In a circle of radius 5 , AB and AC are 2 chords such that AB = AC = 6cm. Find BC

Question

In this question I can easily calculate the perpendicular distances of AB and AC from the centre O by the perpendicular bisector property. However , after this I am left with a quadrilateral BOFC (F is midpoint of AC) in which I can't really find the perpendicular distance of BC from the centre neither can I find any arc lengths .

How do I proceed ? Do I try to brutally get the angles using trigonometry or coordinate geometry can be useful here? Do pardon my confusing figure Those 6cm chords have been bisected into 3cm and 3cm I can get 4cm as the distance but can't proceed

Answer 1

Basic approach. Drop a perpendicular from $A$ to $\overline{BC}$ at point $M$ (also the midpoint of $\overline{BC}$). Pythagoras gives us

$$ OM^2+BM^2 = OB^2 = 25 $$

but also

$$ AM^2+BM^2 = AB^2 = 36 $$

and substitute $AM = AO - OM = 5 - OM$. (Note that $M$ is between $A$ and $O$, and does not fall outside $\overline{AO}$ as you've shown in your diagram.)

Answer 2

Continue $AO$ until it intersects $BC$ in $M$. Since $AB=AC$, in the isosceles triangle, the $AM$ is bisector, median, and height. Then just write $\sin \angle CAM$ in the two right angle triangles $\triangle CAM$ and $\triangle AOF$: $$\sin\angle CAM=\frac{OF}{AO}=\frac{BM}{AB}$$ You are given $AB=6$, $AO=5$, and you calculated $OF=4$. Find $BM$, and $BC=2BM$.

Answer 3

Also using that $AO$ is a perpendicular bisector of $BC$. Let $M$ be the midpoint of $BC$.

Consider the area of $\triangle AOC$ in two ways:

• With base $AC$ and height $OF$:

  $$\text{Area} = \frac12 \cdot 6\cdot 4 = 12\text{ cm$^2$}$$

• With base $AO$ and height $BM = BC/2$:

  $$\begin{align*} \text{Area} = \frac12 \cdot 5 \cdot \frac{BC}2 &= 12\\ BC&= 9.6\text{ cm} \end{align*}$$