Let a be the radius of the ascribed semi circle.
b = a-½(7)
b = (a-3.5) units.
15² = b²+c²
c² = 225+(a-3.5)²
c² = 225-(a²-7a+12.25)
c = √(225-a²+7a-12.25)
c = √(212.75+7a-a²) units.
Therefore;
a...
a = 1+2+1
a = 4 units.
a is the side length of the square.
tanb = 4/2
b = atan(2)°
tanc = 4/1
c = atan(4)°
d = 180-b-c
d = (180-atan(2)-atan(4))°
Observing Sine Rule.
(1/sin(180-atan(2)-atan(4))...
Notice!
Radius of the circle, r is;
r = √(50)
r = 5√(2) units.
a² = 2²+6²
a² = 4+36
a = √(40)
a = 2√(10) units.
tanb = 6/2
b = atan(3)°
c = ½(a)
c = ½(2√(10))
c = √(10) units....
AB = BC = AC = 6 cm.
cos60 = 2/BE
BE = 4 cm.
(DE)² = 4²-2²
DE = 2√(3) cm.
(CE)² = (2√(3))²+4²
(CE)² = 12+16
(CE)² = 28
CE = √(28)
CE = 2√(7) cm.
EF = (2√(7)-4) cm.
EF = 1.29150262...
Calculating angle x (angle CBD).
Let the two equal lengths of the plane shape be 1 unit.
a = 180-30-24
a = 180-54
a = 126°
a is angle ABC.
It implies;
Observing sine rule.
(1/sin30) = (b/sin24)...
Radius, r of the six congruent inscribed circles each is;
3r = 7
r = (7/3) units.
Radius, R of the ascribed circle is;
R = ½(6r)
And r = (7/3) units.
R = ½*6*(7/3)
R = 7 units.
Therefore, Area O...
a = ½(10+6)
a = 8 units.
a is the radius of the ascribed half circle.
b = a-6
b = 8-6
b = 2 units.
b is OB.
Observing similar triangles side lengths ratio.
b - (10-c)
(6-c) - b
Cross Multiply....
Shaded Area is;
4(area quarter circle with radius 2 units - area triangle with height and base 2 units each) + 2(area square with side 2 √(2) units - area semi circle with radius 2 units)
= 4(¼*4π...
Let r be the radius of the ascribed semi circle.
It implies;
r² = 14²+a²
a² = r²-196
a = √(r²-196) cm.
a is OA.
b = a+r
b = (√(r²-196)+r) cm.
b is AB.
Therefore, calculating r.
½*14*(√(r²-196)+r...
Notice, the triangle is isosceles.
It implies;
½(12x+22)+(6y-5) = 90
6x+11+6y-5 = 90
6x+6y = 84
x+y = 14 --- (1).
½(12x+22)+(10y-41) = 90
6x+11+10y-41 = 90
6x+10y = 120
3x+5y = 60 --- (2).
Solvi...
