How do you find this radius?

Question

Question:

If the area and perimeter of an equilateral triangle have the same numerical value, what is the radius of the inscribed circle?

Please show or explain how this question is done?

Answer 1

The area of an equilateral triangle is bh/2, and the height (because of the 60° angle) is (b√3)/2. This makes the area: (b²√3)/4. The perimeter is 3b.

Since the area and perimeter are equal,
(b²√3)/4 = 3b. Multiplying by 4,
(b²√3) = 12b. Dividing by b√3,
b = (12 / √3), or 4√3.

The center inscribed circle is 30° up from the "base" side.
Since the side is 4√3, halfway across is 2√3. Dividing by √3 to find the height to the center, it's just 2.

r = 2.

Answer 2

The base and the area have the same numeric value.

We know that Area = 1/2 Base * Height

The Height of an equilateral triangle is:

Height = sin(60 degrees) * Base

Plug this into the equation, and we get:

Area = 1/2 * Base ^2 * sin(60)

Perimeter = 3 * Base

We're told that the area = perimeter (ignoring units)

3 * Base = 1/2 * Base ^2 * sin(60)

Simplify by dividing each side by Base

3 = 1/2 Base * sin(60)

Solve for Base

Base = 6 / sin(60) [approximately 6.928]

Now, the question is: What is the radius of the inscribed circle? The inscribed circle of any equilateral triangle will be at the intersection of the bisectors of the angles, and tangent to the midpoint of each of the sides.

What is that radius? Consider the bisector of one of the base angles intersecting with the line from the top angle perpendicular to the base. From trig, the height of that point divided by one half the base is the tangent of 30 degrees (the bisector of the 60 degree angle).

tan(30) = Radius / (1/2 Base)

tan(30) = sin(30) / cos(30)

Substituting the value of Base and the trig identity we get:

sin(30) / cos(30) = Radius / (1/2 (6 / sin(60)))

Simplify the constants

sin(30) / cos(30) = Radius / (3 / sin(60))

Rearrange the right side

sin(30) / cos(30) = sin(60) * Radius / 3

As a trig identity, sin(60) = cos(30), so substitute

sin(30) / cos(30) = cos(30) * Radius / 3

Multiply both sides by cos(30)

sin(30) = cos(30)^2 * Radius / 3

We know that sin(30) = 0.5, and from Pythagorean theorum,

cos(30) = sqrt(.75)

Making these substitutions:

0.5 = (sqrt(0.75))^2 * Radius / 3

Simplifying:

0.5 = 0.75 * Radius / 3

0.5 = (0.75/3) * Radius

0.5 = 0.25 * Radius

0.5 / 0.25 = Radius

Radius = 2

I think this answers the question.

Answer 3

3a = a^2/4 *sqrt3 so a= 12/sqrt3 =4 sqrt 3
The inscribed circle has its center at the point of intersection of the bisectors of the angles.
This point ,with a side forms a triangle with 2 angles of 30 deg.
so
tan30 = r/(2sqrt3
sqrt3/3 =r/2sqrt3 so r=2

Answer 4

First, i anticipate the polygon is familiar with n factors. The inscribed radius is frequently referred to as the "apothem", on an identical time as the circumscribed radius is merely the "radius". i visit apply "a" to communicate over with the apothem length and "r" for the radius. the attitude between the radius and the apothem is ?/n. The apothem makes an excellent attitude with the edge, so the apothem and radius type an excellent triangle with the edge. we can see that cos(?/n) = a / r and this would be an user-friendly formula in terms of merely n pertaining to the two lengths.

Answer 5

First, throw everything you know at this, shake hard, and see what falls out...

A = hb/2

p = 3b

A = p (Area and perimeter have same numeric value)

C=п/d

Using the law of sines,

sin(60°)/h = sin(90°)/(p/3)

h = sin(60°)p/3

sin(30°)/r = sin(90°)/(h-r)

I think you should be able to knock out the answer from here. This is trigonometry, right?

Ok, plot spoiler below.

sin(30°) = 1/2, so the equation above yields 3r = h.

Since A=hb/2, h=2A/b and 3r = 2A/b.

r = 2A/(b3)

Since p = 3b and p = A,

r = 2p/p = 2