Mathematics Question and Solution
a² = 3²+4²
a = √(25)
a = 5 units.
a is the hypotenuse of the ascribed right-angled triangle.
tanb = 3/4
b = atan(3/4)°
tanc = 4/3
c = atan(4/3)°
Let d be the radius of the inscribed circle.
Calculating d.
3d+4d+5d = 3*4
12d = 12
d = 1 unit.
e = 2d
e = 2 units.
e is the diameter of the inscribed circle.
f = 3-1-x
f = (2-x) units.
g = (1+x) units.
Calculating x.
4 ~ (1+x)
5 ~ (2-x)
Cross Multiply.
4(2-x) = 5(x+1)
8-4x = 5x+5
9x = 3
x = ⅓ units.
It implies;
g = 1+⅓
g = ⅓(4) units.
h = 4-1-y
h = (3-y) units.
j = (1+y) units.
Calculating y.
5 ~ (3-y)
3 ~ (1+y)
Cross Multiply.
5(1+y) = 3(3-y)
5+5y = 9-3y
8y = 4
y = ½ units.
It implies;
j = 1+½
j = ½(3) units.
k = j-g
k = ½(3)-⅓(4)
k = ⅙(9-8)
k = ⅙ units.
Let l be the required red length.
Calculating l.
l² = 2²+(⅙)²
l² = 4+(1/36)
l² = (145/36)
l = √(145/36)
l = ⅙√(145) units.
l = 2.0069324298 units in decimal.
