Mathematics Question and Solution
Calculating area inscribed circle and tan x.
Notice.
Length of arc of the quarter circle is 4π cm.
(90/360)*2πr = 4π
¼(2πr) = 4π
2πr = 16π
r = 8 cm.
r is the radius of the ascribed quarter circle.
Let y be the radius of the inscribed circle.
a = r-y
a = (8-y) cm.
Calculating x.
a² = y²+y²
(8-y)² = 2y²
64-16y+y² = 2y²
y²+16y-64 = 0
(y+8)² = 64+8²
(y+8)² = 128
y = -8±√(128)
y = -8±8√(2)
It implies;
y ≠ -8-8√(2)
y = (8√(2)-8) cm.
y = 8(√(2)-1) cm.
y = 3.31370849898 cm.
Again, y is the radius of the inscribed circle.
Therefore, the area of the inscribed circle is;
πy²
= π(8√(2)-8)²
= (192-128√(2))π
= 64(3-2√(2))π cm²
Calculating angle x.
b² = 8²+3.31370849898²-2*8*3.31370849898cos45
b = 6.12293491784 cm.
It implies, angle x is;
(6.12293491784/sin45) = (3.31370849898/sinx)
sinx = 0.38268343236
x = asin(0.38268343236)
x = 22.5°
Therefore, tan x is;
tan(22.5)
= 0.41421356237
= √(2)-1
