At 2:30, drop a perpendicular from F to BC and label the intersection as point G. Note that ΔBFG and ΔBAC are similar with sides of ΔBFG half as long as the corresponding sides of ΔBAC. Therefore, BG = 6 and FG = 8. Bisect
Once I had the sides of the small triangle, I got the area as 8*6=48. Then I split it into three triangles with r as the height. 5r + 5r + 6r = 48 so r = 3.
Ok, take the origin at the bottom left. The circle then has the following tangent lines: y=0, y=16-4*x/3, and y=4*x/3. (I'm not going through how to get these - I think it should be obvious to everyone). Triangle corners are at (0, 0), (12, 0) and (6, 8). To get this last one, set 16-4*x/3 = 4*x/3 and solve. The side lengths are thus 12, 10, and 10 (Pythagorean theorem). Now by Heron's formula we get the area of the triangle: semi-perimeter = (12+10+10)/2 = 16 triangle area = sqrt(16*(16-12)*(16-10)*(16-10)) = 48 The radius of the circle is now triangle area divided by semi-perimeter, or r = 48/16 = 3. So the area of the circle is just pi times radius squared, or Area = 9*pi Q.E.D.
Right triangle FBB' is a special right triangle 6-8-10: A = ½b.h = ½*6*8 = 24 cm² Radius of circle is also height of triangles A₁ and A₂ : A = A₁+A₂ = ½6r + ½10r = 8r r = A/8 = 3 cm Area of red circle: Ac = πr² = 9π cm² ( Solved √ )
12=4*3 ; 16=4*4---> 4*5=20. El círculo rojo está inscrito en un triángulo isósceles de base =12=2*6 ; altura =2*4=8 y lados =2*5=10 ---> Área del triángulo isósceles =12*8/2=48 =(12r/2)+10r=16r---> r=3--> Área círculo rojo =9π. Gracias y saludos.
Let the vertices of the initial triangle be A, B, and C, counterclockwise from the top. Let the vertices of the rotated triangle be D, E, and B, counterclockwise from bottom right. Let F be the intersection point between CA and EB, let O be the center of the circle, and let M, N, and P be the points of tangency between circle O and CF, FB, and BC respectively. As AB = DE = 16 = 4(4) and BC = EB = 12 = 3(4), ∆ABC and ∆DEB are both 4:1 ratio 3:4:5 Pythagorean triple right triangles and CA = BD = 5(4) = 20. Draw FP. As ∠FBC = ∠BCF, ∆CFB is an isosceles triangle, so FP bisects ∆CFB and forms two congruent right triangles ∆BPF and ∆FPC. As ∠BCA and ∠PCF are the same angle and ∠FPC = ∠ABC = 90°, ∆FPC (and thus ∆BPF) is similar to ∆ABC. As BC/PC = 12/6 = 2, the larger triangle has a 2:1 ratio of its dimensions compared to the smaller triangles, so FP = 16/2 = 8 and CF = FB = 20/2 = 10. As CM and PC are tangent to circle O and intersect at C, by the two tangent theorem, they are congruent, so CM = PC = 6 and MF = CF-CM = 10-6 = 4. Similarly, NB = BP = 6 and FN = FB-NB = 10-6 = 4. Draw OM, ON, and OP. As M, N, and P are all points of tangency between circle O and triangle ∆CFB, ∠OMF = ∠ONB = ∠OPC = 90°. As ∠MFO and ∠CFP are the same angle and ∠OMF = ∠FPC = 90°, ∆OMF is a similar triangle to ∆FPC and ∆ABC and thus also a 3:4:5 Pythagorean triple right triangle. As MF = 4 and is the longer leg of the triangle, then OF is the hypotenuse at length 5 and OM is the short leg at length 3. As OM is a radius of circle O, r = 3. Circle O: A = πr² = π(3)² = 9π sq units
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It can be easily solved using coordinate geometry by taking right angle vertex at origin
Respected Sir, Good evening......Nicely solved...
Good videos being made available to viewers.Thanks a lot sir.vous enseignez mon apprentissage.merci beaucoup monsieur
At 2:30, drop a perpendicular from F to BC and label the intersection as point G. Note that ΔBFG and ΔBAC are similar with sides of ΔBFG half as long as the corresponding sides of ΔBAC. Therefore, BG = 6 and FG = 8. Bisect
You could solve this in your head using 3:4:5 similar triangles and tangent theorem.
Once I had the sides of the small triangle, I got the area as 8*6=48. Then I split it into three triangles with r as the height. 5r + 5r + 6r = 48 so r = 3.
tan α = 16/12 --> α = 53,13°
r = ½.12 tan α/2 = 3 cm
A = πr² = 9π cm² ( Solved √ )
Great video 🎉
Thank you 😁
Ok, take the origin at the bottom left. The circle then has the following tangent lines: y=0, y=16-4*x/3, and y=4*x/3. (I'm not going through how to get these - I think it should be obvious to everyone). Triangle corners are at (0, 0), (12, 0) and (6, 8). To get this last one, set 16-4*x/3 = 4*x/3 and solve. The side lengths are thus 12, 10, and 10 (Pythagorean theorem).
Now by Heron's formula we get the area of the triangle:
semi-perimeter = (12+10+10)/2 = 16
triangle area = sqrt(16*(16-12)*(16-10)*(16-10)) = 48
The radius of the circle is now triangle area divided by semi-perimeter, or r = 48/16 = 3. So the area of the circle is just pi times radius squared, or
Area = 9*pi
Q.E.D.
Right triangle FBB' is a special right triangle 6-8-10:
A = ½b.h = ½*6*8 = 24 cm²
Radius of circle is also height of triangles A₁ and A₂ :
A = A₁+A₂ = ½6r + ½10r = 8r
r = A/8 = 3 cm
Area of red circle:
Ac = πr² = 9π cm² ( Solved √ )
Yes, it’s right there!
THX for the problem. One remark : the original rotation of the triangle is NOT of 90°
@@johngutwirth7706 You're absolutely right. That was a slip of the tongue.
GREAT PROBLEM ANS SOLUTION BUT I WAS NOT ABLE TO DO THIS HEHEHE
12=4*3 ; 16=4*4---> 4*5=20.
El círculo rojo está inscrito en un triángulo isósceles de base =12=2*6 ; altura =2*4=8 y lados =2*5=10 ---> Área del triángulo isósceles =12*8/2=48 =(12r/2)+10r=16r---> r=3--> Área círculo rojo =9π.
Gracias y saludos.
Let the vertices of the initial triangle be A, B, and C, counterclockwise from the top. Let the vertices of the rotated triangle be D, E, and B, counterclockwise from bottom right. Let F be the intersection point between CA and EB, let O be the center of the circle, and let M, N, and P be the points of tangency between circle O and CF, FB, and BC respectively.
As AB = DE = 16 = 4(4) and BC = EB = 12 = 3(4), ∆ABC and ∆DEB are both 4:1 ratio 3:4:5 Pythagorean triple right triangles and CA = BD = 5(4) = 20.
Draw FP. As ∠FBC = ∠BCF, ∆CFB is an isosceles triangle, so FP bisects ∆CFB and forms two congruent right triangles ∆BPF and ∆FPC. As ∠BCA and ∠PCF are the same angle and ∠FPC = ∠ABC = 90°, ∆FPC (and thus ∆BPF) is similar to ∆ABC. As BC/PC = 12/6 = 2, the larger triangle has a 2:1 ratio of its dimensions compared to the smaller triangles, so FP = 16/2 = 8 and CF = FB = 20/2 = 10.
As CM and PC are tangent to circle O and intersect at C, by the two tangent theorem, they are congruent, so CM = PC = 6 and MF = CF-CM = 10-6 = 4. Similarly, NB = BP = 6 and FN = FB-NB = 10-6 = 4.
Draw OM, ON, and OP. As M, N, and P are all points of tangency between circle O and triangle ∆CFB, ∠OMF = ∠ONB = ∠OPC = 90°.
As ∠MFO and ∠CFP are the same angle and ∠OMF = ∠FPC = 90°, ∆OMF is a similar triangle to ∆FPC and ∆ABC and thus also a 3:4:5 Pythagorean triple right triangle. As MF = 4 and is the longer leg of the triangle, then OF is the hypotenuse at length 5 and OM is the short leg at length 3. As OM is a radius of circle O, r = 3.
Circle O:
A = πr² = π(3)² = 9π sq units
r=3
The easiest way I know of to find the answer is to watch the video :D
😂
r=12÷2cot[½tan-¹(16/12)]=3
Che palle!!!!! Ma possibile che dobbiate spiegare i fondamenti e dobbiate fare tutti i passaggi ? I video sono noiosossimi