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this is the best way to solve this problem we should always try to use basic methods which we never forgot in future. Good problem and nice solution.
Thanks a lot.
2.(10-R) = 10 / cos 45°20 - 2R = 10√2R = 10 - 5√2 = 2,929 cm ( Solved √ )
Área ABC =10*10/2=50 =2*(10-r)r+r²→ 10-5√2 =2,92893... ud.Gracias y un saludo.
[2.(10-R)]² = 2. 10²4 . (10²-20R + R²) = 200R² - 20R + 50 = 0 R = 2,929 cm ( Solved √ )
The radius is 5[2-sqrt(2)]. I checked that the radius is more than zero. This should indicate that I have done a sanity check!!!
this is the best way to solve this problem we should always try to use basic methods which we never forgot in future. Good problem and nice solution.
Thanks a lot.
2.(10-R) = 10 / cos 45°
20 - 2R = 10√2
R = 10 - 5√2 = 2,929 cm ( Solved √ )
Área ABC =10*10/2=50 =2*(10-r)r+r²→ 10-5√2 =2,92893... ud.
Gracias y un saludo.
[2.(10-R)]² = 2. 10²
4 . (10²-20R + R²) = 200
R² - 20R + 50 = 0
R = 2,929 cm ( Solved √ )
The radius is 5[2-sqrt(2)]. I checked that the radius is more than zero. This should indicate that I have done a sanity check!!!