Can you find the Radius of the circle? | (3 Methods) |

[@Piyushkumar100-j3n]

Nice explaination❤

[@PreMath]

Glad you liked it Thanks for the feedback ❤️

[@Marcus-y1m]

strategical solutions , Very amazing . Thanks Sir

[@PreMath]

Most welcome dear🌹 Thanks for the feedback ❤️

[@trumpetbob15]

It was interesting to see Euclid's Theorem right after your second method that basically shows a proof of it using the circle. I've never thought about it and why it works like that before.

[@PreMath]

Excellent! Thanks for the feedback ❤️

[@jimlocke9320]

For a fourth method, we note that ΔABC and ΔACE are similar by angle - angle (common angle

[@PreMath]

Excellent! Thanks for the feedback ❤️

[@hongningsuen1348]

Point E is redunctant for solution as shown by the following method using trigonometric ratios: 1. Right-angled triangle ABC has sides ratios of 5:12:13. (15 and 36 given, 39 found by Pythagoras theorem) For angle BAC, sin (BAC) = 5/13, cos(BAC) = 12/5 2. Angle BOC = 2 x angle BAC (exterior angle of isosceles triangle AOC) Hence Sin (BOC) = 2 sin (BAC) cos (BAC) (double angle formula for sin function) = (2)(5/13)(12/13) = 120/169 3. Radius = OC = BC/sin (BOC) = (15)/(120/169) = 169/8.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@quigonkenny]

Rectangle ABCD: A = hw 540 = 15AB AB = 540/15 = 36 First method: Triangle ∆OBC: OB² + BC² = CO² (36-r)² + 15² = r² 1296 - 72r + r² + 225 = r² 72r = 1521 r = 1521/72 = 169/8 cm Second method: Extend AB to the circumference at E. Let BE = x. Extend CB to the circumference at F. As ∠OBC = 90°, BF = BC = 15, as if a radius of a circle intersects a chord perpendicularly, it bisects that chord. For any two intersecting chords of a circle, the products of their lengths on either side of the intersection point are the same. Thus: AB•BE = CB•BF 36x = 15(15) = 225 x = 225/36 = 25/4 AE = AB + BE 2r = 36 + 25/4 = 169/4 r = (169/4)/2 = 169/8 cm Third method: Let G be the point where DC intersects with the circumference. Extend CB to F. As ∠OBC = 90°, BF = BC = 15, as if a radius of a circle intersects a chord perpendicularly, it bisects that chord. Draw radius OP so that OP intersects GC perpendicularly at H. By the same rule as above, GH = HC, as OP bisects the chord GC. As ∠OHC = ∠HCB = ∠CBO = 90°, ∠BOH must also equal 90°, and OHCB is a rectangle. Therefore GC = 2OB, as OB = HC = GH. Draw diameter GF. As G, F, and C, are point on the circumference and ∠GCF is a 90° angle connecting them all, G and F must be opposite ends of a diameter of the circle. Thus GF and O are collinear. Triangle ∆GCF: GC² + CF² = GF² (2(36-r))² + (2(15))² = (2r)² 4(36-r)² + 4(15)² = 4(r²) (36-r)² + 15² = r² 1296 - 72r + r² + 225 = r² 72r = 1521 r = 1521/72 = 169/8 cm

[@PreMath]

Excellent job! Thanks for sharing ❤️

[@raya.pawley3563]

Thank you

[@PreMath]

You are very welcome! Thanks Raya❤️

[@phungpham1725]

4th approach: 1/ DC intersects the circle at point F. Let DF = x By using the tangent theorem: DF.DC= sq AD--> x= sq AD/DC= 225/36= 6.25 If we drop the height FH to the diameter we have AH= x= 6.25 Notice that the AFCE is a isosceles trapezoid so AH=BE= 6.25 --> AE= 2r = 36+6.25 r = 21.125 cm

[@PreMath]

Excellent! Thanks for sharing ❤️

[@krishnaramachandran7722]

When I looked at the rectangle and the tangent DA to the circle, I thought the first method would be the 4th method.

[@alster724]

Straightforward problem

[@PreMath]

Excellent! Thanks for the feedback ❤️

[@spelunkerd]

For some reason I always choose the longest path to the solution, or at least the only one needing a calculator. In this case I found angle BAC from arctan (15/36). From the inscribed angle theorem I knew angle EOC is double that angle, exactly the same as angle BOC. I calculated OC to be 15/(sin(angle BOC)).

[@himo3485]

Pythagorean Theorem : AB=540/15=36cm r²=(36-r)²+15² r²=1296-72r+r²+225 72r=1521 r=21.125cm Chords Theorem : AB=540/15=36cm 15*15=36*(2r-36) 225=72r-1296 72r=1521 r=21.125cm

[@PreMath]

Excellent! Thanks for sharing ❤️

[@MrPaulc222]

After looking at the diagram for a while and doing a few calculations in my head I came up with 15*15 = 36(2r - 36) as intersecting chords. 225 = 72r - 1296 1521 = 72r They both divide by 3 so 24r = 507 Again: 8r = 169 So r= 169/8 That is 21 and 1/8 so 21.125 This shows me that even if something looks slightly complex, it is sometimes possible to work without a calculator

[@PreMath]

Excellent job! Thanks for the feedback ❤️

[@AbdulJaleel-rd5ul]

For a triangle inscribed in a circle , intersecting chord theorem and Euclid's formula is same.

[@CloudBushyMath]

Cool👍

[@PreMath]

Glad to hear that! Thanks for the feedback ❤️

[@ChuzzleFriends]

A = lw 540 = 15l l = 36 So, the length of rectangle ABCD is 36 cm. Draw the radius of ⊙O containing vertex B. Label the point on the circle E. This forms a diameter AE of the circle. Extend side CB to another point F on the circle, such that the new segment, CF, is a chord of the circle. By the Perpendicular Chord Bisector Theorem, the diameter bisects the chord. Because BC = 15 (by the Parallelogram Opposite Sides Theorem), BF = 15 as well. Use the Intersecting Chords Theorem. 36 * (2r - 36) = 15 * 15 72r - 1296 = 225 72r = 1521 r = 1521/72 = 169/8 So, the radius of the circle is 169/8 centimeters (fraction), or 21.125 centimeters (decimal).

[@aljawad]

I reached the same conclusion when solving the problem using the equation of the circle.

[@santiagoarosam430]

540/15=36=AB r^2-(36-r)^2=15^2 ; r=169/8=21,125. Gracias y un saludo cordial.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@hongningsuen1348]

In case the problem is an MCQ in exam, a quick solution using calculator can be found in 30 seconds: 1. angle BAC = arc tan (BC/AB) = arc tan (15/36) = 22.62 2. angle BOC = 2 x angle BAC = 45.24 (exterior angle of triangle) 3. radius = OC = BC/sin BOC = 15/0.71 = 21.125 = 169/8.

[@hongningsuen1348]

A more formal version of this method: 1. Triangle ABC has sides ratios of 5:12:13. For angle BAC, sin (BAC) = 5/13, cos(BAC) = 12/5 2. Angle BOC = 2 x angle BAC (exterior angle of isosceles triangle AOC) Hence Sin (BOC) = 2 sin (BAC) cos (BAC) (double angle formula for sin function) = (2)(5/13)(12/13) = 120/169 3. Radius = OC = BC/sin (BOC) = (15)/(120/169) = 169/8.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@LionelMaxwellCrKohli01237]

Angle at centre is twice than angle at circumference. Simple.

[@AmirgabYT2185]

r=21,125

[@PreMath]

Excellent! Thanks for sharing ❤️

[@unknownidentity2846]

Let's find the radius: . .. ... .... ..... First of all we calculate the length of AB: A(ABCD) = AB*AD ⇒ AB = A(ABCD)/AD = (540cm²)/(15cm) = 36cm Now let AE be the diameter of the circle with B located on AE. According to Thales theorem the triangle ACE is a right triangle, so we can apply the right triangle altitude theorem: BC² = AB*BE ⇒ BE = BC²/AB = AD²/AB = (15cm)²/(36cm) = (25/4)cm = 6.25cm Now we are able to calculate the radius R of the circle: R = AE/2 = (AB + BE)/2 = (36cm + 6.25cm)/2 = (42.25cm)/2 = 21.125cm Best regards from Germany

[@PreMath]

Excellent job! Thanks for sharing ❤️

[@PREPARE_GURU]

Sir, I want to know that from where you get these type of questions

[@PreMath]

Hello dear, we don't have such magic book! It takes enormous amount of time for research and planning to prepare quality videos! Thanks for asking❤️

[@wackojacko3962]

So @ 10:12 CA² = AB × AE CE² = EB × AE CB² = AB × EB 🙂

[@PreMath]

Excellent! Thanks for the feedback ❤️

[@sergeyvinns931]

АВ=540:15=36; АО=ОС=R, OB=36-R, R^2=15^2+(36-R)^2, R^2=225+1296-72R+R^2, 72R=1521, R=21,125!

[@PreMath]

Excellent! Thanks for sharing ❤️

[@giuseppemalaguti435]

AB=36...15^2+(36-r)^2=r^2...r=1521/72=169/8

[@PreMath]

Excellent! Thanks for sharing ❤️

[@marcgriselhubert3915]

Let'a use an orthonormal, center P (the center of the rectangle), first axis parallel to (AB). O is the intersection of (AB) and of the mediatrice Delta of [A,C]. VectorAC(36; 15) is orthogonal to Delta and Delta contains P, the equation of Delta is then 36.x +15.y = 0, or 12.x + 5.y = 0. The equation of (AB) is y = -15/2, then we have O(-15/2; 25/8). Then VectorOC(119/8; 15) and R^2 = OC^2 = 14161/64 + 225 = 28561/64, so R = 169/8.

[@devondevon4366]

21.125 Line AB = 540/15 = 36 Since AO = the radius= r then OB = 36- r Let's construct a triangle BC0 BC=DA =15 OC= r Hence, r^2 = (36-r)^2 + 15^2 r^2 = 1,296 + r^2 -72r + 225 r^2 = 1521 + r^2 - 72 r 0 = 1521 - 72r 72r = 1521 r = 1521/72 r = 21.125 Answer

[@PreMath]

Excellent! Thanks for sharing ❤️

[@Waldlaeufer70]

4th method: Tangent-chord theorem: L = 540 cm² / 15 cm = 36 cm L * x = 15² x = 15² / 36 = 225 / 36 = 6.25 cm d = L + x = 36 cm + 6.25 cm = 42.25 cm r = d / 2 = 42.25 cm / 2 = 21.125 cm

[@PreMath]

Excellent! Thanks for sharing ❤️

[@kalavenkataraman4445]

21.125

[@PreMath]

Excellent! Thanks for sharing ❤️

[@soli9mana-soli4953]

R^2 = 15^2 + (36 - R)^2 15*15 = 36*X

[@conconstantin7454]

Exactly what I did as well

[@PreMath]

Excellent! Thanks for sharing ❤️

[@gelbkehlchen]

Solution: AB = 540/15 = 36. Pythagoras: (AB-r)²+15² = r² ⟹ (36-r)²+15² = r² ⟹ 1296-72r+r²+225 = r² |-r²+72r ⟹ 1521 = 72r |/72 ⟹ r = 1521/72 = 169/8

[@LuisdeBritoCamacho]

1) AB = 540 / 15 ; AB = 36 cm 2) OA = R 3) OB = (36 - R) 4) OC = OA = R 5) Pink Rectangle Area = 540 Sq cm 6) Let the Point E be the a Vertical Line passing through O and intersecting Line CD. 7) OC^2 = BC^2 + OB^2 8) R^2 = 15^2 + (36 - R)^2 9) R^2 = 225 + 1.296 - 72R + R^2 10) 1.521 - 72R = 0 11) 72R = 1.521 12) R = 1.521 / 72 ; R = 507 / 24 ; R = 169 / 8 cm ; R = 21,125 cm 13) ANSWER : The Radius of the Circle is equal to 169/8 Cm or equal to 21,125 Cm.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@grantschiff7544]

I know 15, so make a quick measuring tape out of paper and done.

[@PreMath]

Thanks for the feedback ❤️

[@himadrikhanra7463]

21..?

[@prossvay8744]

ABCD is rectangle so AB=CD ; AD=BC=15 AB=CD=540/15=36cm Let OB=x OA=OC=OE=R R+x=AB=36 so x=36-R Connect O to C In ∆OBC OB^2+BC^2=OC^2 x^2+15^2=R^2 R^2-x^2=225 R^2-(36-R)^2=225 So R=169/8cm=21.125cm.❤❤❤ Best regards.

[@PreMath]

Excellent job! Thanks for sharing ❤️

[@jimwhalen5675]

First present the problem clearly and cogentlynwith proper.description

[@PreMath]

Thanks for the feedback ❤️

[@riteshkumar8152]

First comment

[@PreMath]

Excellent! Thanks dear❤️

[@joiceroosita5317]

Its too easy..

[@PreMath]

Thanks ❤️

[@devondevon4366]

21.125

[@PreMath]

Excellent! Thanks for sharing ❤️