Can you find the radius? | (Two Methods) | #math #maths | #geometry

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Can you find the radius? | (Two Methods) | #math #maths | #geometry

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[k.amanda.m]

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

[jamestalbott]

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!

[phungpham]

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

[prossvay]

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.

[netravelplus]

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

[unknownidentity]

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

[murdock]

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

[giuseppemalaguti]

AD=a...x+y=24...arctg a/x+arccos7, 5/r=90...arctg x=y=12, a=9, r=12, 5

[LuisdeBritoCamacho]

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.

[Nothingx]

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

[Waldlaeufer]

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

[marcgriselhubert]

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.

[ChuzzleFriends]

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.

[marcelowanderleycorreia]

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

[LeonardEwata-kelle-vqkk]

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

[MrPaulc]

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

[robertlynch]

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.

[michaelgarrow]

This is similar to the formula I used to make arches the proper height when I was building custom homes in San Diego.

[santiagoarosam]

AD=√[15²-(24/2)²] =9→ r²-(r-9)²=12²→ r=25/2.
Gracias y saludos cordiales.

[emachinetool]

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?