Can you find the radius of the Semicircle? | (Geometry skills explained) |

[@damienduquennoy5231]

I suggest to place the 14 unit segment between the two 30 segments. Doing so you have a quarter circle with an horizontal segment of 7 units. let call h the distance between the segment and the horizontal axis. You have then 7² + h² = R² And in the other triangle h² + (R-7)² = 30² You substract the two equations and get R² - 7R - 450 = 0 Delta is 1849 = 43² R1 is (7-43)/2, negative and R2 is (7+43)/2 = 25. Probably faster. Damien

[@soli9mana-soli4953]

Interesting!!👍

[@andreasproteus1465]

Excellent!

[@michaelkouzmin281]

Just another (algebraic) solution: 1. let a = angle AED = angle DEC; => angle BEC = 180-2a; 2. Let us apply the Law of Cosine to triangle AED: 30^2=2r^2-2r^2*cos(a) let t = cos(a) = 1-450/r^2; 0

[@soli9mana-soli4953]

Good!👍

[@johnbrennan3372]

Works out nicely using your method.

[@Ibrahimfamilyvlog2097l]

Good sharing sir❤❤❤❤

[@PreMath]

So nice of you ❤️

[@mathbynisharsir5586]

Very Very useful video sir 👍

[@PreMath]

Thanks and welcome❤️

[@manuelantoniobahamondesa.3252]

Excelente+.

[@ABhaim]

AD = DC (given) AE = EC (radii) ED = ED therefore, triangles AED and CED are congruent (side-side-side) angles AED = DEC (similar angles in congruent triangles) let AED = DEC = x, therefore CED = 180-2x (semi-circle angle) let AE=ED=EC=EB=r using Theorem of Al-Kashi, and by applying some trig identities: 14^2 = r^2 + r^2 - 2*r*r*cos(180-2x) = 2r^2(1 - cos(180-2x)) = 2r^2(1 + cos(2x)) = 2r^2(1 + 2(cos(x))^2 - 1) = 4r^2(cos(x))^2 14 = 2rcos(x) -> cos(x) = 7/r by using Theorem of Al-Kashi on either AED or CED, and by what we know of cos(x), we can calculate r: 30^2 = 900 = r^2 + r^2 - 2*r*r*cos(x) = 2r^2 - 14r r^2 - 7r - 450 = 0 (r - 25)(r + 18) = 0 r = 25 Q.E.D

[@PreMath]

Thanks❤️

[@misterenter-iz7rz]

arcsin 15/r+arcain 15/r+arcsin 7/r=90, so 2 arcsin 15/r=90-arcsin 7/r, 2 15/r sqrt(r^2-225)/r=sqrt(r^2-49)/r, 30sqrt(r^2-225)=sqrt(r^2-49)r, 900 (r^2-225)=(r^2-49)r^2, r^4-(49+ 900 )r^2+900*225=0, r^2=324 or 625, r=18 rejected by check back, or 25.

[@geraldgiannotti8364]

This is the best solution I have seen. Very straight forward. Thank You!

[@bartholomewrichards1663]

I did this the easy way.chords are directly proportional to angles emqnating from the radius, find angles by proportion s=r theta.you have the chords. Then sine law.easy

[@pi5355]

Total length of cchords=30+30+14=74 For chord 30 length angle at centre is approximately 30/74(180) 30^2=2RR-2RRcos(5400/74) 900=1.414R^2 R=25.2approximately

[@quigonkenny]

Draw radii ED and EC. As they are all radii of circle E, EA = EB = EC = ED = r. As EA = EC, CD = DA = 30, and ED is shared, ∆DEA and ∆CED are congruent. Draw AC. As AC is a chord and ED is a radius, ED bisects and is perpendicular to AC at its midpoint M, and AM = MC. Also, all current angles at M are 90°, as AC and ED are perpendicular. So as AM = MC, CD = DA, and MD is shared, ∆DMC and ∆AMD are congruent. Similarly, as EA = ED and EM is shared, ∆EMA and ∆CME are congruent. By Thales' Theorem, for any three points on the circumference of a circle, if two of the points are the ends of a diameter, the angle between them at the third point is 90°. As ∠EAM = ∠BAC and ∠AME = ∠ACB = 90°, ∆AME and ∆ACB are similar triangles. Triangle ∆AME: EM/EA = CB/BA EM/r = 14/2r = 7/r EM = (r)7/r = 7 As EM = 7, MD = r-7. Triangle ∆DMC: (r-7)² + MC² = 30² MC² = 900 - (r²-14r+49) MC² = 851 - r² + 14r MC = √(851-r²+14r) Complete the circle by mirroring arc AB about diameter AB. Extend radius ED to become diameter DF. By Intersecting Chords Theorem, if two chords of a circle intersect, then the products of the lengths of each chord on either side of the point of intersection are equal. Note that as EF = r, MF = 7+r, and remember that AM = MC. MD(MF) = AM(MC) = MC² (r-7)(r+7) = √(851-r²+14r)² r² - 49 = 851 - r² + 14r 2r² - 14r - 900 = 0 r² - 7r - 450 = 0 r² + 18r - 25r - 450 = 0 (r+18)(r-25) = 0 r = -18 ❌ | r = 25 ✓

[@giuseppemalaguti435]

arccos7/r+arccos15/r+arccos15/r=180(gli angoli opposti sono supplementari)...2arccos15/r=180-arccos7/r..apploco cos..2*15^2/r^2-1=-7/r...r^2-7r-450=0...r=25

[@PreMath]

Thanks❤️

[@rudychan8792]

Looks Complicated. But, it works. I was using Sinus Theorm: A + 2A + (A+B) + B = 360° 2A + B = 180° >> B = 180°- 2A .:| 2R = 14 / cos B = 30 /cos A 14×cosA = 30×cosB = -30×cos2A* ↔️ 60p" + 14p - 30 = 0 30p" + 7p - 15 = 0 Cos A = (-7 + 43) ÷ (2×30) = 0,6 !! ↘️ 2R = 30/ cosA = 30/0,6 = 50, ➡️ R = 25 units Great Puzzle ! 😽

[@marcgriselhubert3915]

P, Q, R are respectively the orthogonal projections of E on (BC), (CD), (DA). angles BEP, PEC are x degrees; angles CEQ, QED, DER, REA are y degrees. So angle PED is x + 2y degrees = 90 degrees. (as naturally 2x +4y = 180 degrees) So, we have: x = 90 - 2y degrees. In triangle BEP: sin (x) = 7/R, and in triangle CEQ: sin (y) = 15/R. We know that sin (x) = sin (90 - 2y) = cos (2y) = 1 - 2 (sin (y))^2, so we have: 7/R = 1 - (15/R)^2. This equation gives (we multiplicate by R^2): 7R = R^2 - 450 or R^2 - 7R -450 = 0 This is a second degree equation, the discriminant is (7)^2 _ (4) (1) (-450) = 1849, and sqrt (1849) = 43. So, the solutions are: R = (7 - 43)/ 2 = -18 or R = (7 + 43)/ 2 = 25. The first solution beeing negative is physicaly impossible, so we get finally: R = 25.

[@marcgriselhubert3915]

Sorry, I now see that this has already been proposed by soli9mana-soli4953.

[@wackojacko3962]

Computers can solve this problem maybe ...but @ 2:11 that tiny bit of creativity connecting Points A and C is what sets humans apart from mindless tasks computers perform. 🙂

[@projectfuturistic8774]

I used cosine rule for the triangles got the same quadratic equation thnx for sharing sir ❤

[@soli9mana-soli4953]

With trigonometry we could solve in this way: Tracing the radii perpendicular to each chord and radii ED and EC we get 4 right triangle with angle in E equal to alpha and 2 right triangle with angle in E equal to beta. Knowing that 4 α + 2 β = 180° => β = 90 - 2 α for each right triangle with angle alpha we have: sen α = 15/r (r = radius) for each right triangle with angle beta we have: sen β = 7/r but we know that sen β = sen ( 90 - 2 α) = cos 2 α = 1 - 2sen²α then solving the system: 1 - 2sen²α = 7/r sen α = 15/r we get the equation: r² - 7r - 450 = 0 r = 25 too difficult? 🙃

[@WK-5775]

Bellissimo, l'ho fatto più o meno cosí anch'io. Nota che in inglese il seno si scrive "sin".

[@soli9mana-soli4953]

@@WK-5775 Non in inglese, ma in latino, deriva da sinus👍

[@WK-5775]

@@soli9mana-soli4953 Sed tu angle, non latine scripsisti...

[@soli9mana-soli4953]

@@WK-5775 Google scripsit, in veritate... 😅

[@ai2657]

Sir I used cosine law easily we can do DEC=2x Ans EBC=2x now we can equate

[@sandanadurair5862]

Another simple method using property of cyclic quadrilateral join the diagonals AC and BD. the property is AC.BD = AB.CD + AD.BC ..................eqn 1 AB = 2r ; BC = 14; CD = AD=30 AC = SQRT(4r^2 - 14^2) from right triangle ABC BD = SQRT(4r^2 - 30^2) from right triangle ABD substituting above values in eqn.1 and simplifying we get r.(r^2-7r-450) =0 hence r = 0 ; r = -18; r = 25 (valid)

[@PreMath]

Thanks❤️

[@sujathasankar8931]

This is how I solved it

[@Alishbafamilyvlogs-bm4ip]

Nice❤❤❤❤❤❤

[@PreMath]

Thanks 🤗

[@DanMusceac]

The easy way: put the 14 chord between the two 30 chords and you have a simetric problem simple to resolv.

[@kamilshere9186]

21.2

[@comdo777]

asnwer=35cm

[@prossvay8744]

R=25 units

[@hatac]

He keeps saying "We know that..." My head goes; wait what? I know nothing! lol. I am learning though.

[@alster724]

The problem became easy when I reached the h² comparison

[@PreMath]

Thanks❤️

[@ritamkundu7267]

Very tedious steps.