Mathematics Question and Solution
Calculating Area Ascribed Rectangle.
Let x be the width of the ascribed rectangle.
Let y be the length of the ascribed rectangle.
a = 1+2+1
a = 4 units.
b = (x-2) units.
c = (y-2) units.
4² = (x-2)²+(y-2)²
16 = x²-4x+4+y²-4y+4
8 = x²+y²-4x-4y
x²+y² = 8+4(x+y) --- (1).
d²+1² = 2²
d = √(3) units.
e = (x-1) units.
f = 2d+2e
f = (2x+2√(3)-2) units.
f = (2x+1.46410161514) units.
f is the diagonal of the ascribed rectangle.
It implies;
f² = x²+y²
(2x+1.46410161514)² = x²+y² --- (2).
Equating (1) and (2).
8+4(x+y) = (2x+1.46410161514)²
8+4(x+y) = 4x²+5.85640646056x+2.14359353946
4(x+y) = 4x²+5.85640646056x-5.85640646054
x+y = x²+1.46410161514x--1.46410161514
y = x²+0.46410161514x-1.46410161514 --- (3).
Calculating x.
Substituting (3) in (1).
x²+(x²+0.46410161514x-1.46410161514)² = 8+4(x+x²+0.46410161514x-1.46410161514)
x²+(x²+0.46410161514x-1.46410161514)² = 8+4x+4x²+1.85640646056x-5.85640646056
(x²+0.46410161514x-1.46410161514)² = 3x²+5.85640646056x+2.14359353944
x⁴+0.46410161514x³-1.46410161514x²+0.46410161514x³+0.21539030918x²-0.67949192432x
-1.46410161514x²-0.67949192432x+2.14359353946 = 3x²+5.85640646056x+2.14359353944
x⁴+0.92820323028x³-2.7128129211x²-1.35898384864x+2.14359353946 =3x²+5.85640646056x+2.14359353944
x⁴+0.92820323028x³-5.7128129211x²-7.2153903092x = 0
x³+0.92820323028x²-5.7128129211x-7.2153903092 = 0
Therefore;
x = 2.50402 units.
Again, x is the width of the ascribed rectangle.
Calculating y.
Using (3).
y = x²+0.46410161514x-1.4641016151
And x = 2.50402 units.
y = (2.50402*2.50402)+(2.50402*0.46410161514)-1.46410161514
y = 5.9681342716 units.
Again, y is the length of the ascribed rectangle.
It implies, area ascribed rectangle is;
x*y
= 2.50402*5.9681342716
= 14.9443275788 square units.
