# Math Help - need help for circles properties question

1. ## need help for circles properties question

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

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

i nid help on how to get the answer.

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

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?

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

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

4. Hello, tempq1!

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

Find:

a) the radius of the circle

b) the area of $\Delta AOB$

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

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

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

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

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

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

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