Mathematics Question and Solution
Let a be the big inscribed circle radius.
Let b be the bigger inscribed circle radius.
Let c be the biggest inscribed circle radius.
It implies;
a+b = 12 --- (1).
a+c = 22 --- (2).
b+c = 26 --- (3).
Therefore, resolving (1), (2) and (3) simultaneously to get a, b and c.
Subtracting (2) from (3).
(b+c = 26)-(a+c = 22)
It implies;
b-a = 26-22
b-a = 4
a = b-4 --- (4).
Substituting (4) in (1) to get b.
a+b = 12
And at (4), a = b-4
b-4+b = 12
2b = 12+4
b = ½(16)
b = 8 units.
Again, b is the radius of the bigger inscribed circle.
Calculating a, using (4).
a = b-4
And b = 8 units.
a = 8-4
a = 4 units.
Again, a is the radius of the big inscribed circle.
Calculating c, using (3).
b+c = 26
And b = 8 units.
8+c = 26
c = 26-8
c = 18 units.
Again, c is the radius of the biggest inscribed circle.
Therefore, d, the width of the ascribed rectangle is;
d = 2c
And c = 18 units.
d = 2*18
d = 36 units.
Again, d is the width of the ascribed rectangle.
Calculating e, the length of the ascribed rectangle.
f = c-b
c = 18 units, b = 8 units.
f = 18-8
f = 10 units.
g²+f² = 26²
g² = 26²-10²
g² = 676-100
g = √(576)
g = 24 units.
It implies;
e = b+c+g
e = 8+18+24
e = 50 units.
Again, e is the length of the ascribed rectangle.
Therefore, area ascribed rectangle is;
de
= d*e
= 36*50
= 360*5
= 1800 square units.
