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

[@yalchingedikgedik8007]

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

[@PreMath]

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

[@raghvendrasingh1289]

​@@PreMathanother method sir triangles CAP and CDB are similar (as per diagram) because angle CAP is equal to angle CDB ( angles of same segment) and angle CPA is equal to angle CBD ( each being right angle) hence CA/CD = CP/CB 17/2 R = 15/17 R = 289/30

[@sagardeshpande2092]

Superb

[@ramazanakcan4878]

Gelek spas ji bo bersivên hêja.

[@michaelkouzmin281]

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]

Excellent! Thanks for sharing ❤️

[@albertomontori2863]

this was the first method i was thinking 😂

[@anatoliy3323]

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

[@PreMath]

Very good! Thanks for the feedback ❤️

[@raya.pawley3563]

Thank you

[@PreMath]

You are very welcome! Thanks for the feedback ❤️

[@ОльгаСоломашенко-ь6ы]

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]

Excellent! Thanks for sharing ❤️

[@devondevon4366]

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]

Excellent! Thanks for sharing ❤️

[@marioalb9726]

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

[@PreMath]

Excellent! Thanks for sharing ❤️

[@hongningsuen1348]

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]

Excellent! Thanks for sharing ❤️

[@marioalb9726]

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 √ )

[@marcgriselhubert3915]

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]

Excellent! Thanks for sharing ❤️ Thanks for the feedback ❤️

[@georgebliss964]

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]

Excellent! Thanks for sharing ❤️

[@techeteri]

Cool!

[@Waldlaeufer70]

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]

Excellent! Thanks for sharing ❤️

[@AdemolaAderibigbe-j8s]

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]

Excellent! Thanks for sharing ❤️

[@PrithwirajSen-nj6qq]

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]

Excellent! Thanks for sharing ❤️

[@sergeyvinns931]

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]

Excellent! Thanks for sharing ❤️

[@MrPaulc222]

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]

Excellent! Thanks for sharing ❤️

[@jamestalbott4499]

Thank you!

[@PreMath]

You are very welcome! Thanks for the feedback ❤️

[@lasalleman6792]

Or use formula: Circumradius = side/2sin of opposite angle. Here I use AB as side and ACB for angle. I get 9.6385

[@PreMath]

Thanks for the feedback ❤️

[@himo3485]

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]

Excellent! Thanks for sharing ❤️

[@adept7474]

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

[@PreMath]

Excellent! Thanks for the feedback ❤️

[@giuseppemalaguti435]

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

[@PreMath]

Excellent! Thanks for sharing ❤️

[@wackojacko3962]

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]

Good point! Thanks for the feedback ❤️

[@michaelstahl1515]

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]

Excellent! Thanks for the feedback ❤️

[@prossvay8744]

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]

Excellent! Thanks for sharing ❤️

[@tellerhwang364]

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]

289/30 typo! Thanks ❤️

[@AmirgabYT2185]

r=289/30≈9,63

[@PreMath]

Excellent! Thanks for sharing ❤️

[@jphilsol6459]

hello all, I find the same result with a third method using trigonometry, with inscribed angle theorem.

[@PreMath]

Good job! Thanks for the feedback ❤️

[@unknownidentity2846]

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]

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]

Excellent! Thanks for sharing ❤️

[@unknownidentity2846]

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

[@ptbx6986]

Don't you have to prove line CP is perpendicular to line AB before applying Perpendicular Bisector Theorem?

[@josephsalinas6725]

Eu calculei assim: S=abc/4R

[@nenetstree914]

289/30

[@PreMath]

Excellent! Thanks for sharing ❤️

[@devondevon4366]

9.63333

[@marcelowanderleycorreia8876]

Another solution is use the law os sines.

[@PreMath]

Thanks for the feedback ❤️

[@LuisdeBritoCamacho]

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]

Bravo👍 Thanks for sharing ❤️🙏