Can you find the radius of the Semicircle? | (Geometry skills explained) |
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- Опубликовано: 9 сен 2024
- Learn how to find the radius of the Semicircle. Chords are given inside the semicircle. Important Geometry and Algebra skills are also explained: Thales' Theorem; Pythagorean Theorem; similar triangles; congruent triangles. Step-by-step tutorial by PreMath.com
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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
Interesting!!👍
Excellent!
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
Good!👍
Works out nicely using your method.
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.
Good sharing sir❤❤❤❤
So nice of you ❤️
Excelente+.
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)
Thanks❤️
This is how I solved it
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
Thanks❤️
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 ! 😽
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
Thanks❤️
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
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 ✓
Very Very useful video sir 👍
Thanks and welcome❤️
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
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. 🙂
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? 🙃
Bellissimo, l'ho fatto più o meno cosí anch'io.
Nota che in inglese il seno si scrive "sin".
@@user-gd9vc3wq2h Non in inglese, ma in latino, deriva da sinus👍
@@soli9mana-soli4953 Sed tu angle, non latine scripsisti...
@@user-gd9vc3wq2h Google scripsit, in veritate... 😅
I used cosine rule for the triangles got the same quadratic equation thnx for sharing sir ❤
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.
Sorry, I now see that this has already been proposed by soli9mana-soli4953.
Sir I used cosine law easily we can do
DEC=2x
Ans EBC=2x now we can equate
He keeps saying "We know that..." My head goes; wait what? I know nothing! lol. I am learning though.
asnwer=35cm
21.2
Nice❤❤❤❤❤❤
Thanks 🤗
The easy way: put the 14 chord between the two 30 chords and you have a simetric problem simple to resolv.
R=25 units
The problem became easy when I reached the h² comparison
Thanks❤️
Very tedious steps.
Do all these solutions rely on two segments being equal?
Real world problems very rarely have convenient dimensions.
Angle AEB = 180 degrees, so, using successive approximation with various values of r, you can add the three segment angles and make them total 180 degrees.