Calculating Area of the inscribed Square.
tana = 18/9
a = atan(2)°
Let the side length of the square be b.
sin(atan(2)) = b/c
c = (b/0.894427191) units.
cos(atan(2)) = d/b
d = 0.44...
Length NP is;
√(20²+5²) - (9+3)
= √(425) - 13
= (5√(17)-13) cm.
= 7.6155281281 cm.
Red area is;
½(Area rectangle of length 2r and width r - area semi circle of radius r) - area triangle of height 2r and base r + area sector of radius r and angle (2tan–¹(1/2))° + area triangle...
Sir Mike Ambrose is the author of the question.
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Sir Mike Ambrose is the author of the question.
Yellow Area exactly in square units is;
½*√(3)*(4√(3)-4)
= ½(12-4√(3))
= 2(3-√(3)) square units.
= 2.5358983849 square units.
Let the inscribed square side length be a.
b²+1² = a²
b = √(a²-1) units.
c = 2b
c = 2√(a²-1) units
d = 1+7
d = 8 units.
Observing similar plane shape (right-angled) side length rati...
Mazana Job Shared The Questions.
Number 1.
a = 180-30-45
a = 105°
It implies;
(4/sin105) = (b/sin30)
b = 2.0705523608 units.
c = 45+90
c = 135°
Therefore, the required length,...
Let the side length of the regular octagon be a.
b = ⅛*180(8-2)
b = 135°
b is the single interior angle of the regular octagon.
2c² = a²
c = √(a²/2)
c = ½(√(2)a) units.
c = 0.7071067812a...
r2 will be;
Let it be x.
Therefore;
(4+x)²=(8-x)²+(4-x)²
16+8x+x²=64-16x+x²+16-8x+x²
x²-32x+64=0
(x-16)²=-64+256
x = 16±8√(3)
x ≠ 16+8√(3)
x = 16-8√(3)
Therefore;
x = r2 = 8(2-√...
The radius, r if the small circle will be;
(5+r)²=2(5-r)²
25+10r+r²=50-20r+2r²
Therefore
r²-30r+25=0
(r-15)²=-25+225
r = 15±√(200)
r = 15±10√(2)
It implies;
r ≠ 15+10√(2)
r = 15...
