By OnlineEdumath   |  30th September, 2024
Calculating length OB. a² = 6²+2² a = √(40) a = 2√(10) units. a is AC. tanb = 6/2 b = atan(3)° √(40)² = 2√(50)²-2√(50)²cosc 40 = 100-100cosc cosc = 6/10 c = acos(3/5)° c is angle A...
By OnlineEdumath   |  30th September, 2024
Silver area is; 2(area rectangle with length 6 cm and width 3 cm - area quarter circle with radius 3 cm - area sector with radius 6 cm and angle 30° + area sector with radius 6 cm and angle 60°...
By OnlineEdumath   |  30th September, 2024
Sir Mike Ambrose is the author of the question. Let the length of the two congruent rectangles be x. Let their width be y. Therefore area rectangle is; x * y = xy square units. Area bl...
By OnlineEdumath   |  30th September, 2024
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...
By OnlineEdumath   |  30th September, 2024
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...
By OnlineEdumath   |  30th September, 2024
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...
By OnlineEdumath   |  30th September, 2024
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...
By OnlineEdumath   |  29th September, 2024
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
By OnlineEdumath   |  29th September, 2024
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...
By OnlineEdumath   |  29th September, 2024
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...
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