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

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

• I Actually added one , maybe it's an error Feb 6 at 4:55
• Due to certain restrictions about showing images for new users, only linked URL's are allowed (e.g., to avoid certain possibly very inappropriate for general viewing images to be shown). However, I've now edited your question to include your figure. Feb 6 at 4:57
• You're welcome. I've just edited your question again now to include your second figure. Feb 6 at 5:00
• Done . Won't edit it now . It crashed so I thought that it once again wouldn't show so I just added it... Feb 6 at 5:04
• I apologize for it Feb 6 at 5:39

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.)

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 $$BC=2BM$$.

• Can I use coordinate geometry for this ? Feb 6 at 5:40
• You don't need to. I think it would complicate things. Feb 6 at 5:51

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*}