the diagram,the chord AB is perpendcular to the diameter CD,AB=70cm and CE=25cm.Find
b)the area of triangle AOB,where O is the centre of the circle.

http://i275.photobucket.com/albums/j...ngming/1-1.jpg

ans is:
a)37cm
b)420cm^2

i nid help on how to get the answer.

Have you learnt the Cartesian equation of a circle? (ie x^2 + y^2 = r^2) Or are you studying circle geometry?
• May 1st 2010, 06:12 AM
bjhopper
Quote:

Originally Posted by tempq1
In the diagram,the chord AB is perpendcular to the diameter CD,AB=70cm and CE=25cm.Find
b)the area of triangle AOB,where O is the centre of the circle.

http://i275.photobucket.com/albums/j...ngming/1-1.jpg

ans is:
a)37cm
b)420cm^2

i nid help on how to get the answer.

Hi Tempq,

Use the property of intersecting chords AE xBE =CE xDE It's very easy then.

bjh
• May 1st 2010, 07:44 AM
Soroban
Hello, tempq1!

Quote:

$\displaystyle \text{In circle }O:\:\text{chord }AB \perp \text{diameter }CD,\;AB=70\text{cm},\;CE=25\text{cm.}$

Find:

a) the radius of the circle

b) the area of $\displaystyle \Delta AOB$

http://i275.photobucket.com/albums/j...ngming/1-1.jpg

We have: .$\displaystyle AB = 70 \quad\Rightarrow AE = 35$
. . and: .$\displaystyle CE = 25$

Draw radius $\displaystyle OA = r$
. . Then: .$\displaystyle EO \:=\:r - 25$

In right triangle $\displaystyle AEO\!:\;(r-25)^2 + 35^2 \:=\:r^2$

. . $\displaystyle r^2 - 50r + 625 + 1225 \:=\:r^2 \quad\Rightarrow\quad 50r \:=\:1850$

Therefore: .$\displaystyle \boxed{r \:=\:37\text{ cm}}$

In $\displaystyle \Delta AOB\!:\;\begin{Bmatrix}\text{base} &=& AB &=& 70 \\ \text{height} &=& EO &=& 12\end{Bmatrix}$

Therefore: .$\displaystyle \Delta AOB \;=\;\tfrac{1}{2}bh \;=\;\tfrac{1}{2}(70)(12) \;=\;\boxed{420\text{ cm}^2}$