Can you find Radius of the circumscribed circle? | (Isosceles Triangle) |
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- Опубликовано: 5 июл 2024
- Learn how to find the Radius of the circumscribed circle. Important Geometry and Algebra skills are also explained: Pythagorean Theorem; Intersecting Chords theorem; Perpendicular bisector theorem; Isosceles Triangle. Step-by-step tutorial by PreMath.com
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Thanks PreMath
Thanks prof.
Very good
Two methods are nice
With glades
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Always welcome dear🌹
Glad to hear that!
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Superb
Just another solution:
1. 2 ways to calculate the area of a triangle inscribed in a circumference:
A=sqrt(p(p-a)(p-b)(p-c)) where p= (a+b+c)/2 = Heron's formula
A= abc/(4R)
2. "Filling the blanks"
p=(17+17+16/2) = 25;
A=sqrt(25(25-17)(25-17)(25-16))= sqrt(25*8*8*9) =120 sq units;
R=abc/(4)= 17*17*16/(4*120) = 289/30 sq units.
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this was the first method i was thinking 😂
The first method is more descriptive while the second one is simpler. Thank you for your math lesso, Professor
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Cool!
R=a*b*c/(4*S). Треугольник равнобедренный - площадь найти легко. S=15*16/2=120. R=17*17*16/(4*120)=289/30.
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Radius of circumcircle = product of 3 sides/(4 x area of trianlge)
Get area of triangle by Heron's formula = 120
radius = (17)(17)(16)/(4)(120)= 289/30.
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Cosine rule for isosceles triangle:
c² = 2a².(1 - cos α)
cos α = 1 - ½ c²/a² = 1 - ½16²/17²
α = 56,145°
Cosine rule again:
c² = 2R² ( 1 - cos 2α )
R² = ½.c² / ( 1 - cos 2α )
R² = ½ 16² / (1 - cos 2α )
R = 9,63 cm ( Solved √ )
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r=289/30≈9,63
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Gelek spas ji bo bersivên hêja.
Cosine rule for isosceles triangle:
c² = 2a² (1 - cos α )
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r=17*17*16/4√(25*8*8*9)=289*4/5*8*3=289/30
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289/30 or 9.6333
Draw a perpendicular line through the circle' center to form two congruent triangles with sides 17, 8, and 15
since the triangle is an isosceles.
Draw a line from the vertex of the yellow triangle to the circle's center. This is 'r'.
Draw a line from the circle's center to the triangle's base, forming a new triangle A P R.
AP = the circle's radius
PR = 15-r
AR= 8
Using Pythagorean 8^2 + ( 15-r)^2 = r^2
64 + 225-30r+r^2 = r^2
289 -30r =0
289=30r
r = 289/30
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Thank you!
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You got a like again . I chosed the first way of solution and got the same result. I´m proud to sea your second way becaused I forgot the Intersect chords theorem. I `m sure I`ll never
verget it .
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Or use formula: Circumradius = side/2sin of opposite angle. Here I use AB as side and ACB for angle. I get 9.6385
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Cool! Deconstructed the triangle too extract the radius. Prime example of reverse engineering. Only problem is one method gives an equal amount of units but the 2nd method an approximation. One engineer would design a normal toilet seat with a lid on top. Another engineer would design with lid on the bottom. 🙂
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Use 16 as the base, the way the image shows it.
h is sqrt(17^2 - 8^2) so sqrt(289 - 64) = sqrt(225) = 15
intersecting chords: 8*8 = 15x
x = 64/15
d = 15 + 64/15 = 289/15
r is half that, so 289/30 = r
r is 9 and 19/30 = 9.633333....
I went for intersecting chords but did it as 64 = 15x.
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Height CP = sq.rt of 17^2 - 8^2 by Pythagoras.
CP = 15.
Draw perpendicular from O onto CB at point E.
Triangles CPB & COE are similar.
Thus CB /CP = CO /CE.
CO = r & CE = 8.5.
17 / 15 = r / 8.5.
r = 17 x 8.5 / 15.
r = 9.63.
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The formula for the circumradius of a triangle with sides of lengths a, b, and c is (a*b*c) / sqrt((a + b + c)(b + c - a)(c + a - b)(a + b - c)) so for a triangle with sides a = 16, b = 17, c =17, Circumradius: R = 9.633
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Half base = 8 units
8² + x² = 17²
x² = 17² - 8² = (17 + 8) (17 - 8) = 25 * 9 = 225
x = 15
Intersecting chord theorem:
x * y = 8 * 8
15 y = 64
y = 64/15
d = 15 + 64/15 = 225/15 + 64/15 = 289/15
r = 1/2 * 289/15 = 289/30 ≈ 9.63 units
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Let's use an orthonormal, center P, middle of [A, B] and first axis (PB).
In triangle PAC: PC^2 = AC^2 - AP^2 = 289 -64 = 225 = 15^2, so PC = 15
Then A( -8;0) B(8, 0) and C(0; 15).
The equation of the circle is x^2 + y^2 +a.x +b.y + c = 0
A is on the circle, so: 64 -8.a + c = 0
B is on the circle, so: 64 +8.a + c = 0
C is on the circle, so: 225 +15.b + c = 0
It is easy to obtain: a = 0; b = -161/15; c = - 64
The equation of the circle is x^2 + y^2 -(161/15).y -64 = 0
or x^2 + (y -(161/30))^2 = (161/15)^2 + 64 = 83521/900
So the radius of the circle is sqrt(83521/900) = 289/30.
(We also have the coordinates of O: O(0; 161/30)
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R=a*b*c/4A, A (area triangle)=\/р*(р-а)*(р-b)*(p-c)=\/25*8*8*9=5*8*3=120, R=16*17*17/480=4624/480=289/30=9,63(3).
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Area of the isosceles 🔺
= 1/2*16*√(17^2- 8^2)
=120 sq units
Radius =17*17*16/4*120
= 17*17/30= 9.63 units (approx)
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3 method: R = abc/4S = 17×17×16/4×8×15 = 289/30.
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16/2=8
√[17^2 - 8^2] = √225 = 15
r^2 = 8^2 + (15 - r)^2
r^2 = 64 + 225 - 30r + r^2
30r = 289 r = 289/30
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cos(x)=(17^2+17^2-16^2)/2(17^2)=56.14°
cos(2(56.14°)=(r^2+r^2-16^2)/2(r^2)
so r=9.63 units.❤❤❤
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hello all, I find the same result with a third method using trigonometry, with inscribed angle theorem.
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9.63333
1.△CAP~△CDB(AA)
(1)circumferential angle
→CAP=CDB
(2)APC=DBC=90
2.AC:CD=CP:CB (CD=2R)
→17:2R=15:17→R=289/30😊
option:
R=AC·BC/2CP😊
289/30 typo!
Thanks ❤️
289/30
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Let's find the radius:
.
..
...
....
.....
Let M be the midpoint of AB. Since the triangle ABC is an isosceles triangle (AC=BC), the triangles ACM and BCM are congruent right triangles and we can apply the Pythagorean theorem:
AC² = CM² + AM²
AC² = CM² + (AB/2)²
17² = CM² + (16/2)²
17² = CM² + 8²
289 = CM² + 64
225 = CM²
⇒ CM = 15
The center O of the circumscribed circle is the point where the perpendicular bisectors of all three sides of the triangles intersect. Therefore O is located on CM and the triangles AOM and BOM are also congruent right triangles. By applying the Pythagorean theorem again we can obtain the radius R of the circumscribed circle:
AO² = OM² + AM²
AO² = (CM − CO)² + AM²
R² = (CM − R)² + AM²
R² = CM² − 2*CM*R + R² + AM²
2*CM*R = CM² + AM²
2*CM*R = AC²
⇒ R = AC²/(2*CM) = 17²/(2*15) = 289/30
Best regards from Germany
It is more fun to have another solution😊!
Label the angle ACB as 2 alpha so, sin (alpha) = 8/17 and cos(alpha)= 15/17 and
-> sin (2alpha)=2 sin (alpha) (cos alpha)=240/289
1/ CP intersects the circle at point D.
Area of the triangle ACD=1/2 xACxCDxsin alpha=1/2 x17x2rx8/17=8r--> area of the quadrilateral ACBD=16r
2/ Focus on the triangle ACD, just build another isoceles triangle by extend AD (to the left) a segment AD’ = AD
We have: the area of D’CD= area of (ACBD)=1/2 sq(2r).sin (2alpha)
--> 1/2 sq(2r).sin 2pha=16r
--> sqr . 240/289=16r
--> r =289/30=9.63 units😅
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Don't you have to prove line CP is perpendicular to line AB before applying Perpendicular Bisector Theorem?
STEP-BY-STEP RESOLUTION PROPOSAL :
01) Let's define the Middle Point between Point A and Point B, as Point D.
02) AD = BD = 8
03) Finding the Height (h) of given Isosceles Triangle (ABC) :
04) h^2 = 17^2 - 8^2 ; h^2 = 289 - 64 ; h^2 = 225 ; h = 15
05) h = CD = 15
06) OD = 15 - R
07) OB = R
08) BD = 8
09) OD^2 + BD^2 = OB^2
10) (15 - R)^2 + 64 = R^2 ; R^2 = 225 - 30R + R^2 + 64 ; 289 - 30R = 0 ; 289 = 30R ; R = 289 / 30 lin un ; R ~ 9,6(3) lin un
Thus,
OUR ANSWER : The Radius is equal to 289/30 Linear Units or approx. equal to 9,6(3) Linear Units.
Best Regards form Cordoba Caliphate - Universal Islamic Institute for Study of Ancient Knowledge, Thinking and Wisdom. Department of Mathematics and Geometry.
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Another solution is use the law os sines.
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