Can you find Radius of the circumscribed circle?

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Can you find Radius of the circumscribed circle? | (Isosceles Triangle) |

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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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Can you find Radius of the circumscribed circle? | (Isosceles Triangle) | #math #maths | #geometry
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Пікірлер: 71

@yalchingedikgedik8007 Ай бұрын

Thanks PreMath Thanks prof. Very good Two methods are nice With glades ❤❤❤❤

@PreMath Ай бұрын

Always welcome dear🌹 Glad to hear that! Thanks for the feedback ❤️

@sagardeshpande2092 20 күн бұрын

Superb

@ramazanakcan4878 Ай бұрын

Gelek spas ji bo bersivên hêja.

@michaelkouzmin281 Ай 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 Ай бұрын

Excellent! Thanks for sharing ❤️

@albertomontori2863 Ай бұрын

this was the first method i was thinking 😂

@anatoliy3323 Ай бұрын

The first method is more descriptive while the second one is simpler. Thank you for your math lesso, Professor

@PreMath Ай бұрын

Very good! Thanks for the feedback ❤️

@raya.pawley3563 Ай бұрын

Thank you

@PreMath Ай бұрын

You are very welcome! Thanks for the feedback ❤️

@user-sk9oi9jl2g Ай 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 Ай бұрын

Excellent! Thanks for sharing ❤️

@hongningsuen1348 Ай 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 Ай бұрын

Excellent! Thanks for sharing ❤️

@marioalb9726 Ай 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 √ )

@techeteri Ай бұрын

Cool!

@marioalb9726 Ай бұрын

Cosine rule for isosceles triangle: c² = 2a² (1 - cos α )

@PreMath Ай бұрын

Excellent! Thanks for sharing ❤️

@giuseppemalaguti435 Ай бұрын

r=17*17*16/4√(25*8*8*9)=289*4/5*8*3=289/30

@PreMath Ай бұрын

Excellent! Thanks for sharing ❤️

@devondevon4366 Ай 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 Ай бұрын

Excellent! Thanks for sharing ❤️

@AmirgabYT2185 Ай бұрын

r=289/30≈9,63

@PreMath Ай бұрын

Excellent! Thanks for sharing ❤️

@Waldlaeufer70 Ай 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 Ай бұрын

Excellent! Thanks for sharing ❤️

@michaelstahl1515 Ай 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 Ай бұрын

Excellent! Thanks for the feedback ❤️

@LuisdeBritoCamacho Ай 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 Ай бұрын

Bravo👍 Thanks for sharing ❤️🙏

@lasalleman6792 Ай 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 Ай бұрын

Thanks for the feedback ❤️

@jamestalbott4499 Ай бұрын

Thank you!

@PreMath Ай бұрын

You are very welcome! Thanks for the feedback ❤️

@georgebliss964 Ай 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 Ай бұрын

Excellent! Thanks for sharing ❤️

@marcgriselhubert3915 Ай 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 Ай бұрын

Excellent! Thanks for sharing ❤️ Thanks for the feedback ❤️

@MrPaulc222 Ай 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 Ай бұрын

Excellent! Thanks for sharing ❤️

@PrithwirajSen-nj6qq Ай 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 Ай бұрын

Excellent! Thanks for sharing ❤️

@himo3485 Ай 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 Ай бұрын

Excellent! Thanks for sharing ❤️

@user-yx9kr8ur5q Ай 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 Ай бұрын

Excellent! Thanks for sharing ❤️

@adept7474 Ай бұрын

3 method: R = abc/4S = 17×17×16/4×8×15 = 289/30.

@PreMath Ай бұрын

Excellent! Thanks for the feedback ❤️

@prossvay8744 Ай 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 Ай бұрын

Excellent! Thanks for sharing ❤️

@sergeyvinns931 Ай 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 Ай бұрын

Excellent! Thanks for sharing ❤️

@wackojacko3962 Ай 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 Ай бұрын

Good point! Thanks for the feedback ❤️

@tellerhwang364 Ай 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 Ай бұрын

289/30 typo! Thanks ❤️

@unknownidentity2846 Ай 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 Ай 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 Ай бұрын

Excellent! Thanks for sharing ❤️

@unknownidentity2846 5 күн бұрын

@@phungpham1725 You are absolutely right. Today I finally took the time to understand your solution. Indeed a very interesting approach.👍 Best regards from Germany