Notice.
Radius of the ascribed semi circle is 5 units.
a = (5-R) units.
(5-R)² = b²+R²
b² = 25-10R+R²-R²
b² = 25-10R
b = √(25-10R) units.
c = ½(5)+b
c = (2.5+√(25-10R)) units.
d...
Sir Mike Ambrose is the author of the question.
Area of the curve is;
y = -⅓(x³)+2x²
Point p coordinate is;
(⅓(32), 4)
Therefore Area Red is;
Area under the curve at the points (4, 0)...
Please, move the above question right/left one time to review the solution.
Thank you.
Area inscribed blue square is;
9 square units.
Please, move the above question left/right one time to review the solution.
Thank you.
Coloured area region is;
22.85 cm² to 2 decimal places.
Let a be the side length of the inscribed square.
Calculating a.
b = (6-a) units.
Therefore;
6 - (6-a)
8 - a
Cross Multiply.
6a = 48-8a
14a = 48
7a = 24
a = (24/7) units.
Aga...
Let a and b be the height and base of the triangle respectively.
It implies;
½(ab) = 64
ab = 128 --- (1).
a²+b² = 16²
a = √(256-b²) --- (2).
Therefore, substituting (2) in (1) to get...
Notice.
a+b = 28 units.
It implies;
b = (28-a) units.
Calculating a.
20² = a²+b²
And b = 28-a
400 = a²+(28-a)²
400 = a²+784-56a+a²
2a²-56a+384 = 0
a²-28a+192 = 0
(a-14)² = -1...
Let a be the radius of the half circle.
b = (3-a) units.
It implies, calculate a.
a² = 2²+b²
a² = 4+(3-a)²
a² = 4+9-6a+a²
6a = 13
a = (13/6) units.
Again, a is the radius of the half...
Area red is;
3(½(area rectangle with length 10 units and width 5 units - area semi circle with radius 5 units)) + area triangle with height 5 units and base 10 units + area triangle with two equ...
Notice.
The ascribed regular pentagon is nonagon.
Its single interior angle is;
⅑*180(9-2)
= 20*7
= 140°
a = ½(360-(2*140))
a = ½(360-280)
a = ½(80)
a = 40°
b = ½((180(5-2))-(3*...
