Question

If the length of a side of an equilateral triangle inscribed in a circle of radius 4cm is $4\sqrt{m}$, find the value of m.

Answer 1

Hint: Use the fact that the angle subtended by a chord at the centre is equal to the twice the angle subtended in the alternate segment. Hence prove that $\angle AOD=60{}^\circ $. Use the fact that in a right angle triangle since of an angle is the ratio of the opposite side to the hypotenuse. Hence prove that $AD=OA\sin 60{}^\circ $. Hence find the length of the side of the equilateral triangle. Compare this length to $4\sqrt{m}$ and hence form an equation in m. Solve for m and hence find the value of m.

Complete step-by-step answer:

Given: ABC is an equilateral triangle inscribed in a circle of radius 4cm. $AB=4\sqrt{m}$
To determine: The value of m
Construction: Draw OF perpendicular AB.
We know that the angle subtended by a chord of a circle at the centre of the circle is twice the angle subtended in the alternate segment.
The angle subtended by chord AB at O is $\angle AOB$
The angle subtended by chord AB in the alternate segment is $\angle ACB$
Hence from the above theorem, we have
$\angle AOB=2\times \angle ACB$
Since ABC is an equilateral triangle, we have
$\angle ACB=60{}^\circ $
Hence, we have
$\angle AOB=120{}^\circ $
Now in triangle AOB, we have
AO = OB(radii of the same circle)
Hence triangle AOB is an isosceles triangle. We know that in an isosceles triangle, the altitude drawn from the vertex opposite to the base is also an angle bisector.
Hence, we have OD is the bisector of angle O
Hence, we have
$\angle AOD=\dfrac{1}{2}\angle AOB=\dfrac{1}{2}\left( 120{}^\circ \right)=60{}^\circ $
Now in triangle AOD, we have
AO is the hypotenuse and AD is the side opposite to O
Hence, we have
$\sin \left( \angle AOD \right)=\dfrac{AD}{AO}$
Multiplying both sides by AO and substituting the value of $\angle AOD$, we get
$AD=OA\sin 60{}^\circ $
Now, we have
OA = 4cm and $\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}$
Hence, we have
$AD=4\times \dfrac{\sqrt{3}}{2}=2\sqrt{3}$
Since the perpendicular from the centre to the chord bisects the chord, we have D is the midpoint of AB
Hence, we have AB = 2AD
Hence, we have $AB=4\sqrt{3}$
But given that AB $=4\sqrt{m}$
Hence, we have
$4\sqrt{m}=4\sqrt{3}$
Dividing both sides by 4, we get
$\sqrt{m}=\sqrt{3}$
Squaring both sides, we get
$m=3$
Hence the value of m is 3

Note: Alternative solution: Best method:
We know that in a triangle, by the law of sines
$\dfrac{a}{\sin A}=2R$
Hence, we have
$a=2R\sin A$
Here R = 4cm and $A=60{}^\circ $
Hence, we have
$a=2\times 4\times \sin 60{}^\circ =8\times \dfrac{\sqrt{3}}{2}=4\sqrt{3}$
Hence the length of the side of the equilateral triangle is $4\sqrt{3}$, which is the same as obtained above. Hence following a similar procedure as above, we get m = 3