a = ½(2+7)
a = 4.5 units.
a is the radius of the ascribed half circle.
Let b be the radius of the inscribed circle.
c = (4.5-2b) units.
d = (4.5-b) units.
e = 4.5-2-b
e = (2.5-b) uni...
Let the side length of the regular heptagon be 1 unit.
Calculating angle x.
a = ⅐*180(7-2)
a = ⅐(900)°
a is the single interior angle of the regular heptagon.
b = ½(360-2*⅐(900))
b = 18...
Calculating x.
a = ⅙*180(6-2)
a = 120°
a is the single interior angle of the green regular hexagon.
b = ½(x) units.
c = ⅙(x) units.
It implies;
d ~ b
b ~ c
Therefore;
d ~ ½(...
Let the side length of the ascribed square be 1 unit.
Therefore, the side length of the inscribed green regular triangle 1 unit.
a = 90-60°
a = 30°
b² = 2-2cos30
b = 0.51763809021 units....
sina = 4/(4+1)
a = asin(4/5)°
b² = 1²+4²-2*1*4cos(asin(4/5))
b = ⅕√(305) units.
b = 3.49284983931 units.
(3.49284983931/sin(asin(4/5))) = (4/sinc)
c = asin(0.9161573349)
c = 113.62937773...
a² = (3+5)²+x²
a² = a²-64
a = √(64+x²) cm.
b = (8-x) cm.
It implies;
5 ~ 8
(8-x) ~ √(64+x²)
Cross Multiply.
8(8-x) = 5√(64+x²)
(64-8x)² = 25(64+x²)
4096-1024x+64x² = 1600+25x²
2...
Let single side of the square be x.
Notice;
PB=¼(AB)=¼(x).
Calculating single side length of square ABCD.
It will be;
Area triangle ADP of height x and base (x-¼(x)) + area triangle...
Sir Mike Ambrose is the author of the question.
Let the bigger inscribed square be 2 units.
BC = 1+2+x = (3+x) units.
Calculating x.
(3+x)²=1²+8-4√(2)cos135
3+x=√(13)
x = (√(13)-3)...
Notice.
The ascribed right-angled triangle is not drawn to scale.
Let a be the side length of the inscribed square.
Calculating Area Inscribed Square.
tanb = 3√(3)/(9√(3))
b = atan(⅓)°...
Calculating Length x (AD).
Let a be the diameter of the green inscribed circle.
Therefore;
x ~ a
(b+x) ~ 2a
Cross Multiply.
2ax = ab+ax
2ax-ax = ab
ax = ab
x = b units.
Theref...
