The radius of the circumcircle of an equilateral triangle of √3 unit side, is:
1 unit
This problem asks us to find the radius of the circumcircle for a specific equilateral triangle. We are given the side length of the equilateral triangle as $\sqrt{3}$ units.
The circumcircle of a triangle is the circle that passes through all three vertices of the triangle. Its center is called the circumcenter, and its radius is called the circumradius. In an equilateral triangle, the circumcenter coincides with the centroid, incenter, and orthocenter.
For an equilateral triangle with side length 'a', the circumradius (R) is given by the formula:
$\qquad R = \frac{a}{\sqrt{3}}$
Alternatively, if 'h' is the altitude (height) of the equilateral triangle, the circumcenter divides the altitude in a 2:1 ratio, with the longer part being the circumradius. The altitude of an equilateral triangle is $h = \frac{\sqrt{3}}{2}a$. Thus, $R = \frac{2}{3}h = \frac{2}{3} \times \frac{\sqrt{3}}{2}a = \frac{\sqrt{3}}{3}a = \frac{a}{\sqrt{3}}$.
We are given the side length of the equilateral triangle, $a = \sqrt{3}$ units.
Using the formula $R = \frac{a}{\sqrt{3}}$, we substitute the given value of 'a':
$\qquad R = \frac{\sqrt{3}}{\sqrt{3}}$
Now, we simplify the expression:
$\qquad R = 1$
So, the radius of the circumcircle of the equilateral triangle with $\sqrt{3}$ unit side is 1 unit.
Comparing our result with the given options, we find that 1 unit matches one of the choices.
| Property | Formula (side 'a') |
|---|---|
| Altitude (h) | $\frac{\sqrt{3}}{2}a$ |
| Area (A) | $\frac{\sqrt{3}}{4}a^2$ |
| Circumradius (R) | $\frac{a}{\sqrt{3}}$ or $\frac{\sqrt{3}}{3}a$ |
| Inradius (r) | $\frac{a}{2\sqrt{3}}$ or $\frac{\sqrt{3}}{6}a$ |
In an equilateral triangle, the circumcenter, incenter, centroid, and orthocenter are all the same point. This point is equidistant from all sides (inradius) and all vertices (circumradius).
The relationship between the circumradius (R) and the inradius (r) in an equilateral triangle is particularly simple: $R = 2r$. We can see this from the formulas:
Dividing R by r, we get $\frac{R}{r} = \frac{a/\sqrt{3}}{a/(2\sqrt{3})} = \frac{a}{\sqrt{3}} \times \frac{2\sqrt{3}}{a} = 2$. So, $R = 2r$. The circumcenter is also the centroid, which divides each median (and altitude/angle bisector) in a 2:1 ratio. The circumradius is the larger part (2), and the inradius is the smaller part (1) of the altitude.
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A. 36°
B. 60°
C. 84°
D. 15°
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B. The exterior angle of a triangle = the sum of interior opposite angles.
C. The sum of 2 sides is greater than the 3rd side.
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