Calculating R, radius of the inscribed circle.
a² = 2*3²
a = 3√(2) units.
a is the diagonal of the square.
b² = 2R²
b = √(2)R units.
c = a-b
c = (3√(2)-√(2)R) units.
It implies;...
The single side length of the square, ABCD is;
2√(5) cm
Area Blue is;
Area square with length 2√(5) cm - area triangle with height 2√(5) cm and base √(5) cm - area triangle with height 4cm...
Calculating x, side length of the regular heptagon.
(1/a)+(1/b) = ⅓
3a+3b = ab
ab-3a = 3b
a(b-3) = 3b
a = 3b/(b-3) --- (1).
c = ⅐*180(7-2)
c = ⅐(900)°
c is the single interior angle of the...
Let a be the side length of the square.
b = (a-3) m.
c = (a-5) m.
It implies;
10² = (a-3)²+(a-5)²
100 = a²-6a+9+a²-10a+25
2a²-16a+34-100 = 0
2a²-16a-66 = 0
a²-8a-33 = 0
a²-11a+3a-3...
a² = 64
a = 8 m.
a is the side length of the square.
Calculating radius of the circle.
b = ½(8+c) m.
b is the radius of the circle.
d = ½(8-c) m.
Therefore;
(½(8+c))² = (½(8-c))²+...
Let a be the radius of the quarter circle.
b = (a-7) units.
c = (a-14) units.
It implies;
a² = (a-7)²+(a-14)²
a² = a²-14a+49+a²-28a+196
a²-42a+245 = 0
Resolving the above quadratic...
Please, move the above question left/right one time to review detailed explanation of the solution.
Thank you.
The derived used equations are;
3y = 5x ------- (1)
y = ⅓(5x) ------ (2)
(y...
Sir Mike Ambrose is the author of the question.
Area red exactly as a single fraction is;
Area triangle red with two lengths ⅓(24-8√(3)) cm and ⅓(18-8√(3)) cm, and angle 60°
= ½(⅓(24-8√(3)))...
Please, move the above question left/right one time to review the solution.
Thank you.
Area Square OBCD is;
½(3+2√(2)) square units.
Let a be the radius of the inscribed circle.
b = a+2a
b = 3a units.
c² = 2(3a)²
c = 3√(2)a units.
d = 180-45
d = 135°
Calculating a.
5² = (3√(2)a)²+a²-2*3√(2)a²cos135
25 = 19a²+6...
