Let the length of the two congruent rectangle be 1 unit.
Let the width be x.
Therefore;
1 ~ (1+x)
x ~ 1
Cross Multiply.
1/x = (x+1)/1
x²+x-1 = 0
(x+½)²=1+¼
x = -½±½√(5)
Theref...
Let the radius of the inscribed circle be r.
Therefore the radius of the ascribed circle, R will be;
R = √(4²+r²)
R = √(16+r²) units.
Therefore;
Area yellow is;
Area circle with rad...
Let the side length of the large square be 4 units.
Therefore;
The side length of the blue inscribed square is;
4-(4√(3)/3)
= ⅓(12-4√(3)) units.
Therefore;
Side blue square : Side l...
The width of the rectangle is;
R units.
The length of the rectangle is;
= √((R+6)²-(R-6)²)
= √(R²+12R+36-R²+12R-36)
= √(24R) units.
Therefore;
√(24R) * R = 243
24R³ = 243²
R³ = 2...
Please, move the above question left/right one time to review the solution.
Thank you.
Area Shaded/blue or purple is;
= 8√(2)sin15 cm²
= 2.92820323028 cm²
Shaded area is;
Area sector with radius 10 units and angle acos(0.7) - Area triangle with height √(51) units and base 7 units - area sector with radius 3 units and angle 81.37307344° - area sect...
Calculating the area of the inscribed regular pentagon.
Let the single side length of the inscribed regular pentagon be x.
12²=2y²-2y²cos108
y = √(144/(2-2cos108))
y = 7.416407865 cm.
Ther...
Please, move the above question left/right one time to review the solution.
Thank you.
AO : OB is;
= 5 : 3
Sir Mike Ambrose is the author of the question.
Calculating letter k considering yellow area (trapezoid).
It implies;
Half the sum of two parallel length (54+4k²)/9 units and (54+10k²)/9 units...
Shaded area is;
Area semi circle with radius 7cm + Area rectangle with length 14 cm and width 7 cm - Area semi circle with radius 7cm.
= ½(49π) + (14*7) - ½(49π)
= 98 cm²
Or
Shaded Area...
