Can you find Radius of the circumscribed circle? | (Isosceles Triangle) |
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9 күн бұрынLearn 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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Can you find Radius of the circumscribed circle? | (Isosceles Triangle) | #math #maths | #geometry
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@yalchingedikgedik8007 7 күн бұрын
Thanks PreMath Thanks prof. Very good Two methods are nice With glades ❤❤❤❤
@PreMath 7 күн бұрын
Always welcome dear🌹 Glad to hear that! Thanks for the feedback ❤️
@ramazanakcan4878 3 күн бұрын
Gelek spas ji bo bersivên hêja.
@michaelkouzmin281 8 күн бұрын
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.
@PreMath 8 күн бұрын
Excellent! Thanks for sharing ❤️
@albertomontori2863 7 күн бұрын
this was the first method i was thinking 😂
@anatoliy3323 7 күн бұрын
The first method is more descriptive while the second one is simpler. Thank you for your math lesso, Professor
@PreMath 5 күн бұрын
Very good! Thanks for the feedback ❤️
@techeteri Күн бұрын
Cool!
@raya.pawley3563 7 күн бұрын
Thank you
@PreMath 7 күн бұрын
You are very welcome! Thanks for the feedback ❤️
@user-sk9oi9jl2g 7 күн бұрын
R=a*b*c/(4*S). Треугольник равнобедренный - площадь найти легко. S=15*16/2=120. R=17*17*16/(4*120)=289/30.
@PreMath 7 күн бұрын
Excellent! Thanks for sharing ❤️
@hongningsuen1348 8 күн бұрын
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.
@PreMath 8 күн бұрын
Excellent! Thanks for sharing ❤️
@marioalb9726 8 күн бұрын
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 √ )
@AmirgabYT2185 8 күн бұрын
r=289/30≈9,63
@PreMath 8 күн бұрын
Excellent! Thanks for sharing ❤️
@marioalb9726 7 күн бұрын
Cosine rule for isosceles triangle: c² = 2a² (1 - cos α )
@PreMath 7 күн бұрын
Excellent! Thanks for sharing ❤️
@h4xe7reac75 8 күн бұрын
First perhaps?
@PreMath 8 күн бұрын
Excellent! Thanks dear❤️
@giuseppemalaguti435 8 күн бұрын
r=17*17*16/4√(25*8*8*9)=289*4/5*8*3=289/30
@PreMath 8 күн бұрын
Excellent! Thanks for sharing ❤️
@michaelstahl1515 7 күн бұрын
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 .
@PreMath 7 күн бұрын
Excellent! Thanks for the feedback ❤️
@devondevon4366 6 күн бұрын
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
@PreMath 5 күн бұрын
Excellent! Thanks for sharing ❤️
@wackojacko3962 7 күн бұрын
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. 🙂
@PreMath 7 күн бұрын
Good point! Thanks for the feedback ❤️
@marcgriselhubert3915 8 күн бұрын
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)
@PreMath 7 күн бұрын
Excellent! Thanks for sharing ❤️ Thanks for the feedback ❤️
@Waldlaeufer70 7 күн бұрын
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
@PreMath 7 күн бұрын
Excellent! Thanks for sharing ❤️
@PrithwirajSen-nj6qq 7 күн бұрын
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)
@PreMath 7 күн бұрын
Excellent! Thanks for sharing ❤️
@prossvay8744 8 күн бұрын
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.❤❤❤
@PreMath 8 күн бұрын
Excellent! Thanks for sharing ❤️
@MrPaulc222 7 күн бұрын
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.
@PreMath 7 күн бұрын
Excellent! Thanks for sharing ❤️
@himo3485 8 күн бұрын
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
@PreMath 7 күн бұрын
Excellent! Thanks for sharing ❤️
@jamestalbott4499 7 күн бұрын
Thank you!
@PreMath 7 күн бұрын
You are very welcome! Thanks for the feedback ❤️
@lasalleman6792 7 күн бұрын
Or use formula: Circumradius = side/2sin of opposite angle. Here I use AB as side and ACB for angle. I get 9.6385
@PreMath 7 күн бұрын
Thanks for the feedback ❤️
@adept7474 7 күн бұрын
3 method: R = abc/4S = 17×17×16/4×8×15 = 289/30.
@PreMath 7 күн бұрын
Excellent! Thanks for the feedback ❤️
@sergeyvinns931 8 күн бұрын
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).
@PreMath 8 күн бұрын
Excellent! Thanks for sharing ❤️
@georgebliss964 8 күн бұрын
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.
@PreMath 8 күн бұрын
Excellent! Thanks for sharing ❤️
@user-yx9kr8ur5q 8 күн бұрын
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
@PreMath 8 күн бұрын
Excellent! Thanks for sharing ❤️
@tellerhwang364 8 күн бұрын
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😊
@PreMath 8 күн бұрын
289/30 typo! Thanks ❤️
@jphilsol6459 6 күн бұрын
hello all, I find the same result with a third method using trigonometry, with inscribed angle theorem.
@PreMath 5 күн бұрын
Good job! Thanks for the feedback ❤️
@nenetstree914 8 күн бұрын
289/30
@PreMath 7 күн бұрын
Excellent! Thanks for sharing ❤️
@devondevon4366 6 күн бұрын
9.63333
@marcelowanderleycorreia8876 7 күн бұрын
Another solution is use the law os sines.
@PreMath 7 күн бұрын
Thanks for the feedback ❤️
@ptbx6986 3 күн бұрын
Don't you have to prove line CP is perpendicular to line AB before applying Perpendicular Bisector Theorem?
@unknownidentity2846 7 күн бұрын
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
@phungpham1725 7 күн бұрын
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😅
@PreMath 7 күн бұрын
Excellent! Thanks for sharing ❤️
@LuisdeBritoCamacho 8 күн бұрын
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.
@PreMath 7 күн бұрын
Bravo👍 Thanks for sharing ❤️🙏
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