Can you find the radius? | (Two Methods) |

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Can you find the radius? | (Two Methods) |

Рет қаралды 13,738

3 ай бұрын

Learn how to find the radius of the Circle. Important Geometry and Algebra skills are also explained: Intersecting Chords theorem; perpendicular bisector theorem; Pythagorean theorem. Step-by-step tutorial by PreMath.com.
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• Can you find the radiu...
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Can you find the radius? | (Two Methods) | #math #maths | #geometry
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Пікірлер: 50

@k.amanda.m 3 ай бұрын

I always amazed to these beautiful solutions.. likee wow!!

@PreMath 3 ай бұрын

Glad you like them! Thanks for the feedback ❤️

@jamestalbott4499 3 ай бұрын

Following PreMath, there is a theme of using Algebra & proportions, re-enforcing the basics and compelling you to think "outside the box". Each problem continues to answer the question "Why and how would I apply this?". His classroom must be an incredible place to be! Thank you!

@PreMath 3 ай бұрын

Glad to hear that! You are very welcome! Thanks for the feedback ❤️🙏

@phungpham1725 3 ай бұрын

1/ Aternative 3rd approach: AMD is a 3-4-5 triple. Just drop the perpendicular OH to chord DM. We have HM= 15/2 and the triangle OHM is also a 3-4-5 triple( angle HOM= angle DMA) So HM/0M= 3/5--> OM= 5/3 . HM --> r= 5/3 . 15/2= 75/6= 12.5 units

@PreMath 3 ай бұрын

Excellent! Thanks for sharing ❤️

@unknownidentity2846 3 ай бұрын

Let's find the radius: . .. ... .... ..... Since ADM is a right triangle, we can apply the Pythagorean theorem: DM² = AD² + AM² DM² = AD² + (AB/2)² 15² = AD² + (24/2)² 15² = AD² + 12² 3²*5² = AD² + 3²*4² ⇒ AD² = 3²*3² ⇒ AD = 3*3 = 9 Let N be the midpoint of CD. Then we are able to obtain the radius R of the circle by applying the intersecting chords theorem: DN*CN = (R + ON)*NM (CD/2)*(CD/2) = (R + R − NM)*NM (CD/2)² = (2*R − NM)*NM (AB/2)² = (2*R − AD)*AD AM² = (2*R − AD)*AD 12² = (2*R − 9)*9 144 = (2*R − 9)*9 16 = 2*R − 9 25 = 2*R ⇒ R = 25/2 = 12.5 Best regards from Germany

@PreMath 3 ай бұрын

Excellent! Thanks for sharing ❤️

@prossvay8744 3 ай бұрын

AM=BM=24/2=12 DM^2=AM^2+AD^2 15^2=12^2+AD^2 AD=√225-144=9 Connect D to C let R is Radius of circle 12^2+(R-9)^2=R^2 So R=25/2units=12.5 unitd.❤❤❤ Thanks sir.

@PreMath 3 ай бұрын

Excellent! You are very welcome! Thanks for sharing ❤️

@netravelplus 3 ай бұрын

Excellent. We are blessed to have a wonderful teacher in you, Sir!

@PreMath 3 ай бұрын

So nice of you dear❤️ Thanks for the feedback ❤️

@murdock5537 3 ай бұрын

φ = 30°; DO = MO = r; ∆ DPM → DM = 15; DP = 12; sin⁡(DPM) = sin⁡(3φ) = 1 → PM = 9 → MDP = DNM = δ/2 → sin⁡(δ/2) = 3/5 → cos⁡(δ/2) = 4/5 → cos⁡(δ) = cos^2(δ/2) - sin^2(δ/2) = 7/25 → DOM = δ → 225 = 2r^2(1 - cos⁡(δ)) → r = 25/2

@PreMath 3 ай бұрын

Excellent! Thanks for sharing ❤️

@LuisdeBritoCamacho 3 ай бұрын

Let's go adventuring!! 1) Close the Rectangle [ABCD] 2) OM = R 3) M' is the Point of interception of Line OM and Line CD. 4) OM' = OM - MM' ; OM = R - 9. MM' = AD and AD^2 = 15^2 - 12^2 ; AD^2 = 225 - 144 ; AD^2 = 81 ; AD = 9 Now : 5) (X - 9)^2 + 12^2 = X^2 ; X^2 - 18X + 81 + 144 = X^2 ; 18X = 225 ; X = 225/18 ; X = 75/6 ; X = 25/2 6) My Best Answer is : The Radius of the Circle is equal to 25/2 Linear Units or 12,5 Linear Units.

@PreMath 3 ай бұрын

Excellent! Thanks for sharing ❤️

@marcgriselhubert3915 3 ай бұрын

We use an orthonormal center O and first axis (AB). We have A(-12; R) D(-12; sqrt(R^2 -144)) M(0;R) VectorDM(-12; sqrt(R^2 -144) -R) So, DM^2 = 144 + (R^2 -144 +R^2 -2.R.sqrt(R^2 -144) = 2.R^2 -2.R.sqrt(R^2 -144) and we have the equation 2.R^2 -2.R.sqrt(R^2 -144) = 15^2 = 225 2.R.sqrt(R^2 -14) = 2.R^2 -225. We square: 4.R^2.(R^2 -144) = 4.R^4 -900.R^2 + 225^2 , we simplify: (900 -576).R^2 = 225^2 or 324.R^2 = 225^2 Finally R = 225/sqrt(324) = 225/18 = 25/2.

@Waldlaeufer70 3 ай бұрын

3rd method: Triangle 9-12-15 Tangent-chord theorem: 9x = 12^2 x = 144/9 = 16 d(circle) = 16 + 9 = 25 r = 12.5 units

@phungpham1725 3 ай бұрын

4th method: Build the diameter MOM”= d ---> the triangle MDM’ is a right triangle and similar to DAM. So, d/DM= DM/DA--> d/15=15/9 d=25-> r=12.5

@Waldlaeufer70 3 ай бұрын

Very nice solution! :)

@phungpham1725 3 ай бұрын

@@Waldlaeufer70thank you! Yours is very nice, too!

@ChuzzleFriends 3 ай бұрын

Draw segment CD. This is a side of quadrilateral ABCD while also being a chord of ⊙O. Sides AD & BC are both perpendicular to side AB. By the Lines Perpendicular to a Transversal Theorem, they are parallel. So, ABCD is a parallelogram by the Opposite Sides Parallel & Congruent Theorem. You can then use the Parallelogram Opposite Angles Theorem to prove ABCD is also a rectangle by the Rectangle Corollary. By the Parallelogram Opposite Sides Theorem (ABCD is a rectangle), CD = 24. Draw radius MO. By the Circle Theorem, segments AB & MO are perpendicular. So, radius MO is also perpendicular to chord CD. Label the intersection of segments CD & MO as E. So, by the Perpendicular Chord Bisector Theorem, CE = DE = 12. Find EM. Use knowledge of Pythagorean Triples. (EM, 12, 15) 3 * (3, 4, 5) (3 * 3, 3 * 4, 3 * 5) (9, 12, 15) EM = 9. Draw radius CO. Apply the Pythagorean Theorem on △CEO. a² + b² = c² (r - 9)² + 12² = r² r² - 18r + 81 + 144 = r² -18r + 225 = 0 -18r = -225 r = 25/2 = 12.5 So, the radius of the circle is 12.5 units.

@Nothingx303 3 ай бұрын

Thank you 😊 ❤ sir I used the first method but I like the second method too 😊😊

@marcelowanderleycorreia8876 3 ай бұрын

I solves using the law of cossines plus similarity of triangles. But yous first method is "the killer"!! Congratulations professor!!

@giuseppemalaguti435 3 ай бұрын

AD=a...x+y=24...arctg a/x+arccos7,5/r=90...arctg a/y+arctg√(a^2+y^2)/2r=90..risultano x=y=12,a=9,r=12,5

@MMmaths8800 3 ай бұрын

Amazing sir

@PreMath 3 ай бұрын

Glad to hear that! Many many thanks dear ❤️

@MrPaulc222 3 ай бұрын

I went for intersecting chords: AD = 9 due to 9,12,15. DC = 24. The chords are 12*12 = 9(2r-9) 144 = 18r - 81 18r = 225 2r = 25 so r = 12.5

@robertlynch7520 3 ай бұрын

You missed a "you can do it in your head" 3rd method. Line DM can be divided in half; the perpendicular from that point intersects the center of the circle. As does the endpoint M. Forms a similar triangle to the 9:12:15 (3:4:5) triangle because of complimentary angles. So … ½ (DM = 15) • (5 ÷ 3) = 12.5 … which is the radius. Tada.

@PreMath 3 ай бұрын

Super! You are awesome! Thanks for the feedback ❤️

@LeonardEwata-kelle-vq8kk 3 ай бұрын

Thanks so much this gave me a hard tym in schl buh you introduced me to a new and easy formula thanks

@emachinetool 3 ай бұрын

I have a question on the numbers in diagram. OM = OD = r radius. Triangle MOD is an isosceles triangle, therefore sides are in ratio 1, 1, Sq rt 2. Since MD = 15 units, r must be 15/(square rt 2) units or about 10.7 units. I think numbers are inconsistent?