Mathematics Question and Solution
Let a be the radius of the circle and also the width of the rectangle.
b = (a-x) units.
c = (a-2x) cm.
d = a+c
d = (2a-2x) cm.
d is the length of the rectangle.
12² = (2a-2x)²+(a-x)²
144 = 4a²-8ax+4x²+a²-2ax+x²
144 = 5a²-10ax+5x² --- (1).
2x - (a-x)
(a-x) - (2a-2x)
Cross Multiply.
(a-x)² = 4x(a-x)
a-x = 4x
a = 5x --- (2).
Substituting (2) in (1).
144 = 5(5x)²-10(5x)x+5x²
144 = 125x²-50x²+5x²
144 = 80x²
36 = 20x²
18 = 10x²
x² = 9/5
x = √(9/5)
x = ⅕(3√(5)) cm.
x = 1.3416407865 cm.
Therefore, a (radius of the circle and also width of the rectangle) is;
a = 5x
And x = 1.3416407865 cm.
a = 5*1.3416407865
a = 6.7082039325 cm.
Recall.
d = (2a-2x)
And a = 6.7082039325 cm, x = 1.3416407865 cm.
d = 2*6.7082039325-2*1.3416407865
d = 10.733126292 cm.
Again, d is the length of the rectangle.
Therefore;
Area rectangle is;
ad
= 6.7082039325*10.733126292
= 72 cm²
Perimeter of the rectangle is;
a+a+d+d
= 2(6.7082039325+10.733126292)
= 2*17.4413302245
= 34.882660449 cm.
