(15, 8, 17) is a primitive Pythagorean tripel. Start with the even side. 8/2 = 1*4 is the only way to build the product with (positive) integers that the first factor is less than the second. Set u := 1, v := 4, t := (v-u) = 3, w := (v+u) = 5. Then: - a := t * w = 3 * 5 = 15 - The side lengths in terms of t, u, v, w - b := 2 * u * v = 2 * 1 * 4 = 8 - c := t * v + w * u = 3 * 4 + 5 * 1 = 17 - r_i := t * u = 3 * 1 = 3 - The radius of the inner circle - r_c := w * v = 5 * 4 = 20 - The radiusses of the outer circles - r_a := t * v = 3 * 4 = 12 - r_b := w * u = 5 * 1 = 5 - A = t * u * v * w = 3 * 1 * 4 * 5 = 60 - The area of the triangle in terms of t, u, v, w And finally, what we searched for: - A_i = π (r_i)² = 9π
28.28cm2 approx. U can also make three radii inside the circle by construction and make 3 triangles inside the triangle abc itself. then u can put the whole of triangle abc = sum of 1/2 x 8 x r + 1/2 x 15 x r + 1/2 x 17 x r also triangle abc = 1/2 x 8 x 15 which is 60cm2 now put 60 = 8r/2 + 15r/2 + 17r/2 to get 60 = 20r and r is 3cm therefore pir2 22/7 x 3 x 3 = 28.28cm2 approx
Poncelet's Theorem: The sum of the legs = The sum of hypotenuse + diameter of the inscribed circle 8 + 15 = 17 + diameter 23 = 17 + diameter 23 - 17 = diameter 6 = diameter diameter = 2 (radius) 6 = 2 ( radius ) 3 = radius Area of circle: π r² -> π (3)² = 9π Sorry for the bad english, it's not my native language
R = (a + b - c)/2
Equivalently, R = a*b /(a+b+c). Easy to remember: radius of inscribed circle in right triangle is twice its area divided by the perimeter.
(15, 8, 17) is a primitive Pythagorean tripel. Start with the even side. 8/2 = 1*4 is the only way to build the product with (positive) integers that the first factor is less than the second. Set u := 1, v := 4, t := (v-u) = 3, w := (v+u) = 5.
Then:
- a := t * w = 3 * 5 = 15 - The side lengths in terms of t, u, v, w
- b := 2 * u * v = 2 * 1 * 4 = 8
- c := t * v + w * u = 3 * 4 + 5 * 1 = 17
- r_i := t * u = 3 * 1 = 3 - The radius of the inner circle
- r_c := w * v = 5 * 4 = 20 - The radiusses of the outer circles
- r_a := t * v = 3 * 4 = 12
- r_b := w * u = 5 * 1 = 5
- A = t * u * v * w = 3 * 1 * 4 * 5 = 60 - The area of the triangle in terms of t, u, v, w
And finally, what we searched for:
- A_i = π (r_i)² = 9π
Awesome 🙌 thank you 🙏🏻
28.28cm2 approx. U can also make three radii inside the circle by construction and make 3 triangles inside the triangle abc itself. then u can put the whole of triangle abc = sum of 1/2 x 8 x r + 1/2 x 15 x r + 1/2 x 17 x r
also triangle abc = 1/2 x 8 x 15 which is 60cm2
now put 60 = 8r/2 + 15r/2 + 17r/2
to get 60 = 20r
and r is 3cm
therefore pir2
22/7 x 3 x 3 = 28.28cm2 approx
Wow Amazing 🦾👌🏾
Simply use r=delta/s
Perfect 👌🏾 🎊🎉🥳
Poncelet's Theorem:
The sum of the legs = The sum of hypotenuse + diameter of the inscribed circle
8 + 15 = 17 + diameter
23 = 17 + diameter
23 - 17 = diameter
6 = diameter
diameter = 2 (radius)
6 = 2 ( radius )
3 = radius
Area of circle:
π r² -> π (3)² = 9π
Sorry for the bad english, it's not my native language
Thank you for simplifying it💪🏻🥳🔥🎊
17 = (8 - r) + (15 - r) = 23 - 2r
r = (23 - 17)/2 = 3
circle area = 9π
Exactly!!
@@ApexmathematicsInMotion78 🙂🙏
Simple Way/Formula:
2s = 8+15+17 = 40 ➡️ s = 20
A = r•s = (1/2)•8•15 = 60
r = 60/20 = 3
O = π•r² = 9π 😉
Much-much Easier 👍
Well done friend, much appreciated 🙌🎉🎊
r²+17r=60
r²+17r-60=0
(r+20)(r-3)=0
r=-20 rejected
r=3 ok
Area(circ)=π3²
=9π
Perfectly summarised 🙌🔥🔥
15-8=7)(15+8=23÷17×7=9.47)(17-9.47÷2=3.63)(8×8-3.63×3.63=50.8231squroth=7.129÷2.4142135624=2.9629×2.9529×3.14159268=27.3942652986 area circile
Wow cool 🎉🎊👌🏾 🔥🔥
9π
Perfect 👌🏾
8 * 15 = 120
120 / 2 = 60
8R + 15R + 17R = 120
40R = 120
R = 120 / 40
R = 3
A (circle) = 9Pi Square Units.
Perfect Summary 🔥🔥
9π
Perfect 👌🏾