You can also use the fact that the lines between the vertices and the mid-points of the opposite sides divide each other in the ratio 2:1. In this case the line lengths are 4 sqrt(3) so r = 4/sqrt(3)
The general formula to find the inradius of the incircle in any triangle also works. In this case, it simplifies to _a √3 / 6_ where _a_ is the side of the equilateral triangle. Thus, the formula for the area of the incircle in an equilateral triangle is _π a² / 12._
I approached this slightly differently. Where you made the triangle AOD, I observed that the total area of the whole triangle is made up of six of these identical triangles; AOD, AFO, BOD, BOE, COF, COE. Each one has an area equal to 0.5×4×r = 2r, and thus the total area of ABC is 12r. So then you can easily find the area of ABC via calculating the height CD, set the two exoressions for the area of ABC as equal to each other, rearrange for r, and then calculate the circle area.
we just can calculate the area of triangle first and then divide this triangle in three part from center and height of each part will be radius and side will base and just using formula area(it will be 1/3 of that) 1/2 * base * height we can get r easily (please reply sir)
Можно воспользоваться теоремой о биссектрисах треугольника. Делится точкой пересечения в отношении 2:1, считая от вершины. Поделить высоту, которая является биссектрисой на 3. Это и будет радиус вписанной окружности.
Clone triangle ABC, rotate it 180° around O. Now you've got an regular hexagram, with six equilateral triangles surrounding a regular hexagon, all with side length 8/3, and the hexagon has the green circle inscribed. Radius of the circle is equal to the minimal radius of the hexagon, equals 4/sqrt(3) in this case. Formula for area of a circle resulted in 16 * pi / 3
In an equilateral triangle, the height is also the median, and the bisector of the the 3 corner angles. So r=OB=CD /3=1/3 x 8sqrt3/2=4sqrt3/3 Area of the inscribed circle= pi .( 16x3)/9= 16pi /3 sq units
El centro del círculo coincide con el baricentro del triángulo equilátero y dista de la base un tercio de la altura “h” → h=(8√3)/2 =4√3 → Radio r=h/3=(4√3)/3 → Área verde =πr² =π16x3/9 =π16/3 =16.7552 Gracias y un saludo cordial.
Equilateral triangle ABC has area = 1/2 × 8 × 8 × cos(60°) = 8√3 Each of triangles AOB, BOC, and AOC have base = 8, height = r, and area = 1/2 × 8 × r = 4r. So area of triangle ABC = 3 × 4r = 12r = 8√3 → r = 4/√3 Area of circle = πr² = 16π/3
As the triangle is equilateral, the triangle FOA will have the angles 30:60:90. The hypotenuse of such a triangle is 2 times the shortest side. The shortest side is the radius of the circle, so the length OA is 2r. This makes the whole length of the line AOE 3r which is the height of the triangle. Using Pythagoras, the height of the triangle is also the sq root of (8^2 - 4^2) = sq root of (64-16) =the sq root of 48. So the radius is 1 third of the square root of 48. The area is pi * r^2 so the area of the circle is 48/9*pi.
Nice, quite simple and straight forward, the kind of question I can do in front of my friends and have them in awe of my knowledge 😂😂. Love it, thanks again 👍🏻
I started by finding 2r + X = 4 root 3. Then manipulated (R+X)^2 = R^2 + 4^2 , Getting the R^2 term to drop out, then the X^2 terms to cancel leaving 4 root 3 times X = 16. Then X = 16/ 4root 3. Rationalize the denominator X= ( 4 root 3) / 3. Then you have Height = 4 root 3 which is equal to 2R + X. Substitute and solve you get R = (4 root 3)/3. Using the formula A= Pi X R^2 you get Pi X 48/9 or 16.755 sq. un.
The height of the equilateral triangle is three times the radius. Considering the height AE and the triangle AOD which has angles of 30°, 60° and 90°, therefore the hypotenuse is twice the minor cathetus. But the minor cathetus OD is the radius, so along the height AE we have AO which is twice the radius and OE which is the radius
You can make a 120 degree sector by drawing lines through the centre to 2 of the midpoints and use cosine rule to find the radius (knowing the line connecting the midpoints has to be 4 because it creates a mini equilateral triangle)
Once we found the height h, the radius of the inscribed circle is always ⅓ of the height length . The centroid of an equilateral triangle which are also perpendicular to the bases, and splits the medians into two segments measuring ⅓ of the length ( which is the radious of the circle ) and ⅔ of the length, respectively.
The radius is 4/sqrt(3). Reason: Half of the side is 4. If we draw a line from a corner to the center of the circle, it bisects the angle at the corner. Together with a perpendicular that bisects the side, we get a right triangle between a triangle-corner, the circle-center and the midpoint of the side, which is a 30-60-90 triangle. In such a triangle the ratio between the sides are: hypothenuse =2, longer side = sqrt(3), shorter side = 1. So in order to get from the longer to the shorter side, you need to divide the longer side (4) by sqrt(3). The radius is the short side and is therefore 4/sqrt(3).
Both cases trigonometry was used whether it was in the heading that Two methods : one with trigonometry Other without trigonometry ++ But there is proof to compute in -radius only with the help of geometry
I got the same answer but used a different trig construction. I got: _r_ tan(60) = 4. This solves to r = 4 / tan(60) ~ 2.3094... which also gets the area of the circle being A ~ 16.755... square units.
Because that triangle is equilateral triangle so that h=aV3/2 (a=8 is edge of triangle) in equilateral triangle triangle height line also median line h= 8V3/2=4V3 R= 1/3 xh=1/3x4V3=4V3/3 A= R²"3,14=(4V3/3)²x3,14=16/3x4,14=16,74
The Traingle CBA equality we make two lengths hypotunese hieght CD divide Triangle According to Theorem Phythagorean h=2√(12) sins rule degree 60⁰=h/8 , 60=√(3)/2, h=4√(3) r/4=4/h , Area of Green shaded circle (16/3)π
Here's a method using only Pythagoras, without trig functions: Draw CD and determine the altitude of the triangle: 8^2 - 4^2 = 64 - 16 = 48; √48 = √((16)(3)) = 4√3: Now consider triangle AOD, or any of the six similar triangles: OD = r, and AO = 4√3 - r. Again invoking Pythagoras: 4^2 + r^2 = (4√3 - r)^2 = 48 - (8√3)(r) + r^2; subtract r^2 from both sides and it drops out: 4^2 = 48 - (8√3)(r); multiply out: 16 = 48 - (8√3)(r); collect terms and rearrange: (8√3)(r) = 32; solve for r: r = 32/(8√3) = 4/√3. Then the area of the circle is π(4/√3)^2 = 16π/3. Just what I needed to get he day off to a good start. Cheers. 🤠
This type of triangle is devided by the centre of the circle O in 3 equal triangles (AOB,BOC,COA) with a side=8 and h1=OF=OD=OE= R Area of the white triangle is A= 3.8.R/2=12R and A= 8.h/2= 4h , so R=h/3 R^2=h^2/9 h^2=8^2- 4^2=64-16=48 so R^2= h^2/ 3^2 = 48/9 The green circle area A= 3,14.R^2= 3,14. 48/9 =16,76
"With or Without You" "I cant live..." 🎵 U2 With Heron's or Trinity: A = (1/4)•(√3)•a" s = (3/2)•a A = s•r ➡ r = A/s = (1/6)•(√3)•a O = π•r" = (π/12)•a" ↔ a = 8 O = (16/3)•π = 16,7552^ "Take an Easy way out"
Via Heron's Formula: Area of triangle = radius of circle x semi-perimeter Semi-perimeter = (a+b+c)/2 Semi-perimeter = (3x8)/2 Semi-perimeter = 12 Area of triangle = 0.5 bh Area of triangle = 0.5 (8)(4√3) Area of triangle = 16√3 Radius of circle = (16√3)/12 Radius of circle = (4/3)√3 units Area of circle = π (radius)^2 Area of circle = π ((4/3)√3)^2 Area of circle = 16π/3 sq.units
By the way we need to remember some some distances of equatorial triangle like height,distance from mid point to a arm and a corner. The distance from mid point to arm is a/2×3½ so we got the radius now put it into circle's area = pie × (8/2×3½)²=16.76
When you know that in an equilateral triangle the inradius is 1/3 the length of an altitude (because each altitude is also a median of the triangle), then the solution of this problem is straight forward.
Approximations are weird for me because of values like 3½ and Pi ... I guess one question is what is really going on for instance when approaching Pi...a non-terminating non-repeating decimal. 🙂
A pretty deep question. We now accept as "numbers" things we can only approach by a limit process, never actually get to. "God made the integers, the rest is the work of Man" -- Dedekind (I think)
If we are expected to have memorized that sin(60) = sqrt(3)/2, why aren't we expected to know that the sides of a 30-60-90 triangle are x-x*sqrt(3)-2x? If you know that, then you know that if you draw a line from O to a corner you're creating a 30-60-90 triangle with long leg of length 4. Then the radius is 4 = x*sqrt(3) or x =4/sqrt(3).
If your picture does not contain one of the letters a, b or c and you nonetheless say that a^2 + b^2 = c^2 is the Pythagoras Theorem you only show, that you don't know, what you are talking about. Sorry for this unpleasant statement.
You can also use the fact that the lines between the vertices and the mid-points of the opposite sides divide each other in the ratio 2:1. In this case the line lengths are 4 sqrt(3) so r = 4/sqrt(3)
Centroid fact!
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Those lines are called as medians. Centroid divides each median in the ratio 2:1.
The general formula to find the inradius of the incircle in any triangle also works.
In this case, it simplifies to _a √3 / 6_ where _a_ is the side of the equilateral triangle.
Thus, the formula for the area of the incircle in an equilateral triangle is _π a² / 12._
I approached this slightly differently.
Where you made the triangle AOD, I observed that the total area of the whole triangle is made up of six of these identical triangles; AOD, AFO, BOD, BOE, COF, COE.
Each one has an area equal to 0.5×4×r = 2r, and thus the total area of ABC is 12r.
So then you can easily find the area of ABC via calculating the height CD, set the two exoressions for the area of ABC as equal to each other, rearrange for r, and then calculate the circle area.
we just can calculate the area of triangle first and then divide this triangle in three part from center and height of each part will be radius and side will base and just using formula area(it will be 1/3 of that) 1/2 * base * height we can get r easily
(please reply sir)
Можно воспользоваться теоремой о биссектрисах треугольника. Делится точкой пересечения в отношении 2:1, считая от вершины. Поделить высоту, которая является биссектрисой на 3. Это и будет радиус вписанной окружности.
Clone triangle ABC, rotate it 180° around O. Now you've got an regular hexagram, with six equilateral triangles surrounding a regular hexagon, all with side length 8/3, and the hexagon has the green circle inscribed. Radius of the circle is equal to the minimal radius of the hexagon, equals 4/sqrt(3) in this case. Formula for area of a circle resulted in 16 * pi / 3
DE = EF = FD = 4. Extend DO till edge of circle at H via G. GH*HD = 2*2 = 4. GH = 4/sqrt12. So Area = 16.755
In an equilateral triangle, the height is also the median, and the bisector of the the 3 corner angles. So r=OB=CD /3=1/3 x 8sqrt3/2=4sqrt3/3
Area of the inscribed circle= pi .( 16x3)/9= 16pi /3 sq units
El centro del círculo coincide con el baricentro del triángulo equilátero y dista de la base un tercio de la altura “h” → h=(8√3)/2 =4√3 → Radio r=h/3=(4√3)/3 → Área verde =πr² =π16x3/9 =π16/3 =16.7552
Gracias y un saludo cordial.
You are very welcome!
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
Love and prayers from the USA! 😀
Angle(EBD) =60°
angle(EOD) =120°
Angle(EBO) =30°
EB=4cm.
tan30°=1/√3 =r/4
r=4/√3
Green 🟢=π(4/√3)^2
=16π/3 unit
h =4sqrt(3) using pythagoras.
h.(h-2r) = 4.4
As per Tangent secant rule
h^2-2hr = 16
48-8sqrt(4)r = 16
r = 4/sqrt(3)
A = 16pi/3
The solution can be intuited from similar triangles
ACD and AOD are similar
therefore
Tan 30 == AD/CD == OD/AD
r = OD = 4/4 * (4sqrt(3))
Equilateral triangle ABC has area = 1/2 × 8 × 8 × cos(60°) = 8√3
Each of triangles AOB, BOC, and AOC have base = 8, height = r, and area = 1/2 × 8 × r = 4r.
So area of triangle ABC = 3 × 4r = 12r = 8√3 → r = 4/√3
Area of circle = πr² = 16π/3
As the triangle is equilateral, the triangle FOA will have the angles 30:60:90. The hypotenuse of such a triangle is 2 times the shortest side. The shortest side is the radius of the circle, so the length OA is 2r. This makes the whole length of the line AOE 3r which is the height of the triangle. Using Pythagoras, the height of the triangle is also the sq root of (8^2 - 4^2) = sq root of (64-16) =the sq root of 48. So the radius is 1 third of the square root of 48. The area is pi * r^2 so the area of the circle is 48/9*pi.
Nice, quite simple and straight forward, the kind of question I can do in front of my friends and have them in awe of my knowledge 😂😂. Love it, thanks again 👍🏻
Find area through heron's formula of ABC --(1)
Also, area of ABC=r×(8+8+8)/2--(2)
Equating (1) &(2)
you will get r
I started by finding 2r + X = 4 root 3. Then manipulated (R+X)^2 = R^2 + 4^2 , Getting the R^2 term to drop out, then the X^2 terms to cancel leaving 4 root 3 times X = 16. Then X = 16/ 4root 3. Rationalize the denominator X= ( 4 root 3) / 3. Then you have Height = 4 root 3 which is equal to 2R + X. Substitute and solve you get R = (4 root 3)/3. Using the formula A= Pi X R^2 you get Pi X 48/9 or 16.755 sq. un.
The height of the equilateral triangle is three times the radius. Considering the height AE and the triangle AOD which has angles of 30°, 60° and 90°, therefore the hypotenuse is twice the minor cathetus. But the minor cathetus OD is the radius, so along the height AE we have AO which is twice the radius and OE which is the radius
Very elegant solutions. Thanks Professor!❤🥂
Glad to hear that!
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You can make a 120 degree sector by drawing lines through the centre to 2 of the midpoints and use cosine rule to find the radius (knowing the line connecting the midpoints has to be 4 because it creates a mini equilateral triangle)
Once we found the height h, the radius of the inscribed circle is always ⅓ of the height length .
The centroid of an equilateral triangle which are also perpendicular to the bases, and splits the medians into two segments measuring ⅓ of the length ( which is the radious of the circle ) and ⅔ of the length, respectively.
S=16π/3≈16,75
The radius is 4/sqrt(3). Reason: Half of the side is 4. If we draw a line from a corner to the center of the circle, it bisects the angle at the corner. Together with a perpendicular that bisects the side, we get a right triangle between a triangle-corner, the circle-center and the midpoint of the side, which is a 30-60-90 triangle. In such a triangle the ratio between the sides are: hypothenuse =2, longer side = sqrt(3), shorter side = 1. So in order to get from the longer to the shorter side, you need to divide the longer side (4) by sqrt(3). The radius is the short side and is therefore 4/sqrt(3).
Both cases trigonometry was used
whether it was in the heading that
Two methods
: one with trigonometry
Other without trigonometry
++
But there is proof to compute in -radius only with the help of geometry
The centre O of the inscribed circle is, also the centroid of the equilateral triangle ABC which divides CB in 2:1. So, OD=1/3 of CB=1/3 of 4√3.
We directly find the radius of circle in equilateral triangle is=8/(2√3) i.e 4/√3 . Area=16π/3.
M3 area of triangle = semi primeter * in radius inradius =A/S A=16 ROOT3 S=12 IN RADIUS =4*root3 /3 hence area of circle = 16 pi/3
Triangle FED is also equilateral with side l=4 and is inscribed in the circle, so r=l/√3=4/√3
I got the same answer but used a different trig construction. I got:
_r_ tan(60) = 4.
This solves to r = 4 / tan(60) ~ 2.3094... which also gets the area of the circle being A ~ 16.755... square units.
Because that triangle is equilateral triangle so that h=aV3/2 (a=8 is edge of triangle) in equilateral triangle triangle height line also median line h= 8V3/2=4V3
R= 1/3 xh=1/3x4V3=4V3/3
A= R²"3,14=(4V3/3)²x3,14=16/3x4,14=16,74
The Traingle CBA equality we make two lengths hypotunese hieght CD divide Triangle According to Theorem Phythagorean h=2√(12) sins rule degree 60⁰=h/8 , 60=√(3)/2, h=4√(3) r/4=4/h , Area of Green shaded circle (16/3)π
Excellent!
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Here's a method using only Pythagoras, without trig functions:
Draw CD and determine the altitude of the triangle:
8^2 - 4^2 = 64 - 16 = 48; √48 = √((16)(3)) = 4√3:
Now consider triangle AOD, or any of the six similar triangles:
OD = r, and AO = 4√3 - r. Again invoking Pythagoras:
4^2 + r^2 = (4√3 - r)^2 = 48 - (8√3)(r) + r^2; subtract r^2 from both sides and it drops out:
4^2 = 48 - (8√3)(r); multiply out:
16 = 48 - (8√3)(r); collect terms and rearrange:
(8√3)(r) = 32; solve for r:
r = 32/(8√3) = 4/√3.
Then the area of the circle is π(4/√3)^2 = 16π/3.
Just what I needed to get he day off to a good start.
Cheers. 🤠
I think the solution is much simpler. By drawing the two medians you get tan30=r/4, hence r and area of the circle
This type of triangle is devided by the centre of the circle O in 3 equal triangles (AOB,BOC,COA) with a side=8 and h1=OF=OD=OE= R
Area of the white triangle is A= 3.8.R/2=12R and A= 8.h/2= 4h , so R=h/3 R^2=h^2/9 h^2=8^2- 4^2=64-16=48 so R^2= h^2/ 3^2 = 48/9
The green circle area A= 3,14.R^2= 3,14. 48/9 =16,76
By trigonometry, radius is 4 tan 30=4/root 3, area of circle is 16/3 pi=16.755 approximately. 🙂
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Thanks for video.Good luck sir!!!!!!!!!!
Very well explained👍
Thanks for sharing😊😊
"With or Without You"
"I cant live..." 🎵 U2
With Heron's or Trinity:
A = (1/4)•(√3)•a"
s = (3/2)•a
A = s•r ➡ r = A/s = (1/6)•(√3)•a
O = π•r" = (π/12)•a" ↔ a = 8
O = (16/3)•π = 16,7552^
"Take an Easy way out"
Via Heron's Formula:
Area of triangle = radius of circle x semi-perimeter
Semi-perimeter = (a+b+c)/2
Semi-perimeter = (3x8)/2
Semi-perimeter = 12
Area of triangle = 0.5 bh
Area of triangle = 0.5 (8)(4√3)
Area of triangle = 16√3
Radius of circle = (16√3)/12
Radius of circle = (4/3)√3 units
Area of circle = π (radius)^2
Area of circle = π ((4/3)√3)^2
Area of circle = 16π/3 sq.units
By the way we need to remember some some distances of equatorial triangle like height,distance from mid point to a arm and a corner. The distance from mid point to arm is a/2×3½ so we got the radius now put it into circle's area = pie × (8/2×3½)²=16.76
Without trigonometry:
Extend OF by length r to the left to point O'. This forms an equilateral triangle COO' of side 2r.
CO+ OD= 2r+r = h
r = h/3
Area of triangle = r *( semi perimeter)
You solve math very long to find R of circle ,use nature of median line in triangle from top 2/3 and 1/3 from edge to solve
Since O is the centroid then OD= r= 1/3 of CD.
Thank you
When you know that in an equilateral triangle the inradius is 1/3 the length of an altitude (because each altitude is also a median of the triangle), then the solution of this problem is straight forward.
Approximations are weird for me because of values like 3½ and Pi ... I guess one question is what is really going on for instance when approaching Pi...a non-terminating non-repeating decimal. 🙂
A pretty deep question. We now accept as "numbers" things we can only approach by a limit process, never actually get to. "God made the integers, the rest is the work of Man" -- Dedekind (I think)
Area of incircle of an equilateral triangle, equal to 22%7*side square %12. I think so.formula.
Yay! I solved it (16/3) * pi
No r method. For triangle AOD (AD)^2 + (OD)^2 = (AO)^2, 4^2 + r^2 = (2*r)^2 , 16 + r^2 = 4*r^2, r^2=16/3
2*r proof P(ABC)=6P(AOB) 8h/2=6*4*r/2 h=3r AO=h-r AO=3r-r=2r AO=CO=2r 😎
If we are expected to have memorized that sin(60) = sqrt(3)/2, why aren't we expected to know that the sides of a 30-60-90 triangle are x-x*sqrt(3)-2x? If you know that, then you know that if you draw a line from O to a corner you're creating a 30-60-90 triangle with long leg of length 4. Then the radius is 4 = x*sqrt(3) or x =4/sqrt(3).
√[8²-4²]=√48=4√3 r+2r=4√3 3r=4√3 r=4√3/3
area of Green Circle : 4√3/3*4√3/3*π=16π/3
I remember this question had appeared in March 1994 10th class board exams. I was part of this exam and most of us could not solve ot back then 😢
More easy method. Length od is route 2route3
Без подобия треугольников AO можно выразить как h-r и подставить в теорему Пифагора.
5,333pi=16,755
Elegant
r=4/3sqrt3...A=16/3pi
Excellent!
Thanks for sharing! Cheers!
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100th like by me
If your picture does not contain one of the letters a, b or c and you nonetheless say that a^2 + b^2 = c^2 is the Pythagoras Theorem you only show, that you don't know, what you are talking about. Sorry for this unpleasant statement.
In English please.