Mathematics Question and Solution
Calculating Area Triangle ABC and Length AN ÷ Length NC.
Let AB equal x.
a = (3*x)/(3+4)
a = ⅐(3x) units.
a is AT = AN.
b = (4*x)/(3+4)
b = ⅐(4x) units.
b is TB = MB.
c = a+4
c = ⅐(3x)+4
c = ⅐(3x+28) units.
c is AC.
d = b+4
d = ⅐(4x)+4
d = ⅐(4x+28) units.
d is BC.
Calculating x.
(4*x)+(4*⅐(3x+28))+(4*⅐(4x+28)) = ⅐(4x+28)*⅐(3x+28)
4x+⅐(12x)+⅐(112)+⅐(16x)+⅐(112) = ((4x+28)(3x+28))/49
196x+84x+784+112x+784 = 12x²+196x+784
84x+784+112x = 12x²
12x²-196x-784 = 0
3x²-49x-196 = 0
It implies;
x ≠ -3.32367
x = 19.657 units.
Again, x is AB.
Recall.
a = ⅐(3x) units.
And x = 19.657 units.
a = ⅐(3*19.657)
a = 8.424428571 units.
e = 4+a
e = 12.424428571 units.
e is AC.
Recall Again.
b = ⅐(4x) units.
And x = x = 19.657 units.
b = ⅐(4*19.657)
b = 11.232571428 units.
f = 4+b
f = 15.232571428 units.
f is BC.
Therefore;
Area triangle ABC is;
½*ef
= 0.5*12.424428571*15.232571428
= 94.6279978299 square units.
Length AN ÷ Length NC is;
a/r
= 8.424428571/4
= 2.10610714275
≈ 2.11 to 2 decimal places.
