Can you find the Radius of the circle? | (3 Methods) |
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- Опубліковано 22 тра 2024
- 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; Thales' Theorem; Euclid's Theorem. Step-by-step tutorial by PreMath.com.
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Can you find the Radius of the circle? | (3 Methods) | #math #maths | #geometry
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Nice explaination❤
Glad you liked it
Thanks for the feedback ❤️
For a fourth method, we note that ΔABC and ΔACE are similar by angle - angle (common angle
Excellent!
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strategical solutions , Very amazing . Thanks Sir
Most welcome dear🌹
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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.
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I reached the same conclusion when solving the problem using the equation of the circle.
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
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Cool👍
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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).
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
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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.
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.
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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
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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
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For a triangle inscribed in a circle , intersecting chord theorem and Euclid's formula is same.
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
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When I looked at the rectangle and the tangent DA to the circle, I thought the first method would be the 4th method.
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!
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Thank you
You are very welcome!
Thanks Raya❤️
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
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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)).
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.
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Angle at centre is twice than angle at circumference. Simple.
540/15=36=AB
r^2-(36-r)^2=15^2 ; r=169/8=21,125.
Gracias y un saludo cordial.
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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.
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
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Straightforward problem
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Sir, I want to know that from where you get these type of questions
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❤️
AB=36...15^2+(36-r)^2=r^2...r=1521/72=169/8
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540÷15=36, (36-r)^2+15^2=r^2, 36^2-72r+15^2=0, r=(36^2+15^2)/72=(39×39)/72=(13×13)/8=169/8.😊
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So @ 10:12
CA² = AB × AE
CE² = EB × AE
CB² = AB × EB
🙂
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r=21,125
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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.
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21..?
R^2 = 15^2 + (36 - R)^2
15*15 = 36*X
Exactly what I did as well
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21.125
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I know 15, so make a quick measuring tape out of paper and done.
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First present the problem clearly and cogentlynwith proper.description
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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.
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First comment
Excellent!
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Its too easy..
Thanks ❤️
21.125
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