πr² = 9π
r² = 9
r = 3 units.
r is the radius of the inscribed circle.
cosa = 6/7
a = acos(6/7)°
a = 31°
b = 90-a
b = (90-acos(6/7))°
b = 59°
tan(0.5*59) = r/c
c = 3/tan(0.5*59)
c =...
Sir Mike Ambrose is the author of the question.
Let the radius of the semi circle be 4 units.
Area triangle ABC is;
½(2√(2)*4√(2))
= 8 square units.
Area red is;
½((5, 2) (4, 4) (2, 2...
Let the width of the green inscribed rectangle be a.
Let the length of the green inscribed rectangle be b.
c = (5-a) cm.
d = (5-b) cm.
It implies.
(5-a)(5-b) = 2*5
(5-a)(5-b) = 10 -...
Calculating green inscribed area.
Let the green inscribed square side length be a.
b = (15-a) units.
c = (8-a) units.
It implies.
a² = ½(8(15-a))
2a² = 120-8a
a²+4a-60 = 0
Resol...
Let the radius of sector ABC be 2 units.
Therefore;
Calculating TP = PA = RP = BR, let it be x.
2x²=(2-x)²
x²+4x-4=0
x = (2√(2)-2) units.
Calculating RQ = QC, let it be y.
2y+x=2,...
a² = 11²+8²
a = √(121+64)
a = √(185) units.
a² = 4²+b²
√(185)² = 4²+b²
b² = 185-16
b = √(169)
b = 13 units.
b is the length of the 2 congruent rectangles.
c = b-11
c = 2 units.
The...
a² = 4²+(5+3)²
a = √(80)
a = 4√(5) units.
a is the hypotenuse of the ascribed right-angled triangle.
Observing similar plane shape (right-angled) side length ratios to get ?, the required len...
Let a be the radius of the half circle.
b = (a-2) cm.
It implies, calculating a.
a² = 4²+(a-2)²
a² = 16+a²-4a+4
4a = 20
a = 5 cm.
Again, a is the radius of the half circle.
Therefor...
Shaded Area is;
Area circle with radius 5 cm + 4(¼(area square with single side length 10 cm - area circle with radius 5 cm))
= 25π + 4(¼(100 - 25π))
= 25π + 4(25 - ¼(25π))
= 25π + 100 - 2...
Let AB be 1 unit.
a = 180-28-14-64
a = 180-106
a = 74°
a is angle BAC.
It implies;
(1/sin64) = (b/sin74)
b = 0.9350149393 units.
b is BC.
c = 180-14-64-58
c = 180-136
c = 44°
c...
