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*...
Sir Mike Ambrose is the author of the question.
Let the single side length of the equilateral triangle be 6 unit.
Therefore;
Area ABC (Equilateral Triangle) is;
½(36)sin60
= 9√(3) square...
Please, move right/left one time to review the solution of the question above.
Thank you.
Area inscribed circle is;
0.654 square units to 3 decimal places.
Calculating the exact value of AC.
Considering similar scalene triangles CBA and CAD.
Let AC be a.
12 - a
a - 27
Cross Multiply.
a² = 27*12
a = √(27*12)
a = √(3*3*3*2*2*3)
a = 3*...
7² = 3²+5²-2*3*5cosa
49 = 34-30cosa
cosa = (34-49)/30
a = acos(-15/30
a = 120°
a is angle ADC.
b = 180-a
b = 60°
b is angle ADB.
Therefore, the required angle, angle ABC is;
Let it...
Shaded Area exactly in square cm is;
Area rectangle with length 6 cm and width 2√(3) cm - Area triangle with height (10-4√(3)) cm and base (10-4√(3))sin60.
= 6*2√(3) - ½(10-4√(3))²sin60
= 12...
Let the single side length of the inscribed equilateral triangle be 1 unit.
Therefore;
Green Area is;
¼(2√(3)+2-π) square unit.
Large Semi Circle Area is;
½(π(1+(√(3)/2))²
= (π(7+4√...
Calculating angle x.
Let the hypotenuse of the inscribed right-angled triangle of the ascribed quadrilateral be 1 unit.
2a² = 1²
a = √(1/2) units.
a = 0.7071067812 units.
c = (b+0.707106...
