The side of an equilateral triangle is 12 cm. What is the radius of the circle circumscribing this equilateral triangle?
The question asks us to find the radius of the circle that circumscribes an equilateral triangle with a side length of 12 cm.
A circle that circumscribes a triangle passes through all three vertices of the triangle. The radius of this circle is called the circumradius.
For an equilateral triangle with a side length 'a', the radius (R) of the circumscribing circle is given by the formula:
\( R = \frac{a}{\sqrt{3}} \)
Alternatively, the circumradius can also be calculated using the height 'h' of the equilateral triangle. The height of an equilateral triangle with side 'a' is \( h = \frac{a\sqrt{3}}{2} \). The circumcenter (center of the circumscribing circle) is also the centroid, which divides the median (and height) in the ratio 2:1. So, the circumradius R is \( \frac{2}{3} \) of the height h:
\( R = \frac{2}{3} h = \frac{2}{3} \times \frac{a\sqrt{3}}{2} = \frac{a\sqrt{3}}{3} \)
Let's use the first formula \( R = \frac{a}{\sqrt{3}} \) as it directly uses the side length.
Given the side length of the equilateral triangle, \( a = 12 \) cm.
Using the formula \( R = \frac{a}{\sqrt{3}} \):
\( R = \frac{12}{\sqrt{3}} \)
To rationalize the denominator, we multiply the numerator and the denominator by \( \sqrt{3} \):
\( R = \frac{12}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \)
\( R = \frac{12\sqrt{3}}{3} \)
Now, simplify the expression:
\( R = 4\sqrt{3} \)
So, the radius of the circle circumscribing the equilateral triangle is \( 4\sqrt{3} \) cm.
Given side length \( a = 12 \) cm.
Circumradius \( R = 4\sqrt{3} \) cm.
| Given | Formula Used | Calculation | Result |
|---|---|---|---|
| Side length \( a = 12 \) cm | \( R = \frac{a}{\sqrt{3}} \) | \( R = \frac{12}{\sqrt{3}} = \frac{12\sqrt{3}}{3} = 4\sqrt{3} \) | \( R = 4\sqrt{3} \) cm |
| Property | Formula (side 'a') | Formula (height 'h') | Notes |
|---|---|---|---|
| Height (h) | \( h = \frac{a\sqrt{3}}{2} \) | - | Also a median, angle bisector, perpendicular bisector |
| Area (A) | \( A = \frac{\sqrt{3}}{4} a^2 \) | \( A = \frac{1}{2} a h \) | - |
| Circumradius (R) | \( R = \frac{a}{\sqrt{3}} = \frac{a\sqrt{3}}{3} \) | \( R = \frac{2}{3} h \) | Radius of circumscribing circle |
| Inradius (r) | \( r = \frac{a}{2\sqrt{3}} = \frac{a\sqrt{3}}{6} \) | \( r = \frac{1}{3} h \) | Radius of inscribing circle |
| Relation R and r | \( R = 2r \) | \( R = 2r \) | Circumradius is twice the inradius |
In an equilateral triangle, several important centers coincide at a single point:
This unique property simplifies calculations related to circles and points of concurrency in equilateral triangles. The circumradius is the distance from this central point to any vertex, and the inradius is the distance from this point to any side (measured perpendicularly).
In this problem, we found the circumradius using the side length, confirming one of the standard formulas for equilateral triangles.
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