Respuesta :
Answer:
r = 10.91 units.
Step-by-step explanation:
The triangle has a 60° angle, and the two adjacent sides are 12 and "12 times the square root of 3".
So, the area of the triangle, Δ = [tex]\frac{1}{2} \times 12 \times 12\sqrt{3} \times \sin 60^{\circ} = 108[/tex] sq. units.
Now, we have to find the radius of the circle with the same vertex as a center, if the arc in the triangle bisects the area of the triangle.
So, if the radius of the circle is r units, then
[tex]\pi r^{2} \times \frac{60}{360} = \frac{108}{2}[/tex]
⇒ 0.5238 r² = 54
⇒ r² = 103.092
⇒ r = 10.15 units. (Answer)
You can use the law of cosine to find the third side and then can use heron's formula to find the area of the triangle. That will help to find the radius of the specified circle.
The radius of the circle for the specified conditions is 10.16 units approximately.
What is law of cosine?
Let there is triangle ABC such that |AB| = a units, |AC| = b units, and |BC| = c units and the internal angle A is of θ degrees, then we have:
[tex]c = \sqrt{a^2 + b^2 - 2ab\cos(\theta)}[/tex]
What is heron's formula to get the area of a triangle with specified side lengths?
Suppose there is a triangle ABC with side lengths a units, b units and c units. Then its area is given as:
[tex]A = \sqrt{s(s-a)(s-b)(s-c)}[/tex]
where we have [tex]s =\dfrac{a+b+c}{2}[/tex]
What is the area of the region swept out by radius rotated θ degrees in a circle?
Full rotation is of 360 degrees which covers whole area [tex]\pi \times r^2[/tex] squared units.
Thus, θ degrees rotation by the radius will cover:
[tex]A_{\theta}= \dfrac{\pi r^2}{360^\circ} \times \theta[/tex]
Using above conclusions to find the radius of the circle to find the area
Refer to the diagram given below.
Let the radius of the circle be r units.
We have:
AP = AQ = r units
AB = 12√3 units
AC = 12 units
BC = c units
Using cosine law, we have:
[tex]c = \sqrt{a^2 + b^2 - 2ab\cos(\theta)}\\c = \sqrt{144 \times 3 + 144 - 2 \times 12 \sqrt{3}\times 12 \times cos(60)}\\c= 12\sqrt{4-\sqrt{3}} \approx 18.0717 \: \rm units[/tex]
Using heron's formula, we get:
[tex]s = \dfrac{a+b+c}{2} = \dfrac{12 + 12\sqrt{3} + 18.0717}{2} \approx 25.43[/tex]
The area of triangle ABC is
[tex]A = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{11676.76} \approx 108.06[/tex]
The area of the sector swept by radius for 60 degrees in circle is
[tex]A_{\theta}= \dfrac{\pi r^2}{360^\circ} \times \theta = \dfrac{\pi r^2}{360^\circ} \times 60^\circ = \dfrac{\pi r^2}{6}[/tex]
Since this area is said to be half of area of triangle ABC, thus, we have:
[tex]\dfrac{A}{2} = A_\theta\\\\108.06/2 = \dfrac{\pi r^2}{6}\\\\r = \sqrt{3 \times 108.06 \div \pi} \approx 10.1582 \: \rm units[/tex]
(we take positive root since radius is measurement of length thus, being a non negative quantity)
Thus,
The radius of the circle for the specified conditions is 10.16 units approximately.
Learn more about law of cosine here:
https://brainly.com/question/17289163
