Isosceles Triangle In Circle?

Geometry Level 2

In a circle of radius $5 \text{ cm}$ , $AB$ and $AC$ are 2 chords such that $AB=AC=6 \text{ cm}$ . the length of chord $BC$ is $\text{\_\_\_\_\_\_\_\_\_\_ cm}$ .

9.6 10 9.8 9.4 11 8

4 solutions

Shaun Leong
Feb 22, 2016

Let the centre of the circle be $O$ and let $\angle AOB=\angle AOC = \theta$ .

Apply cosine rule on the triangle $AOB$ , $6^2=5^2+5^2-2\cdot5^2\cos \theta \Rightarrow \cos \theta = \dfrac {14}{50}.$

Similarly apply cosine rule on the triangle $BOC$ ,

$\begin{aligned} BC^2&=&5^2+5^2-2\cdot5^2\cos(2\pi-2\theta) \\ BC&=&\sqrt{50-50\cos(2\theta)} \\ BC&=&\sqrt{50-50(2\cos^2 \theta -1)} \end{aligned}$

Using the value of $\cos \theta = \dfrac {14}{50}$ from above, we get $BC=\boxed{9.6}$ .

This problem can be simply done using heron's formula

Nikkil V - 5 years, 3 months ago

After 1st use of law of cosines, note that OA intersects BC at a right angle so a right triangle is formed with hypotenuse OC = 5. Therefore 1/2 BC = 5 * sin(arccos(14/50) = 4.8. BC = 2*4.8 = 9.6

Roger Erisman - 4 years, 4 months ago

Nikkil Nikkichan
Feb 22, 2016

For clarification heron's formula is A=[s(s-a)(s-b)(s-c)]^0.5 where s=(a+b+c)/2

Caeo Tan - 5 years, 3 months ago

Thank you.

Nikkil V - 5 years, 3 months ago

Hari Krishna Sahoo
Mar 1, 2016

Easiest solution, Kudos

Eden Dupont - 5 years, 3 months ago

J Chaturvedi
Feb 29, 2016

Let O be the center of circle and 2L be the length BC. The area of triangle OAB can be found in two ways. If AB is taken as base, the height would be 4(5^2-3^2=4^2) and the area would be 6×4/2=12. If OA is taken as base, the height would be L and the area would be 5L/2. Equating the two, we get 5L/2=12 or L=24/5. Therefore, BC=2L=9.6.

Isosceles Triangle In Circle?

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