You should have labeled the arcs 3x, 4x, and 5x to make sure that we knew from the picture that you were giving us ratios. You explained it in the video, but there are some viewers who'd rather solve the problem themselves before starting your video.
Yes, 3x, 4x and 5x would be better labels. Integers are usually measurements rather than proportions. For solving the problem, we need these measurements in degrees or radians, not units of length. So, 3x + 4x + 5x = 12x = 360° or 2π radians and x = 30° or π/6 radians.
Hello, There appears to be an inconsistency because the perimeter of the circle is 2.pi.r = 12, so if r = 1 you conclude that pi = 6? Could you help me please?
I think this question has a loophole. Area iof circle is pi, which translates to r = 1. Howver total circumference is 2*pi*r = 12, giving a different r of 6/pi. 😅
At 5;45, we can apply Area = (1/2) ab sin(c) to ΔABC, computing the area of one triangle instead of computing areas of 3 triangles and summing their areas. Lengths AB and AC are more straightforward to calculate than BC, so we choose them as a and b. ΔABO is a special 45°-45°-90° isosceles right triangle, its hypotenuse is √2 times either side, so a = (√2)(1) = √2. Drop a perpendicular from O to AC, dividing ΔACO into 2 special 30°-60°-90° right triangles, each with long side (√3)/2 times the hypotenuse, or (√3)/2, total length b = √3. So a = √2 and b = √3. c =
@@phungpham1725 Your method is different. You divided ΔABC into 2 triangles and summed the areas. I computed the area of one triangle ΔABC with one equation, but did need to divide ΔABC into smaller triangles to find side lengths a and b, and I constructed a triangle to find sin(75°).
A = πr² π = πr² r² = 1 r = 1 Arcs AB, AC, & BC are at ratios of 3, 4, & 5 in that order. So, the lengths of the respective arcs are 3k, 4k, & 5k (where k is some constant). So, the circumference of the circle is 12k (or 2π). We can find the measures of the arcs. Measure of arc AB = 3k/12k = 1/4 * 360° = 90° Measure of arc AC = 4k/12k = 1/3 * 360° = 120° Measure of arc BC = 5k/12k = 5/12 * 360° = 150° ∠A is an inscribed angle. By the Measure of an Inscribed Angle Theorem, m∠A = (150°)/2 = 75°. Before calculating the area of △ABC, we need to find the lengths of sides AB & AC. They are adjacent to ∠A. Draw radii AO, BO, & CO. Find m∠AOB & m∠AOC. The measure of a central angle is the measure of the arc it intercepts, and vice versa. So, ∠AOB is a right angle and m∠AOC = 120°. This makes △AOB an isosceles right triangle. Use this formula: c = a√2 AB = 1 * √2 = √2 Draw a radius perpendicular to side AC. This bisects ∠AOC and divides △AOC into two congruent right triangles. The smaller angles measure 60° each, so the triangles are both special 30°-60°-90° right triangles. We can use another special formula: c = 2a 1 = 2a a = 1/2 b = a√3 = 1/2 * √3 = (√3)/2 This takes up half the length of the side AC. So, AC = √3. A = 1/2 * a * b * sinC = 1/2 * √2 * √3 * sin(75°) = 1/2 * √6 * (√6 + √2)/4 = √6 * (√6 + √2)/8 = [√6 * (√6 + √2)]/8 = (6 + √12)/8 = [6 + (√4)(√3)]/8 = (6 + 2√3)/8 = (3 + √3)/4 So, the area of the green triangle is (3 + √3)/4 square units (exact), or about 1.18 square units (approximation).
Это пропорции на которые делится угловое значение дуги, опирающейся на хорду, и соответствующий центральный угол, а не длины дуги, по ним и находят значения этих центральных углов.
1/ Radius= 1 -> the perimeter of the circle= 2 pi 2/ If the ratios of the arc AB, AC and BC are 3-4-5 We have: Arc AB/3=Arc AC/4 =Arc BC/5 =(Arc AB+ArcAC+Arc BC)/(3+4+5)= 2pi/12=pi/6-->180 degree/6=30 degrees So angle AOB=90 degrees angle AOC=120 degrees and angle BOC=150 degrees. --> angleA=75 degrees, angle B=60 degrees=angle C=45 degrees. 3/Drop the height AH to BC We have AHC is a right isosceles 45-90-45 and AHB is a 30-90-60 triangle. By using the cosine law We have AC= sqrt3 -> AH= h=sqrt3/sqrt2 HB=1/sqrt2 The area of the blue triangle= Area of AHC + Area of AHB = 3/4+sqrt3/4= (3+sqrt3)/4 sq units😅
Sorry @PreMath. The question itself is wrong. It should be "Given the Circle Area = π. If arcs AB, BC and CA are 1/4th, 5/12th and 1/3rd of the circle circumference respectively, find out the area of the Triangle ABC.
Sir the relation between arc and radius and centre angle is Arc =r*centre angle (in radians) As r=1 we may say the ratio of three centre angles is 5 ঃ 3ঃ 4 It will save our time. 360/12=30 The angles are 150, 90,120 degrees Area of triangle =1/2(sin 150 +sin 90 +sin 120) =(1/2+1+√3/2)/2 sq units Comment please. [ Edited in consultation with a Math lover Sri Suprajeet Biswas. ]
radius of circle = s / (2 π) = 6 /π A /3 = B/4 = C /5 = (A +.B + C) /12 = 2 π /12 = π /6 sin ( A) = sin. (3 π /6) = 1 sin ( B) = sin ( 4 π./6) = √3/2 sin ( C) = sin ( 5 π /6) = 1/2 HERE BY | ∆ ABC | =( r^2/2) ( 1 + √3/2 + 1/2) = (6 / π) ^2 ( 3 + √3) /4
Let's find the area: . .. ... .... ..... First of all we calculate the radius R of the circle: A = πR² π = πR² 1 = R² ⇒ R = 1 From the given ratios of the arcs we can conclude: ∠AOB = 2π*arc(AB)/[arc(AB) + arc(AC) + arc(BC)] = 2π*3/(3 + 4 + 5) = 2π*3/12 = π/2 ∠AOC = 2π*arc(AC)/[arc(AB) + arc(AC) + arc(BC)] = 2π*4/(3 + 4 + 5) = 2π*4/12 = 2π/3 ∠BOC = 2π*arc(BC)/[arc(AB) + arc(AC) + arc(BC)] = 2π*5/(3 + 4 + 5) = 2π*5/12 = 5π/6 Now we are able to calculate the area of the green triangle. With R=OA=OB=OC we obtain: A(ABC) = A(OAB) + A(OAC) + A(OBC) = (1/2)*OA*OB*sin(∠AOB) + (1/2)*OA*OC*sin(∠AOC) + (1/2)*OB*OC*sin(∠BOC) = (1/2)*R²*[sin(∠AOB) + sin(∠AOC) + sin(∠BOC)] = (1/2)*1²*[sin(π/2) + sin(2π/3) + sin(5π/6)] = (1/2)*(1 + √3/2 + 1/2) = (1/2)*(3/2 + √3/2) = (3 + √3)/4 ≈ 1.183 Best regards from Germany
2πr is not equal to 3+4+5=12 12 is not the magnitude of length of the perimeter, but rather a ratio of relationship of the lengths of segments of arcs adding together to the length of the perimeter. This way arc-AB represents (3/12) of total length the perimeter, arc-AC represent (4/12) the total length of the perimeter , and arc-BC represents (5/12) of the total length of the perimeter; so that => arc-AB + arc-AC + arc-BC = (3/12)perimeter + (4/12)perimeter + (5/12) perimeter = (12/12) perimeter = perimeter ~ 6,28.
Your answer is wrong because using the circumference of the circle as drawn is 12 units and from that the radius of circle is (6/3.141592654) units=r=1.9098559317 units. So, the problem was presented wrong in the first place. Am I right? My answer is 7.05 square units for the green triangle.
Very nice question!!
Glad you think so!
Thanks for the feedback ❤️
You should have labeled the arcs 3x, 4x, and 5x to make sure that we knew from the picture that you were giving us ratios. You explained it in the video, but there are some viewers who'd rather solve the problem themselves before starting your video.
Yes, 3x, 4x and 5x would be better labels. Integers are usually measurements rather than proportions. For solving the problem, we need these measurements in degrees or radians, not units of length. So, 3x + 4x + 5x = 12x = 360° or 2π radians and x = 30° or π/6 radians.
@@jimlocke9320 I actually found the angles in radians but then was wondering if it was OK to use the law of cosines to find the sides.
@@StuartSimon First, I thought it's sides of triangle 😂😂 then radians..
That last sentence is me lol (the original comment, not the replies)
But I also check my answers by going to the end
@@ChuzzleFriends YEEEEEES, that last sentence is me lol 😂😂😅
didnt make sense to me until i listened to hear that 3 4 5 are ratios.
Hello,
There appears to be an inconsistency because the perimeter of the circle is 2.pi.r = 12, so if r = 1 you conclude that pi = 6?
Could you help me please?
3,4,5 are simply ratios and not absolute numbers. They should have been labeled 3k, 4k,5k or something like that for clarity.
I think this question has a loophole. Area iof circle is pi, which translates to r = 1. Howver total circumference is 2*pi*r = 12, giving a different r of 6/pi. 😅
Absolutely
At 5;45, we can apply Area = (1/2) ab sin(c) to ΔABC, computing the area of one triangle instead of computing areas of 3 triangles and summing their areas. Lengths AB and AC are more straightforward to calculate than BC, so we choose them as a and b. ΔABO is a special 45°-45°-90° isosceles right triangle, its hypotenuse is √2 times either side, so a = (√2)(1) = √2. Drop a perpendicular from O to AC, dividing ΔACO into 2 special 30°-60°-90° right triangles, each with long side (√3)/2 times the hypotenuse, or (√3)/2, total length b = √3. So a = √2 and b = √3. c =
I did it the same way but my posting is later! 😊
@@phungpham1725 Your method is different. You divided ΔABC into 2 triangles and summed the areas. I computed the area of one triangle ΔABC with one equation, but did need to divide ΔABC into smaller triangles to find side lengths a and b, and I constructed a triangle to find sin(75°).
@@jimlocke9320It’s OK! All roads lead to Rome😊!By the way, If chose to calculate AB first, it would have been a bit faster.
Superb👍
Excellent!
Glad to hear that!
Thanks for the feedback ❤️
A = πr²
π = πr²
r² = 1
r = 1
Arcs AB, AC, & BC are at ratios of 3, 4, & 5 in that order.
So, the lengths of the respective arcs are 3k, 4k, & 5k (where k is some constant).
So, the circumference of the circle is 12k (or 2π). We can find the measures of the arcs.
Measure of arc AB = 3k/12k = 1/4 * 360° = 90°
Measure of arc AC = 4k/12k = 1/3 * 360° = 120°
Measure of arc BC = 5k/12k = 5/12 * 360° = 150°
∠A is an inscribed angle. By the Measure of an Inscribed Angle Theorem, m∠A = (150°)/2 = 75°.
Before calculating the area of △ABC, we need to find the lengths of sides AB & AC. They are adjacent to ∠A.
Draw radii AO, BO, & CO. Find m∠AOB & m∠AOC.
The measure of a central angle is the measure of the arc it intercepts, and vice versa. So, ∠AOB is a right angle and m∠AOC = 120°.
This makes △AOB an isosceles right triangle. Use this formula:
c = a√2
AB = 1 * √2 = √2
Draw a radius perpendicular to side AC. This bisects ∠AOC and divides △AOC into two congruent right triangles. The smaller angles measure 60° each, so the triangles are both special 30°-60°-90° right triangles. We can use another special formula:
c = 2a
1 = 2a
a = 1/2
b = a√3
= 1/2 * √3
= (√3)/2
This takes up half the length of the side AC. So, AC = √3.
A = 1/2 * a * b * sinC
= 1/2 * √2 * √3 * sin(75°)
= 1/2 * √6 * (√6 + √2)/4
= √6 * (√6 + √2)/8
= [√6 * (√6 + √2)]/8
= (6 + √12)/8
= [6 + (√4)(√3)]/8
= (6 + 2√3)/8
= (3 + √3)/4
So, the area of the green triangle is (3 + √3)/4 square units (exact), or about 1.18 square units (approximation).
Hi, if r = 1 perimeter of the circle is 2 times pi times 1 is equal 2*pi that is not equal 12 . why ????
3,4,5 are ratios in which circumference is divided
not length of arc
listen to the introduction sir gives
Excellent .
Это пропорции на которые делится угловое значение дуги, опирающейся на хорду, и соответствующий центральный угол, а не длины дуги, по ним и находят значения этих центральных углов.
ok. thanks. (Спасибо.)
Thanks Sir
Thanks PreMath
Very nice and enjoyable
❤❤❤❤
Should have put constant k on the arc lengths since they are not the actual lengths given the radius is 1.
Thanks for the feedback ❤️
Thank you!
You are very welcome!
Thanks for the feedback ❤️
1/ Radius= 1 -> the perimeter of the circle= 2 pi
2/ If the ratios of the arc AB, AC and BC are 3-4-5
We have:
Arc AB/3=Arc AC/4 =Arc BC/5 =(Arc AB+ArcAC+Arc BC)/(3+4+5)= 2pi/12=pi/6-->180 degree/6=30 degrees
So angle AOB=90 degrees
angle AOC=120 degrees and angle BOC=150 degrees.
--> angleA=75 degrees, angle B=60 degrees=angle C=45 degrees.
3/Drop the height AH to BC
We have AHC is a right isosceles 45-90-45 and AHB is a 30-90-60 triangle.
By using the cosine law
We have AC= sqrt3
-> AH= h=sqrt3/sqrt2
HB=1/sqrt2
The area of the blue triangle= Area of AHC + Area of AHB = 3/4+sqrt3/4= (3+sqrt3)/4 sq units😅
Excellent!
Thanks for sharing ❤️
Area triangle AOB=1/2, area triangle AOC=\/3/4, area triaqngle BOC=1/4, area triangle ABC= 1/2+\/3/4+1/4=(3+\/3)/4.
АВ=\/2, АС=\/3, ВС=(\/3+1)/\/2.
S=(3+√3)/4≈11,83
I managed AOB = 1/2 cm^2, and that
Nice work!
Thanks for the feedback ❤️
How can the radius be 1, when the circumference is 12?
Same doubt
Can the area of acircle be equal 3.17
R= a*b* c/(4*S), S=0.5* a* b* sin(45), c=√2, S=0.5*c*a*sin(60), b=√3. S=0.5*√2*√3*sin(75)=0.5*√6* sin(45+30)=(√3+3)/4.
Sorry @PreMath. The question itself is wrong. It should be "Given the Circle Area = π. If arcs AB, BC and CA are 1/4th, 5/12th and 1/3rd of the circle circumference respectively, find out the area of the Triangle ABC.
Awesome, many thanks, Sir!
φ = 30°; πr^2 = π → r = 1; 2πr = 2π = 12x → x = π/6 → 3x = π/2 = 3φ → 4x = 2π/3 = 4φ → 5x = 5φ
∆ ABC → BC = a; AC = b; AB = c = √2 → BAO = OBA = 3φ/2; AO = BO = CO = r = 1
cos(4φ) = -cos(6φ - 4φ) = -cos(2φ) = -sin(φ) = -1/2; cos(5φ) = -cos(6φ - 5φ) = -cos(φ) = -√3/2 →
a^2 = 2(1 - cos(5φ)) = 2 + √3 → √(2 + √3) = (√2/2)(√3 + 1) = a; b^2 = 2(1 - cos(4φ)) = 3 → b = √3
BOC = 5φ → CBO = OCB = φ/2 → CBA = 3φ/2 + φ/2 = 2φ → sin(2φ) = cos(φ) = √3/2 →
area ∆ ABC = (1/2)sin(2φ)ab = (√3/4)(√3 + 1)
Excellent!
Thanks for sharing ❤️
Law of sine isnt needed, when you have got. the half of a Square with side length 1!
Sir the relation between arc and radius and centre angle is
Arc =r*centre angle (in radians)
As r=1 we may say the ratio of three centre angles is
5 ঃ 3ঃ 4
It will save our time.
360/12=30
The angles are 150, 90,120 degrees
Area of triangle
=1/2(sin 150 +sin 90 +sin 120)
=(1/2+1+√3/2)/2 sq units
Comment please.
[ Edited in consultation with a Math lover Sri Suprajeet Biswas. ]
Thanks for the feedback ❤️
radius of circle = s / (2 π) = 6 /π
A /3 = B/4 = C /5 = (A +.B + C) /12
= 2 π /12 = π /6
sin ( A) = sin. (3 π /6) = 1
sin ( B) = sin ( 4 π./6) = √3/2
sin ( C) = sin ( 5 π /6) = 1/2
HERE BY
| ∆ ABC | =( r^2/2) ( 1 + √3/2 + 1/2)
= (6 / π) ^2 ( 3 + √3) /4
Let's find the area:
.
..
...
....
.....
First of all we calculate the radius R of the circle:
A = πR²
π = πR²
1 = R²
⇒ R = 1
From the given ratios of the arcs we can conclude:
∠AOB = 2π*arc(AB)/[arc(AB) + arc(AC) + arc(BC)] = 2π*3/(3 + 4 + 5) = 2π*3/12 = π/2
∠AOC = 2π*arc(AC)/[arc(AB) + arc(AC) + arc(BC)] = 2π*4/(3 + 4 + 5) = 2π*4/12 = 2π/3
∠BOC = 2π*arc(BC)/[arc(AB) + arc(AC) + arc(BC)] = 2π*5/(3 + 4 + 5) = 2π*5/12 = 5π/6
Now we are able to calculate the area of the green triangle. With R=OA=OB=OC we obtain:
A(ABC)
= A(OAB) + A(OAC) + A(OBC)
= (1/2)*OA*OB*sin(∠AOB) + (1/2)*OA*OC*sin(∠AOC) + (1/2)*OB*OC*sin(∠BOC)
= (1/2)*R²*[sin(∠AOB) + sin(∠AOC) + sin(∠BOC)]
= (1/2)*1²*[sin(π/2) + sin(2π/3) + sin(5π/6)]
= (1/2)*(1 + √3/2 + 1/2)
= (1/2)*(3/2 + √3/2)
= (3 + √3)/4
≈ 1.183
Best regards from Germany
Radius=1 unit
????..2πr=3+4+5=12...r=6/π...A=π(6/π)^2=36/π=1...???
2πr is not equal to 3+4+5=12
12 is not the magnitude of length of the perimeter, but rather a ratio of relationship of the lengths of segments of arcs adding together to the length of the perimeter. This way arc-AB represents (3/12) of total length the perimeter, arc-AC represent (4/12) the total length of the perimeter , and arc-BC represents (5/12) of the total length of the perimeter; so that => arc-AB + arc-AC + arc-BC = (3/12)perimeter + (4/12)perimeter + (5/12) perimeter = (12/12) perimeter = perimeter ~ 6,28.
@@tony_0088un trabocchetto stupido per vedere chi apre il video e chi no
Your answer is wrong because using the circumference of the circle as drawn is 12 units and from that the radius of circle is (6/3.141592654) units=r=1.9098559317 units. So, the problem was presented wrong in the first place. Am I right? My answer is 7.05 square units for the green triangle.