Let a be the diameter of the semi circle.
a² = b²+14²
b² = a²-196
b = √(a²-196)
It implies;
b² = 6²+6²-2*6*6(-cosc)
a²-196 = 72+72cosc
a²-196-72 = 72cosc
72cosc = a²-268
cosc = (a²-26...
Let BD = 1 unit.
It implies;
CD = 2BD = 2 units.
a = 180-60
a = 120°
a is angle ADB.
b = 180-120-45
b = 15°
b is angle BAD
(1/sin15) = (c/sin120)
c = 3.346065215 units.
c is AB.
d² = 3.34606521...
Let BC be a (square side).
tanb = 1/2
b = atan(½)°
sinatan(½) = c/a
c = 0.4472135955a
d = 2c
d = 0.894427191a
d is CE.
((½)a)² = 2²+e²
e² = (a²/4)-4
e = √(¼(a²-16))
e = ½√(a²-16)
f = e+(a/2)
f...
Calculating PQ.
a = 8+2
a = 10 cm.
a is the diameter, AC of the semi circle.
b = ½(AC)
b = ½(10)
b = 5 cm.
b is the radius of the semi circle.
c = b-2
c = 5-2
c = 3 cm.
5² = 3²+d²
d² = 25-9
d =...
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Area S, exactly in square unit is;
Area triangle with height 2 unit and base (2sin120) unit + Area triangle with height 2 unit and base 2√(2) unit - Area sector with radius 2 unit and angle (30+at...
Notice.
The green circles are congruent (their radius is equal). Let it be r.
Calculating r.
a = (1-r) units.
b = (1+r) units.
c = (1-r-r-r) = (1-3r) units.
It implies;
Observing Pythagoras Rule...
Calculating Red Area.
tana = 5/10
a = atan(½)°
b = 2a
b = 2atan(½)°
tanc = 10/5
c = atan(2)°
d = 2c
d = 2atan(2)°
It implies;
Area Red is;
Area sector with radius 5 units and angle 2atan(2)°...
a² = 2*6²
a² = 72
a = 6√(2) cm.
a is diagonal of the square.
Let r be the radius of the inscribed circle.
It implies;
b² = 2r²
b = √(2)r cm.
Therefore;
4+r+b = a
4+r+√(2)r = 6√(2)
Calculating r....
Area Orange is;
87/10 square unit.
Please, move the above question/figure left, one time to review the solution.
Thank you.
