Sir Mike Ambrose is the author of the question.
Let the two equal lengths of the isosceles triangle be 2 units.
Therefore;
Area triangle ABC is
= 2sin36 square units.
= 1.17557050458 squ...
Sir Mike Ambrose is the author of the question.
Let the single side length of the regular heptagon be 1 unit.
Therefore;
Area Regular Heptagon is;
7/(4tan(180/7))
= 3.633912444 square un...
Calculating a.
b² = a²+1
b = √(a²+1) units.
b is AC.
Therefore;
a - 2a
1 - √(a²+1)
Cross Multiply.
a√(a²+1) = 2a
√(a²+1) = 2
a²+1 = 4
a² = 3
a = √(3) units.
a is BC.
Therefo...
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...
