25/8 Draw a perpendicular line from BC to form two 3-4-5 right triangles, ABP and ACP. BP = 3 and AP=4 label O on line AP as the circle's center. O will be located on AP, given the isosceles triangle, ABC. From O, draw a line to B, then draw another to P to form a triangle, BOP OB = r OP= 4 -r (since AP=4), and BP =3 Using Pythag : r^2 = 3^2 + ( 4-r)^2 r^2 = 9 + 16 + r^2 - 8r 8r = 25 r = 25/8 Answer
4th way: Find h=4 by the Pythagorean theorem as you have. Then, draw a line from the center of the circle on h to intersect line AC at the midpoint, Q, at a right angle. Triangle AOQ is similar to the larger right triangle ACP. Therefore, PC/AP=OQ/2.5. 3/4=OQ/2.5. OQ = 15/8. Now use the Pythagorean theorem to find the radius, OA: (15/8)^2 + (5/2)^2 = OA^2. OA or R = 25/8.
_Outro método:_ A área ABC é facilmente encontrada fazendo [ABC]=2×(4×3)/2=12 [ABC]= abc/4R, onde a,b e c são os lados do triângulo ABC e R é o raio da circunferência. Assim, 12=5×5×6/4R 2=5×5/4R 8R=5×5 *R=25/8*
I'm new to this, but it seems that it should be (4x = 5*5), which solves to x = 6.25. Or am I missing something? Again, I am new to this, so please correct me if I am wrong.
@Ellehsdee The x value is the section of the diameter below the chord.This value is calculated to be 2.25 from the formula for intersecting chords (Euclid) The length of the diameter is calculated to be 4+2.25=6.25. From this, the radius is calculated to be 6.25/2 =3.125.
Just draw diameter from point A intersecting BC in 2 segments of 3 Th height of isosceles triangle is obviously 4 (due to two 345 triangles or 30,60,90 degree angles) Now use intersecting rule 3x3=9 9 ÷ 4= 2.25 2.25 + 4 = 6.25 diameter 6.25 ÷ 2 = 3.125 Radius
The most direct method using no construction is to use Heron's formula to calculate area of the triangle from the 3 sides and then use circumradius formula to find R = abc/4xarea.
You don't need Heron's formula - it's an isosceles triangle. The line from the apex to the base intersects the base at 90 degrees at the midpoint of BC at say, D. So BD = CD = 6/2 = 3. You then have a Pythagorean 3-4-5 triangle, with AD = 4. The area is then BD*AD = 3*4 = 12.
Circumradius R =abc/4 area of triangle a=5 ,b=5 c =6 , area of triangle = square root of s(s-a) (s-b) (s-c) and s=a+b+c÷2. s= 5+5+6÷2. s= 16÷2=8. Area of the triangle = square root of 8×[8-5] ×[8-5]×[8-6] = square root of. 8×3×3×2 =square root of 144 =12 circumradius R= abc÷ 4 area of the triangle R= 5×5×6÷4×12 =150÷48=25÷8= 3.125
Without using heron formula or any construction other than vertical height. After finding height, we can use a simple formula as below. Radius= product of sides of triangle other than base divided by 2 times of height. So, Radius= 5*5/(2*4)= 25/8.
We divide triangle in middle to 2 right triangles and they will be both similiar because it is icoseles so 3-4-5 specaial triangle so height is 4 then we use inside circle cors where from radius to cord it divides cord to 2 which makes base 3 and side with 5 to 5/2 then we measure height for the base triangles which will be |/r²-9 then we add both r and |/r²-9 to find that both equal 4
My method, use of formulas : R = a*b*c/ 4*A and (Heron) A = sqrt(p*(p-a)*(p-b)*(p-c) with half perimeter p=0.5*(a+b+c) and A = triangle area p = 0.5(5+5+6) = 8 ---> A = sqrt(8*3*3*2) = 12 ----> R = 5*5*6 / 4*12 = 25/8
Insted of using sine formula we can solve it in an easy way, viz , Extended BP touches the circumference (say) at Q. Then BP = 2C, Hence, (BP)^2 = (PQ)(AP) Or 3^2 =. 4(PQ) Or. PQ = 9/4 = 2.25 Or AQ = 4+2.25 = 6.25 which is 2r Hence r = 6.25/2 or 3.125
After finding the length of the height from A, I would introduce cartesian coodinates with the origin to be B(0,0) and BC the x-axis, so C has coordinates C(6,0) and A(3,4). Now you have to find the radius though 3 points B(0,0), C(6.0) and A(3,4) with the center O(a,b). Because the triangle is isocele, the point O is on the vertical line through A, so a=3. (4-b)^2=R^2=9+b^2, so b=7/8. Finally R=4-7/8=25/8.
Split the triangle vertically to get a 3/4/5 right triangle. The angle at C thus has sine of 4/5 = 0.8. By the law of sines we then have 5/0.8 = D, the diameter. Therefore the radius is 2.5/0.8 = 25/8. Q.E.D.
In triangle ABC drawn in a circle of radius R where AH is the height related to side BC we have the relationship: AB*AC=AH*2R and from it 5*5=4*2R so R=25/8
D es el punto medio de BC y "s" es la flecha del arco BC---> Potencia de D respecto a la circunferencia =BD*DC=AD*s---> 3*3=4s---> s=9/4---> 2r=4+(9/4)=25/4--->r=25/8. Gracias y saludos.
There is simpler method than all 3 approaches shown. Proceed as in method 1 to get AP = 4. Construct line CQ. Triangle ACQ is right triangle & is similar to triangle ACP having 3:4:5 side ratio with scaling factor = 5/4. So AQ = (5/4)(5) = 25/4. Radius is half that = 25/8.
Both ACP and AQC are right triangles with common angle A. In this reason they are similar ==> |AQ|/|AC| = |AC|/|AP| ==> |AQ| = |AC|*|AC|/|AP| = 5 * 5 / 4 = 25/4 R = |AQ|/2 = (25/4)/2 = 25/8
Find the angles. Simple Pythagorean. Then use this formula: circumradius = a/ 2sin(A) Where a , is a side of the triangle, and A, is the angle opposite of side a. Circumradius R = 3.125
@@maluhadli468 It's a simple Isosceles triangle. You can split the existing triangle into two equal right triangles and simply use the Pythogorean rule to find the angles. The angles at B and C are both 53.13 degrees. Angle A is 73. 74 degrees. The height from line BC is 4. So you basically have two 3,4,5 right triangles joined together. They share the same height, which is 4. Measured from the base line BC
4th method, the easiest 1) From vertex A we lower the perpendicular AP to BC. 2) From the middle of AC to AP, construct a segment NO perpendicular to AC. AN= NC= 5/2= 2.5. AO= R. 3) 5/4=R/2.5; R=5×2.5/4= 3.125
The radius is 25/8. This is definitely an example of Pythagorean triples. Foe the first method, this involves the perpendicular bisector theorem and the intersecting chords theorem. For the second method, this uses the perpendicular bisector theorem and shwos the right scalene triangle that is contained in the equilateral triangle. I like the second method and I think that I need to start visualizing the equivaleny intersecting chids theorem on equilateral triangles. And I think that THIS is anothe property of Pythagorean triples. Also please look up my comment for yesterday's video.
By using herons formula Half perimeter of ∆ABC is 5+5+6/2=8 :.area=√s(s-a)(s-b)(s-c) =√8(8-5)(8-5)(8-6) =√8*3*3*2 =3√2*2*2*2*2 ১ =12√2 1/2*6*x=12√2 3x=12√2 x=4√2 By Circle triangle theorem Radius=4√2*2\3 =8√2/3 =8*1.141/3 =3.77(aprox)
The right angle 3,4,5 triangle is immediately obvious, while knowing that BP x PC (3 x 3 = 9) divided by 4 (= 2.25) gives you the diameter of 4 + 2.25 = 6.25. The radius is then 6.25/2 = 3.125 (or 25/8). Done in 20 seconds. Too easy.
cos A/2 = h÷5 (h hauteur du triangle isocèle de sommet A et de coté 5) cos A/2 = 2,5÷R (R rayon du cercle de cente O et de rayon recherché et 2,5 demi-base du triangle isocèle de sommet O et de base 5) h coté de l'angle droit du triangle rectangle d’hypoténuse 5 et de second coté de l' angle droit qui vaut 3 (demi-base du triangle isocèle de sommet A) Donc h=4 (Pythagore) d'où : 4÷5 = 2,5÷R R = (5 × 2,5)÷4 R = 3,125
Midpoint of BC: calling it M . Extending AM to P where AP is diameter. BC =6 ,given. 6/2=3 BM.MC =3.3.=9 Two chords bisecting : product of their parts are equal. AM.MP =9 AM.AM= 5.5-3.3 =16 so AM=4. So MP=9/4 AP is diameter = 4 +9/4 equals 2.radius. 2r=6+1/4 Radius is 3+1/8. Now I'll check this.
Gjate gjithr jetes time kam pasur respekt te vecante per perdonat qe kishin arsye te shendoshe ne matematike. Por me falni per nje verejtje ne vizatimin e krsaj figure. Te paktn me sy seg BC ta benit me te gjate. Kete e arrinit ta vendosni me afer pike O. Pak a shume edhe nje figure e ndertuar mire te ndihmon ne zgjidhje. Flm
25/8
Draw a perpendicular line from BC to form two 3-4-5 right triangles, ABP and ACP.
BP = 3 and AP=4
label O on line AP as the circle's center. O will be located on AP, given the isosceles triangle, ABC.
From O, draw a line to B, then draw another to P to form a triangle, BOP
OB = r OP= 4 -r (since AP=4), and BP =3
Using Pythag :
r^2 = 3^2 + ( 4-r)^2
r^2 = 9 + 16 + r^2 - 8r
8r = 25
r = 25/8 Answer
Nicely explained sir., thanks.
@@thangavelm.m8003 Glad to be of help
4th way: Find h=4 by the Pythagorean theorem as you have. Then, draw a line from the center of the circle on h to intersect line AC at the midpoint, Q, at a right angle. Triangle AOQ is similar to the larger right triangle ACP. Therefore, PC/AP=OQ/2.5. 3/4=OQ/2.5. OQ = 15/8.
Now use the Pythagorean theorem to find the radius, OA: (15/8)^2 + (5/2)^2 = OA^2. OA or R = 25/8.
_Outro método:_
A área ABC é facilmente encontrada fazendo
[ABC]=2×(4×3)/2=12
[ABC]= abc/4R, onde a,b e c são os lados do triângulo ABC e R é o raio da circunferência.
Assim,
12=5×5×6/4R
2=5×5/4R
8R=5×5
*R=25/8*
The method of intersecting chords will also yield the radius more quickly(4*x=3*3).This yields a diameter of 6.25.,r=3.125
I'm new to this, but it seems that it should be (4x = 5*5), which solves to x = 6.25. Or am I missing something? Again, I am new to this, so please correct me if I am wrong.
@Ellehsdee The x value is the section of the diameter below the chord.This value is calculated to be 2.25 from the formula for intersecting chords (Euclid)
The length of the diameter is calculated to be 4+2.25=6.25. From this, the radius is calculated to be 6.25/2 =3.125.
@pooransingh1882 Ahh, thank you so much for the charity on this.
Just draw diameter from point A intersecting BC in 2 segments of 3
Th height of isosceles triangle is obviously 4 (due to two 345 triangles or 30,60,90 degree angles)
Now use intersecting rule
3x3=9
9 ÷ 4= 2.25
2.25 + 4 = 6.25 diameter
6.25 ÷ 2 = 3.125 Radius
設半徑長為R,畫∠BAC角平分線交BC於D,離A點R長度取得圓心O位置,則AD長為4,OD長為4-R , CD長為3→三角形ODC R^2= (4-R)^2+3^2→ R=25/8
The most direct method using no construction is to use Heron's formula to calculate area of the triangle from the 3 sides and then use circumradius formula to find R = abc/4xarea.
You don't need Heron's formula - it's an isosceles triangle. The line from the apex to the base intersects the base at 90 degrees at the midpoint of BC at say, D. So BD = CD = 6/2 = 3.
You then have a Pythagorean 3-4-5 triangle, with AD = 4. The area is then BD*AD = 3*4 = 12.
@@Istaphobic My point is not using any construction including perpendicular bisector theorem for isosceles triangle.
@@hongningsuen1348 Fair enough, I just personally think Heron's formula is a lot more work.
@@Istaphobic No, because the Heron is general for any triangle ! His formula is correct
@@WahranRai I know it does. I just said in the case of this particular isosceles triangle that it's a lot more work. Lol.
Circumradius R =abc/4 area of triangle a=5 ,b=5 c =6 , area of triangle = square root of s(s-a) (s-b) (s-c) and s=a+b+c÷2. s= 5+5+6÷2. s= 16÷2=8. Area of the triangle = square root of 8×[8-5] ×[8-5]×[8-6] = square root of. 8×3×3×2 =square root of 144 =12 circumradius R= abc÷ 4 area of the triangle R= 5×5×6÷4×12 =150÷48=25÷8= 3.125
Without using heron formula or any construction other than vertical height. After finding height, we can use a simple formula as below.
Radius= product of sides of triangle other than base divided by 2 times of height.
So, Radius= 5*5/(2*4)= 25/8.
We divide triangle in middle to 2 right triangles and they will be both similiar because it is icoseles so 3-4-5 specaial triangle so height is 4 then we use inside circle cors where from radius to cord it divides cord to 2 which makes base 3 and side with 5 to 5/2 then we measure height for the base triangles which will be |/r²-9 then we add both r and |/r²-9 to find that both equal 4
My method, use of formulas :
R = a*b*c/ 4*A and (Heron) A = sqrt(p*(p-a)*(p-b)*(p-c)
with half perimeter p=0.5*(a+b+c) and A = triangle area
p = 0.5(5+5+6) = 8 ---> A = sqrt(8*3*3*2) = 12 ----> R = 5*5*6 / 4*12 = 25/8
😮T😮
One of the Finest Method No other Calculation is required after this Heron's formula is enough for all types of Triangles to calculate Area.....
Insted of using sine formula we can solve it in an easy way, viz ,
Extended BP touches the circumference (say) at Q. Then
BP = 2C, Hence,
(BP)^2 = (PQ)(AP)
Or 3^2 =. 4(PQ)
Or. PQ = 9/4 = 2.25
Or AQ = 4+2.25 = 6.25 which is 2r
Hence
r = 6.25/2 or 3.125
After finding the length of the height from A, I would introduce cartesian coodinates with the origin to be B(0,0) and BC the x-axis, so C has coordinates C(6,0) and A(3,4). Now you have to find the radius though 3 points B(0,0), C(6.0) and A(3,4) with the center O(a,b). Because the triangle is isocele, the point O is on the vertical line through A, so a=3. (4-b)^2=R^2=9+b^2, so b=7/8. Finally R=4-7/8=25/8.
Split the triangle vertically to get a 3/4/5 right triangle. The angle at C thus has sine of 4/5 = 0.8. By the law of sines we then have 5/0.8 = D, the diameter. Therefore the radius is 2.5/0.8 = 25/8. Q.E.D.
In triangle ABC drawn in a circle of radius R where AH is the height related to side BC we have the relationship: AB*AC=AH*2R and from it 5*5=4*2R so R=25/8
D es el punto medio de BC y "s" es la flecha del arco BC---> Potencia de D respecto a la circunferencia =BD*DC=AD*s---> 3*3=4s---> s=9/4---> 2r=4+(9/4)=25/4--->r=25/8.
Gracias y saludos.
A=SQR(S(S-a)(S-b)(S-c))=(abc)/4R; S=(a+b+c)/2=8; A=SQR(8x2x3x3)=12; A=(5x5x6)/4R=12; R=25/8
There is simpler method than all 3 approaches shown. Proceed as in method 1 to get AP = 4. Construct line CQ. Triangle ACQ is right triangle & is similar to triangle ACP having 3:4:5 side ratio with scaling factor = 5/4. So AQ = (5/4)(5) = 25/4. Radius is half that = 25/8.
A theorem says radius = (ABxACxBC)/(4 x Area).
Or: in triang. (ABQ) you can easily find PQ by Euclides 2nd Theorem: PQ = BP^2/AP
This was a fun one, thanks. Seemingly perplexing problem actually able to be solved in a person's head!
Both ACP and AQC are right triangles with common angle A.
In this reason they are similar ==> |AQ|/|AC| = |AC|/|AP|
==> |AQ| = |AC|*|AC|/|AP| = 5 * 5 / 4 = 25/4
R = |AQ|/2 = (25/4)/2 = 25/8
Find the angles. Simple Pythagorean. Then use this formula: circumradius = a/ 2sin(A) Where a , is a side of the triangle, and A, is the angle opposite of side a. Circumradius R = 3.125
What is the value of the angle A how to find explainit
@@maluhadli468 It's a simple Isosceles triangle. You can split the existing triangle into two equal right triangles and simply use the Pythogorean rule to find the angles. The angles at B and C are both 53.13 degrees. Angle A is 73. 74 degrees. The height from line BC is 4. So you basically have two 3,4,5 right triangles joined together. They share the same height, which is 4. Measured from the base line BC
4th method, the easiest
1) From vertex A we lower the perpendicular AP to BC.
2) From the middle of AC to AP, construct a segment NO perpendicular to AC.
AN= NC= 5/2= 2.5. AO= R.
3) 5/4=R/2.5;
R=5×2.5/4= 3.125
The radius is 25/8. This is definitely an example of Pythagorean triples. Foe the first method, this involves the perpendicular bisector theorem and the intersecting chords theorem. For the second method, this uses the perpendicular bisector theorem and shwos the right scalene triangle that is contained in the equilateral triangle. I like the second method and I think that I need to start visualizing the equivaleny intersecting chids theorem on equilateral triangles. And I think that THIS is anothe property of Pythagorean triples. Also please look up my comment for yesterday's video.
Quarto método: teorema de Faure.
Quinto método: S=abc/4R.
Aprendi aqui, professor !
By using herons formula
Half perimeter of ∆ABC is 5+5+6/2=8
:.area=√s(s-a)(s-b)(s-c)
=√8(8-5)(8-5)(8-6)
=√8*3*3*2
=3√2*2*2*2*2
১
=12√2
1/2*6*x=12√2
3x=12√2
x=4√2
By Circle triangle theorem
Radius=4√2*2\3
=8√2/3
=8*1.141/3
=3.77(aprox)
The right angle 3,4,5 triangle is immediately obvious, while knowing that BP x PC (3 x 3 = 9) divided by 4 (= 2.25) gives you the diameter of 4 + 2.25 = 6.25. The radius is then 6.25/2 = 3.125 (or 25/8). Done in 20 seconds. Too easy.
4(2r - 4) = 9 → r = 25/8; or: ABC = BCA = δ → AOC = 2δ → sin(δ) = 4/5 → cos(δ) = 3/5 →
cos(2δ) = cos^2(δ) - sin^2(δ) = -7/25 → 25 = 2r^2(1 - cos(2δ)) → r = 25/8
The easiest way
R=abc/(4S);
S= 6×4/2= 12
R= 5×5×6/(4×12)= 150/48= 3.125
Why so complicate? If you have QP = 9/4 and AP = 4 .... so you have the diameter 9/4 + 4 .... r = diameter/2
If we use the formula r²=a²b²c²÷[(a+b-c)(b+c-a)(c+b-a)(a+b+c)], then it is so brutal
Me thinking about the same thing
cos A/2 = h÷5 (h hauteur du triangle isocèle de sommet A et de coté 5)
cos A/2 = 2,5÷R (R rayon du cercle de cente O et de rayon recherché et 2,5 demi-base du triangle isocèle de sommet O et de base 5)
h coté de l'angle droit du triangle rectangle d’hypoténuse 5 et de second coté de l' angle droit qui vaut 3 (demi-base du triangle isocèle de sommet A)
Donc h=4 (Pythagore)
d'où : 4÷5 = 2,5÷R R = (5 × 2,5)÷4 R = 3,125
I went for 4(2r - 4) = 9
8r - 16 = 9
8r = 25
r = 25/8 or 3.125 if preferred.
Excellent!
Midpoint of BC: calling it M . Extending AM to P where AP is diameter. BC =6 ,given. 6/2=3
BM.MC =3.3.=9 Two chords bisecting : product of their parts are equal. AM.MP =9
AM.AM= 5.5-3.3 =16 so AM=4. So MP=9/4
AP is diameter = 4 +9/4 equals 2.radius. 2r=6+1/4
Radius is 3+1/8.
Now I'll check this.
(2"R-4)*4=3*3 por el Teorema de las cuerdas
(5)^2 =25 (5)^2 =25 (6)^2=36 {25+25+36}=86 360°ABC/86=4.16ABC 4.4^4 2^2.2^2^2^2 1^1.1^1^1^2 1^2 (ABC ➖ 2ABC+1).
r x cos(1/2 A) = 2.5
5 x cos(1/2 A) = 4
so we have: r/5 = 2.5/4
r= 25/8
5×5=25-3×3=16squroth=4)(3×3=9÷4=2.5+4=6.5÷2=3.25×3.25=105626×3.14159268=33.183
The way the teachers explain it to you complicates it even more
7:17
Objection
Why you useing pq as radius?
Gjate gjithr jetes time kam pasur respekt te vecante per perdonat qe kishin arsye te shendoshe ne matematike. Por me falni per nje verejtje ne vizatimin e krsaj figure. Te paktn me sy seg BC ta benit me te gjate. Kete e arrinit ta vendosni me afer pike O. Pak a shume edhe nje figure e ndertuar mire te ndihmon ne zgjidhje. Flm
minute 5.50- is there not an error ?
94=2.25+4=6.25÷2=3.125×3.125×3.14159268=30.6796ful area
2/3 of perpendicular which is 4 ans is 8/3
r^2 = 9 + (4 - r)^2
r^2= 9+ 16 - 8 r + r^2
8r = 25
r = 25/8= 3 + ( 1/8)
How do you conclude the perpendicular passes through centre?
По определению. Перпендикуляр к хорде, проведенный через ее середину является диаметром
3.125×3.125×3.14159268=30.6796 full area
Why not adding 4 and 9/4 and divide by 2
R=5×5×6/(4×12)
I feel fully math boosted now!
Mericans no unusand maff.
Excellent
R=abc/4s.
Ավելի էֆեկտիվ կլիներ օգտագործել լարին ուղղահայաց տրամագծի հատկությունը երբ հաշվել էր բարձրությունը
It is very useful.
Very good
Second method is easier than other methods
S= abc/ 4r ,mais simples
Pata nahi yaar main to sqrt(3^2+x^2)=4-x karke solve kar lunga maths student bhi nahi hun
Thanks easy
Faking difficult, so complex faking maths😅
God taught mathematics and geometry to humans, so who can understand God's mathematics.
25/8=3.25
3.13
25/8
4 cm
It's 3.125
3.12
❤
r=25/8
👍👍👍👍👍👍👍👍👍👍👍👍
Poor delivery
Mind calculation in 20 sec, 25/8
joinOB.Then r^2=9-8r+r^2+16=>r=25/8
25/8