Mathematics Question and Solution
Calculating Shaded Area.
Notice.
ABCD is a square.
Let x be the side length of the square.
a²+x² = 40²
a = √(1600-x²) units.
a is BF.
b = x-a
b = (x-√(1600-x²)) units.
b is AF.
It implies;
cosc = x/40 --- (1).
cosc = (x-√(1600-x²))/10 --- (2).
Equating (1) and (2).
x/40 = (x-√(1600-x²))/10
Cross Multiply.
10x = 40(x-√(1600-x²))
x = 4(x-√(1600-x²))
x = 4x-4√(1600-x²)
4√(1600-x²) = 3x
16(1600-x²) = 9x²
16*1600-16x² = 9x²
16*1600 = 25x²
x² = 16*64
x = √(16*64)
x = 4*8
x = 32 units.
Again, x is the side length of the square.
Recall.
a = √(1600-x²)
And x = 32 units.
a = √(1600-32²)
a =√(576)
a = 24 units.
Recall Again.
b = (x-√(1600-x²))
And x = 32 units.
b = 32-√(1600-32²)
b = 32-24
b = 8 units.
d²+8² = 10²
d² = 100-64
d = √(36)
d = 6 units.
Therefore, shaded area is;
Area square with side length 32 units - Area triangle with height 32 units and base 24 units - Area triangle with height 8 units and base 6 units.
= 32²-½(32*24)-½(8*6)
= 1024-384-24
= 616 square units.
