Mathematics Question and Solution
Calculating the length and width of the rectangle.
Let the length be a.
Let the width be b.
Notice;
½(a) = b
Or
2b = a
It implies;
b²+b² = 4²
2b² = 16
b² = 8
b = 2√(2) cm.
Therefore;
a = 2b
a = 4√(2) cm.
Let the semi circle radius be c.
Calculating c.
d²+2² = c²
d = √(c²-4)
e = d+2
e = √(c²-4)+2
It implies;
c² = (4√(2))²+(√(c²-4)+2)²-2*4√(2)*(√(c²-4)+2)cos45
c² = 32+c²-4+4√(c²-4)+4-8(√(c²-4)+2)
0 = 32+4√(c²-4)-8√(c²-4)-16
0 = 16-4√(c²-4)
16 = 4√(c²-4)
4 = √(c²-4)
16 = c²-4
20 = c²
c = √(20)
c = 2√(5) cm.
Again, c is the radius of the semi circle.
Notice;
d = √(c²-4), and c = 2√(5) units.
Therefore;
d = √((2√(5))²-4)
d = √(20-4)
d = √(16)
d = 4 cm.
tanf = 2/4
f = atan(½)°
g = 2f
g = 2atan(½)°
Therefore, Area of the Shaded Region is;
Area semi circle with radius 2√(5) units - Area triangle with height and base 2√(2) units respectively - Area square with side 2√(2) units - Area sector with angle 2atan(½)° and radius 2√(5) units + Area triangle with height 2√(5) units and base 2√(5)sin(2atan(½)) units.
= ½(2√(5))²*π-½(2√(2))²-(2√(2))²-(2atan(½)*π*(2√(5))²/360)+½(2√(5))²*sin(2atan(½))
= 10π-4-8-⅑(atan(0.5))π+8
= 10π-4-⅑(atan(0.5))π
= ⅑(90π-πatan(0.5)-36) cm²
= ⅑(π(90-atan(0.5))-36) cm²
= ⅑(πatan(2)-36) cm²
= 18.1429743559 cm² exactly in decimal.
