Notice.
4 units is the radius of the inscribed circle.
a² = 2(4)²
a = 4√(2) units.
8² = (4√(2))²+4²-2*4*4√(2)cosb
64 = 32+16-32√(2)cosb
32√(2)cosb = 48-64
cosb = -16/(32√(2))
b = acos(-1/(2√(2)))...
a²+9² = 15²
a = √(15²-9²)
a = 12 units.
a is AF.
c = (9-b) units.
Therefore;
12 ~ b
9 ~ (9-b)
Cross Multiply.
108-12b = 9b
21b = 108
b = ⅐(36) units.
b is the radius of the inscribed half circl...
Calculating x, length DE.
a = ½(6)+x
a = (3+x) units.
Therefore;
Length DE, x is;
a = 8
3+x = 8
x= 8-3
x = 5 units.
Calculating yellow area.
It implies;
2a² = 4²
Where a is the height of the yellow area.
Therefore;
2a² = 4²
2a² = 16
a² = 8
a = √(8)
a = 2√(2) units.
Area Yellow is;
½*a*6
=...
Calculating angle x.
Let the two equal lengths be 1 unit each.
(1/sin(180-24-30)) = (a/sin30)
a = 0.61803398875 units.
b = 1+a
b = 1.61803398875 units.
c = 24+30
c = 54°
(1/sin(180-...
Calculating the required angle, let it be x.
Let a be the side length of the ascribed square.
½(ab) = 1
b = 2/a --- (1).
1+1+c = ½(a²)
c = ½(a²)-2
c = ½(a²-4) square units.
d = a-b...
Calculating x.
Let a be the height of the inscribed isosceles triangle.
b = ½(7)+7
b = 10.5 units.
c² = 10.5²+a²
c = √(a²+110.25) units.
Let d be one of the two equal inscribed angles.
cosd = 1...
Let a be the side length of the red inscribed square.
b = (4-a) cm.
Therefore;
4² = 2b²
4² = 2(4-a)²
16 = 2(16-8a+a²)
8 = 16-8a+a²
a²-8a+8 = 0
(a-4)² = -8+(-4)²
(a-4)² = 16-8
a = 4±√(8)
a ≠ (4+2√...
Calculating x.
sina = 3.5/12.5
a = asin(3.5/12.5)°
b = 2a
b = 2asin(3.5/12.5)°
sinc = 10/12.5
c = asin(10/12.5)°
d = 2c
d = 2asin(10/12.5)°
e = 180-b-d
e = 180-2asin(3.5/12.5)-2asin(10/12.5)
e...
cosa = (2√(5))/(3√(5))
a = acos(⅔)°
a is angle PRT.
tana = 25/b
b = 25/(tan(acos(⅔)))
b = 22.360679775 units.
b = 10√(5) units.
Therefore, area triangle TRP is;
½*25b
= ½*25*10√(5)
= 125√(5) squa...
