CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

In a circle of radius 5 cm, AB and AC are two chords. such that $$AB=AC=6\:cm$$. Find the length of chord BC.


Solution

We know that, if AB and AC are two equal chords of a circle, then the centre of the circle lies on the bisector of 

$$\angle BAC$$.

Here, $$AB=AC=6\:cm$$. So, the bisecor of $$\angle BAC$$ passes through the centre O i.e. OA is the bisector of $$\angle BAC$$.


Since, the internal bisector of an angle divides the opposite sides in the ratio of the sides containing the 
angle. Therefore, M divides BC in the ratio $$1:1$$, i.e. M is the mid-point of BC.

$$\Rightarrow OM\perp BC$$. 

In the right triangle ABM, we have 

$$\Rightarrow AB^{2}=AM^{2}+BM^{2}$$

$$\Rightarrow 36=AM^{2}+BM^{2}$$

$$\Rightarrow BM^{2}=36-AM^{2}$$ ...(i)

In the right triangle OBM, we have

$$\Rightarrow OB^{2}=OM^{2}+BM^{2}$$

$$\Rightarrow 25=(OA-AM)^{2}+BM^{2}$$

$$\Rightarrow BM^{2}=25-(OA_AM)^{2}$$

$$\Rightarrow 25-(5-AM)^{2}$$  .....(ii)

From equations (i) and (ii), we get

$$36-AM^{2}=25-(5-AM)^{2}$$

$$\Rightarrow 11-AM^{2}+(5-AM)^{2}=0$$

$$\Rightarrow 11-AM^{2}+25-10AM+AM^{2}=0$$

$$\Rightarrow 10AM=36$$

$$\Rightarrow AM=3.6$$

Putting $$AM=3.6$$ in equation (i), we get

$$BM^{2}=36-(3.6)^{2}=36-12.96$$

$$\Rightarrow BM=\sqrt{36-12.96}=\sqrt{23.04}=4.8\:cm$$

Hence, $$BC=2\:BM=2\times 4.8\:cm=9.6\:cm$$.

480885_317895_ans_2ab405868f1c451790254b9e08bcac7d.jpg

Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image