How to Draw a Circumscribed Circle Around a Triangle
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According to the Police force of Sines, whatever triangle △ABC has a mutual ratio of sides to sines of reverse angles, namely $$\frac{a}{ sin A} = \frac{b}{ sin B} = \frac{c}{ sin C} .$$
This mutual ratio has a geometric significant: information technology is the diameter (i.east. twice the radius) of the unique circle in which △ABC can be inscribed, called the circumscribed circle of the triangle. Earlier proving this, nosotros need to review some elementary geometry.
A fundamental angle of a circle is an angle whose vertex is the centre O of the circle and whose sides (called radii) are line segments from O to two points on the circle. In Figure 1, \(\bending\)O is a central angle and we say that it intercepts the arc– BC.
An inscribed angle of a circumvolve is an angle whose vertex is a point A on the circle and whose sides are line segments (chosen chords) from A to two other points on the circle. In Figure 2, \(\angle\)A is an inscribed angle that intercepts the arc –BC. Nosotros state here without proof a useful relation between inscribed and primal angles:
Theorem 1. If an inscribed angle \(\angle\)A and a cardinal angle \(\angle\)O intercept the same arc, then \(\angle\)A = \(\frac{ane}{ 2} \angle\)O. Thus, inscribed angles which intercept the aforementioned arc are equal.
Figure 3 shows two inscribed angles, \(\bending\)A and \(\angle\)D, which intercept the same arc– BC as the central angle \(\angle\)O, and hence \(\angle\)A = \(\bending\)D = \(\frac{1}{2}\) \(\angle\)O (and then \(\angle\)O = two \(\bending\)A = 2 \(\bending\)D). We will now prove our exclamation virtually the mutual ratio in the Law of Sines:
Theorem 2. For whatsoever triangle △ABC, the radius R of its circumscribed circle is given by: $$2R = \frac{a}{ sin A} = \frac{b}{ sin B }= \frac{c}{sin C}$$
(Note: For a circle of diameter 1, this means a = sin A, b = sin B, and c = sin C.)
To prove this, let O be the center of the circumscribed circle for a triangle △ABC. And so O can be either inside, outside, or on the triangle, as in Figures 4, 5 & 6 below. In the first two cases, draw a perpendicular line segment from O to \(\overline{AB}\) at the indicate D.
The radii \(\overline{OA}\text{ and }\overline{OB}\) have the aforementioned length R, so △AOB is an isosceles triangle. Thus, from unproblematic geometry nosotros know that \(\overline{OD}\) bisects both the angle \(\angle\)AOB and the side \(\overline{AB}\). So \(\angle\)AOD = \(\frac{1}{ ii} \angle\)AOB and AD = \(\frac{c}{ 2}\) . But since the inscribed angle \(\bending\)ACB and the fundamental bending \(\angle\)AOB intercept the same arc– AB, we know from Theorem 1 that \(\angle\)ACB = \(\frac{1}{ two} \bending{AOB}\). Hence, \(\angle\)ACB =\(\angle\)AOD. So since C =\(\angle\)ACB, we have
$$sin C = sin \angle{AOD} = \frac{AD}{ OA} = \frac{\frac{c}{ ii}}{ R} = \frac{c}{ 2R} ⇒ 2R = \frac{c}{ sin C} ,$$
so by the Constabulary of Sines the result follows if O is inside or outside △ABC. Now suppose that O is on △ABC, say, on the side AB, as in Figure vi. Then AB is a diameter of the circumvolve, then C = 90º by Thales' Theorem. Hence, sin C = 1, and then 2R = AB = c = \(\frac{c}{ i} = \frac{c}{ sin C}\) , and the result again follows by the Law of Sines.
Example one
Observe the radius R of the circumscribed circumvolve for the triangle △ABC whose sides are a = 3, b = four, and c = 5.
Solution: We know that △ABC is a right triangle. So as we run into from Figure 7, sin A = 3/5. Thus, $$2R = \frac{a}{ sin A} = \frac{3}{\frac{ 3}{ v}} = 5 ⇒ \boxed{R = 2.5} .$$
Note that since R = two.v, the bore of the circle is v, which is the aforementioned as AB. Thus, AB must be a diameter of the circle, and so the center O of the circle is the midpoint of AB.
Corollary 1 For whatever right triangle, the hypotenuse is a diameter of the confining circle, i.east. the center of the circumvolve is the midpoint of the hypotenuse.
For the right triangle in the above example, the circumscribed circle is unproblematic to draw; its center can be establish by measuring a distance of two.5 units from A along \(\overline{AB}\). We need a different procedure for astute and obtuse triangles, since for an acute triangle the center of the circumscribed circle will exist inside the triangle, and it volition exist outside for an obtuse triangle. Notice from the proof of Theorem 2 that the center O was on the perpendicular bisector of one of the sides \(\overline{AB}\)). Similar arguments for the other sides would evidence that O is on the perpendicular bisectors for those sides:
Corollary 2. For whatsoever triangle, the middle of its confining circumvolve is the intersection of the perpendicular bisectors of the sides.
Retrieve from geometry how to create the perpendicular bisector of a line segment: at each endpoint utilise a compass to describe an arc with the same radius. Pick the radius big plenty so that the arcs intersect at ii points, as in Figure 8. The line through those 2 points is the perpendicular bisector of the line segment. For the circumscribed circle of a triangle, you need the perpendicular bisectors of only two of the sides; their intersection will be the heart of the circumvolve.
Example ii
Find the radius R of the confining circle for the triangle △ABC where a = 2, b = 3, and c = 4. So describe the triangle and the circumvolve.
Solution: Using the Constabulary of Cosines, we can observe that A = 28.9º, so twoR = a sin A = 2 sin 28.9º = 4.xiv, so \(\boxed{R = ii.07}\) .
In Figure 10 we prove how to draw △ABC: apply a ruler to draw the longest side \(\overline{AB}\) of length c = four, so use a compass to draw arcs of radius 3 and 2 centered at A and B, respectively. The intersection of the arcs is the vertex C.
In Effigy 11 we show how to draw the circumscribed circle: depict the perpendicular bisectors of \(\overline{AB}\) and \(\overline{AC}\); their intersection is the center O of the circle. Use a compass to draw the circle centered at O which passes through A.
Theorem 2 tin be used to derive some other formula for the surface area of a triangle:
Theorem 3. For a triangle △ABC, allow M exist its area and let R exist the radius of its circumscribed circle. Then $$K = \frac{abc}{4R} (\text{and hence }R = \frac{abc}{4K} ) .$$
To prove this, note that past Theorem ii we have $$2R = \frac{a }{sin A} = \frac{b}{ sin B} = \frac{c}{ sin C} ⇒ sin A = \frac{a}{ 2R} , sin B = \frac{b}{ 2R} , sin C = \frac{c}{ 2R}.$$
Substitute those expressions into formula \(\text{Surface area = } K = \frac{a^2 sinB sin C}{2 sinA} \) for the surface area Grand:
$$K = \frac{a^ii sin B sin C}{ two sin A} = \frac{a^2 • \frac{b}{2R} • \frac{c}{2R}}{ 2 • \frac{a}{ 2R}} = \frac{abc}{ 4R}$$
Combining Theorem 3 with Heron'south formula for the area of a triangle, nosotros get:
Corollary 3. For a triangle △ABC, let s = \(\frac{1}{ 2}\) (a+b+ c). So the radius R of its circumscribed circle is $$R = \frac{abc}{ 4\sqrt{southward (s−a) (s−b) (s− c)}} .$$
In improver to a confining circumvolve, every triangle has an inscribed circle, i.e. a circle to which the sides of the triangle are tangent, as in Figure 12.
Let r be the radius of the inscribed circle, and permit D, Due east, and F be the points on \(\overline{AB}, \overline{BC}\), and \(\overline{Ac}\), respectively, at which the circle is tangent. Then \(\overline{OD}\) ⊥ \(\overline{AB}\), \(\overline{OE}\) ⊥ \(\overline{BC}\), and \(\overline{OF}\) ⊥ \(\overline{AC}\). Thus, △OAD and △OAF are equivalent triangles, since they are correct triangles with the same hypotenuse \(\overline{OA}\) and with respective legs \(\overline{OD}\) and \(\overline{OF}\) of the same length r. Hence, \(\angle\)OAD = \(\angle\)OAF, which means that \(\overline{OA}\) bisects the angle A. Similarly, \(\overline{OB}\) bisects B and \(\overline{OC}\) bisects C. Nosotros accept thus shown:
For any triangle, the center of its inscribed circle is the intersection of the bisectors of the angles.
We will utilize Figure 12 to discover the radius r of the inscribed circle. Since \(\overline{OA}\) bisects A, we meet that tan \(\frac{1}{ two} A = \frac{r}{ AD}\) , and then r = AD • tan \(\frac{one}{ 2}\) A. At present, △OAD and △OAF are equivalent triangles, so AD = AF. Similarly, DB = EB and FC = CE. Thus, if we allow s = \(\frac{1}{ 2}\) (a+b+ c), we see that
$$2s = a + b + c = (Ad+DB) + (CE+EB) + (AF +FC)$$ $$= AD + EB + CE + EB + AD + CE = 2(Ad+EB+CE)$$ $$s = AD + EB + CE = Advertizement + a$$ $$AD = due south−a .$$
Hence, r = (s−a) tan \(\frac{i}{ two}\) A. Similar arguments for the angles B and C give us:
Theorem 4. For any triangle △ABC, allow s = \(\frac{1}{ two}\) (a+b+c). And then the radius r of its inscribed circumvolve is $$r = (s−a) tan \frac{ane}{ 2} A = (s−b) tan \frac{ane}{ 2} B = (s− c) tan \frac{1}{ 2}C .$$
We also see from Figure 12 that the area of the triangle △AOB is $$Area(△AOB) = \frac{1}{ 2} base×top = \frac{1}{ 2} c r .$$
Similarly, Area(△BOC) = \(\frac{1}{ ii}\) a r and Surface area(△AOC) = \(\frac{one}{ two}\) b r. Thus, the expanse 1000 of △ABC is
$$G = Area(△AOB) + Area(△BOC) + Expanse(△AOC) = \frac{1}{ 2} c r + \frac{one}{ 2} a r + i two\frac{1}{ 2} b r$$
$$=\frac{one}{ ii} (a+b+ c) r = sr ,$$ so past Heron'south formula we get
$$r = \frac{K}{ south} = \frac{\sqrt{s(s−a) (s−b) (due south− c)}}{ s} = \sqrt{\frac{south(south−a)(s−b)(s− c)}{ s^2}} = \sqrt{\frac{south (southward−a) (due south−b) (s− c)}{ s}} .$$
We have thus proved the post-obit theorem:
Theorem 5. For any triangle △ABC, let s = \(\frac{1}{ two}\) (a+b+c). Then the radius r of its inscribed circumvolve is $$r = \frac{K}{ s} = \sqrt{\frac{due south (southward−a) (s−b) (due south− c)}{ due south}} .$$
Recall from geometry how to bisect an bending: utilise a compass centered at the vertex to draw an arc that intersects the sides of the angle at two points. At those 2 points use a compass to draw an arc with the same radius, large enough so that the two arcs intersect at a bespeak, every bit in Figure 13. The line through that bespeak and the vertex is the bisector of the angle. For the inscribed circumvolve of a triangle, yous need merely two angle bisectors; their intersection will exist the center of the circle.
Example 3
Find the radius r of the inscribed circle for the triangle △ABC where a = 2, b = 3, and c = 4. Draw the circumvolve.
Solution: Using Theorem 2.11 with s = \(\frac{ane}{ 2}\) (a+b+c)= \(\frac{1}{ 2}\) (2+3+four) = \(\frac{9}{ 2}\) , we have
$$r =\sqrt{\frac{s (due south−a) (s−b) (s− c)}{ southward}} = \sqrt{\frac{(\frac{9}{2} −ii) (\frac{ nine}{ 2} −3) (\frac{ 9}{ 2} −4) }{\frac{ 9}{ ii}}} = \sqrt{\frac{5}{ 12}} .$$
Figure 14 shows how to depict the inscribed circle: draw the bisectors of A and B, then at their intersection use a compass to depict a circle of radius r = \(\sqrt{5/12}\) ≈ 0.645.
Source: https://opencurriculum.org/5492/circumscribed-and-inscribed-circles/
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