Let the radius of the ascribed quarter circle be 2 units.
Let r be the radius of the inscribed circle.
Calculating r.
(2-r)² = r²+1²
4-4r+r² = r²+1
3 = 4r
r = ¼(3) units.
r = 0.75 units.
a = 2-r...
Let AC = 1 unit.
(1/sin160) = (CD/sin5)
CD = 0.25482634415 units.
(BC/sin30) = (1/sin80)
BC = 0.50771330594 units.
(BD)² = 0.50771330594²+0.25482634415²-2*0.25482634415*0.50771330594cos55
BD = 0....
Let the inscribed square side be 1 unit.
a² = 1²+1²
a = √(2) units.
Where a is a diagonal of the inscribed square.
It implies;
a = r = √(2) units.
Where r is the radius of the ascribed semi circle...
Let the square side be r.
Therefore r is the length of the red rectangle and also the radius of the inscribed quarter circle.
Let the base of the red area be a.
b = √(16-a²)
Therefore;
r² = (r-a)...
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EF = 2r.
Where r is the radius of the circle.
r = √((21/2)²+6²)
= ½(3√(65)) units.
Therefore EF = 2r
= 2*½(3√(65))
= 3√(65) units.
EF = 24.1867732449 units.
AP = √(20²+10²)
AP = √(500)
AP = 10√(5) cm.
OA (radius of the semi circle) is:
½(AP)
= 5√(5) cm.
tana = 20/10
a = atan(2)°
Where a is angle MAO = angle AMO.
b = (180-2atan(2))°
Whe...
a² = 5²-4²
a = 3 units.
b² = 3²+1²
b = √(10) units.
Where b is the square side length.
Area inscribed square is;
b²
= √(10)²
= 10 square units.
Calculating x.
c² = √(10)²-3²
c = 1 unit.
tand =...
Hypotenuse of the ascribed right-angled triangle is;
√(40²+h²)
Therefore;
40h = 4h+(40*4)+4√(40²+h²)
40h = 4h+160+4√(1600+h²)
10h = h+40+√(1600+h²)
(9h-40)² = 1600+h²
81h²-720h+1600 = 1600+h²...
