+0  
 
+1
20
4
avatar+52 

Triangle XYZ is equilateral. Points Y and Z lie on a circle centered at O, such that X is the circumcenter of triangle OYZ, and X lies inside triangle OYZ. If the area of the circle is 48pi, then find the area of triangle XYZ.

I am not sure what to do. This question is overdo, so any help is appreciated!

 Aug 8, 2023
 #1
avatar
0

Let's start by finding the radius of the circle. Given that the area of the circle is 48pi, we can write:

Area of the circle = pi * r^2
48pi = pi * r^2

Dividing both sides by pi:

48 = r^2

Taking the square root of both sides:

r = sqrt(48)

Since triangle XYZ is equilateral and X is the circumcenter, we know that X is equidistant from Y and Z. Let's call this distance d. In other words, d = XY = XZ.

Now, let's find OD (the distance from O to D), where D lies on YZ and XD ⊥ YZ. Since triangle OYZ is a right-angled triangle (it's not stated but assumed), we can use Pythagorean theorem:

OD^2 + XD^2 = r^2

Substituting r with sqrt(48):

OD^2 + XD^2 = 48

In an equilateral triangle, the median also acts as an altitude and bisects the opposite side. So OD is half the length of side YZ.

Now let's call s as a side of triangle XYZ:

OD = (s/2)

XD can also be represented as (YZ/2) or (s/2).

Plugging these values into our previous equation:

((s/2))^2 + ((s/2))^2 = 48

Simplifying:
(s^2)/4 + (s^2)/4

Combining like terms:

(s^2)/2 = 48

Multiplying both sides by 2:

s^2 = 96

Taking square root of both sides:
s = sqrt(96)

Now that we have found the side of triangle XYZ, we can find its area using Heron's formula or by simply using this formula for equilateral triangles:

Area = (s^2 * sqrt(3))/4

Substituting the value of s:

Area = (96 * sqrt(3))/4

Area = 24sqrt(3)

So, the area of triangle XYZ is approximately 24sqrt(3).

 Aug 8, 2023
 #2
avatar
-1

Since X is the circumcenter of triangle OYZ, it lies on the perpendicular bisectors of all three sides of the triangle. This means that XY = XZ = R, and YZ = 2R/sqrt(3), which is the diameter of the circle.

 

The area of triangle XYZ is given by K = (S)(r)/2, where S is the semiperimeter of the triangle and r is the radius of the inscribed circle. In this case, S = YZ/2 = R/sqrt(3), so the area of triangle XYZ is

K = (R/sqrt(3))(R)/2 = R^2/(2 * sqrt(3)).

Since the area of the circle is piR^2 = 48pi, we have R^2 = 48, so the area of triangle XYZ is

K = 48/(2 * sqrt(3)) = 24/sqrt(3) = \boxed{12sqrt(3)}.

 Aug 8, 2023
 #3
avatar+52 
0

Sorry, I'm a little lost. There are multiple points in your answer which confuse me. 
Why is XY = XZ = radius (I'm assuming R is radius) because X lies on the perpendicular bisectors of all three sides of the triangle? Points Y and Z lie on a circle centered on O, so OY and OZ would be the radius, unless you're referring to another circle? 

What is the inscribed circle? 

How did you get 48/(2 * sqrt(3)) = 12sqrt(3)?

Sorry, and thanks.

choutowne  Aug 8, 2023

2 Online Users

avatar