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[@rabotaakk-nw9nm]

7:17 c=53; a m=7; n=2
b=m²-n²=49-4=45
a=2mn=2•7•2=28 😁

[@devondevon4366]

73 , 45, and 28
Let vertical line of the triangle = a
then from the point of tangency to the top of the triangle is a - 10
and the horizontal = b
then from point of tangency to the vertex of the triange is a- 10
then the hypotenuse = (a + b -20) tangent circle theores
Hence, the perimeter of the triangle = a + b + a+ b -20
Hence, 2 a + 2b - 20 =126
2a + 2b = 146
a + b =73
Hence, c= 53 (126-73) one get 53, it is basically over given Pythagorean triple 45, 28, and 73
a^2 + b^2 + 2ab = 73^2 (square a + b =73)
a^2 + b^2 =53^2 ( Pythag)
53^2 + 2ab = 73^2
2ab = 73^2 - 53^2
2ab =2520
ab = 1260
a = 1260/b
Since a + b = 73, then
1260/b + b -73=0
1260 + b^2 - 73b=0
(b-28)(b-45) = 0
b =28 and b=45

[@jkhazraji7358]

Semiperimeter * radius(r) = Area(A)
Semiperimeter = 126 /2 = 63
A= 63 * 10= 630
Area of triangle = ½a*b= 630 ===> a*b= 1260 ..... (1)
a+b+c=126,
a+b-c=20(2r)
===> a+b = 73 ===> b = 73 - a ......(2)
c = 126 - 73 = 53
from eq. 1 and 2:
a * (73 - a) = 1260 =====> a = (28,45) , b =(45,28)

[@jkhazraji7358]

a+b+c=126,
a+b-c=20(2r)
===> a+b = 73 ===> b = 73 - a
===> c= 126 - 73 = 53
b² + a² = c²
===>(73-a)² + a² = 53²
====> (a,b,c) = ( 28,45, 53) or (45, 28, 53)

[@gaiatetuya92]

円の半径などいらない。
3辺とも互いに素な自然数になるなら原始ピタゴラス数を求めるのと同じだ。
３辺は m^2ーn^2 , 2mn , m^2＋n^2 とおける。( m(>n) , nは自然数 )
周の長さが126より　(m^2ーn^2) ＋ (2mn) ＋ ( m^2＋n^2)＝126
整理すると m(m＋n)＝63
ｍ,ｎは自然数だから ｍ＝7 , m+n＝9 ∴n=2
よって３辺は　45 , 28 , 53

[@imetroangola4943]

*Outra maneira:*
ab/2=(a+b+c)R/2→ab=1260.
Por Pitágoras:
a²+b²=c². Por outro lado,
(a+b)²=a²+b²+2ab. Como a+b+c=126→a+b=126-c e ab=1260. Logo,
(126-c)²=c²+2×1260. Daí,
126²-252c+c²=c²+2520→
252c=15876-2520=13356
c=13356/252→ *c=53.*
Assim,
a+b=126-53→a+b=73 e como ab=1260. Facilmente, resolve pela fórmula de equação do segundo grau. Você vai encontrar *a=28 e b=45.*

[@devondevon4366]

28, 45, 83

[@RealQinnMalloryu4]

(10)^2= 100 {18°A+18°B+90°C} =126°ABC {126°ABC/100} =1.26ABC 1^1.2^13 2^13^1 2^1^1 2^1 (ABC ➖ 2ABC+1).

[@himo3485]

10+10+x+x+y+y=126 2(10+x+y)=126 10+x+y=63 x+y=53
(10+x)²+(10+y)²=(x+y)² 100+20x+x²+100+20y+y²=x²+2xy+y² 200+20x+20y=2xy
200+20(x+y)=2xy 1260=2xy xy=630
x²+y²+1260=2809 x²+y²=1549 　　　　　　　　
(x-y)²=1549-1260=289 x-y=17
2x=70 x=35 y=18
BC=10+x=10+35=45 AB=10+y=10+18=28 AC=x+y=35+18=53

[@michaeldoerr5810]

Hello I am actually a little confused at the 10:43 mark you have 2(a^2-73a+126*10). I think that you forgot to take the 2 out of 126. That would leave you with 2(a^2-73a+63*10) which is equivalent to 2a^2-146a+126*20. I think that that would have made this calculation quicker. I could be wrong.

[@michaeldoerr5810]

I meant the 10:13 mark.

[@MathBooster]

I take 2 out of 20. So, we don't further need to take 2 out of 126.

[@michaeldoerr5810]

​@@MathBoosterI actually tried the simplification my way and I now have learned that factor by grouping can backfire if simplified too much.

[@shankhadeepmukherjee9985]

❤❤❤amazing math

[@user-ll5wl9gq4x]

На этот раз все в принципе завязано на Пифагоровой триаде 28-45-53! Легко также вывести и применить формулу 2r = a + b - c. СПАСИБО за красивые задания!🖕

[@Cricketdoctor_1999]

bro can you explain it...plz😇

[@user-ll5wl9gq4x]

@@Cricketdoctor_1999 Бро! Позже объясню! Извини! Сейчас занят

[@user-ll5wl9gq4x]

@@Cricketdoctor_1999 a = x + r, b = y + r, c = x + y, => a + b + c = (x + y) +2r + (x+y), => a + b + c = c + 2r +c , => 2r = a + b - c