a = ⅕*180(5-2)
a = 108°
a is the single interior angle of the ascribed regular pentagon.
b = 360-2a-90
b = 360-306
b = 54°
c = 90-b
c = 36°
d = 180-2c
d = 108°
Notice.
The insc...
Let a be the radius of the inscribed half circle.
b = (a-1) units.
It implies, calculating a.
a² = (a-1)²+5²
a² = a²-2a+1+25
2a = 26
a = 13 units.
Again, a is the radius of the inscrib...
Sir Mike Ambrose is the author of the question.
The Equation of the curve is;
y = -¾(x²) + 6x
Therefore shaded area is;
Area under the curve at point (4, 0) and ((4+√(6)), 0) - Area recta...
Sir Mike Ambrose is the author of the question.
Area R exactly, in square cm decimal form is;
Area square with side 2√(11) cm - Area triangle with side 2√(11) cm and 4.97194902762 cm, and angle...
a² = 20²+15²
a = √(625)
a = 25 units.
25 - 20
15 - b
Cross Multiply.
25b = 15*20
b = 12 units.
b is the height of the shaded inscribed region, triangle.
Again;
25 - 20
20 - c...
Let a be the radius of the larger circumscribed circle.
Calculating a.
cos30 = 4/r
½√(3) = 4/r
r = 8/√(3)
r = ⅓(8√(3)) units.
Therefore, area circumscribed circle is;
π(8/√(3))²
= ⅓...
Sir Beg is the author of the question.
Notice.
The composite plane shape is not drawn to scale.
2a² = 2²
a² = 2
a = √(2) cm.
a is OC = AC.
2b² = √(2)²
b² = 1
b = 1 cm.
b is OF = CF...
Let a be the radius of the circle and also the width of the rectangle.
b = (a-x) units.
c = (a-2x) cm.
d = a+c
d = (2a-2x) cm.
d is the length of the rectangle.
12² = (2a-2x)²+(a-x)²...
a = r+r
a = 2r units.
a is the diameter of the circle.
b = 3+1
b = 4 units.
c²+(2r)² = 4²
c = √(16-4r²) units.
It implies;
2√(3) - 4
r - √(16-4r²)
Cross Multiply.
4r = 2√(3(1...
Sir Mike Ambrose is the author of the question.
Area yellow exactly in square cm is;
Area triangle with height 10√(13)/9 cm and base 5√(13)/6 cm
= ½(10√(13)/9*5√(13)/6)
= ½((13*50)/(54))
= (...
