By OnlineEdumath   |  29th August, 2024
Calculating Area Shaded. Let r be the radius of the semi circle. Calculating r. ((6r/√(10))-2)²=(r/2)²+(3r/2)² 72r²-48√(10)r+80=50r² 22r²-48√(10)r+80= 0 11r²-2⁴√(10)r+40= 0 Therefore...
By OnlineEdumath   |  28th August, 2024
Calculating area of the inscribed red square. Let a be the side length of the inscribed red square. 2b² = a² b = √(a²/2) b = ½√(2)a units. 10² = a²+(½√(2)a)²-2a*½√(2)acos(45+90) 100 = ½...
By OnlineEdumath   |  28th August, 2024
Let the base of the quadrilateral be 1 unit. a = 180-58-28-38 a = 180-124 a = 56° (1/sin56) = (b/sin(58+28)) b = 1.2032796622 units. (1/sin56) = (c/sin38) c = 0.7426219217 units. d...
By OnlineEdumath   |  28th August, 2024
Let the ascribed quarter circle radius be 1 unit. 2a² = 1² a² = ½ a = ½√(2) units. a = 0.7071067812 units. a is the radius of the blue inscribed quarter circle. Area blue inscribed quarte...
By OnlineEdumath   |  28th August, 2024
Sir Mike Ambrose is the author of the question. Let the single side length of the small square be 1 unit. Therefore; Area large square is; ((1/tan15)-1)² = 7.46410161514 square units....
By OnlineEdumath   |  28th August, 2024
Sir Mike Ambrose is the author of the question. The width of the rectangle is; (5/2)+(11/2)  = 8 cm Area Rectangle is; 12*8 = 96 cm² Area Green is; ½(7*12)sin((90-atan(5√(3)/11)))...
By OnlineEdumath   |  27th August, 2024
Let the square side length be 1 unit. tan15 = a/1 a = 0.2679491924 units. b = 1+a b = 1.2679491924 units. c = 1-a c = 0.7320508076 units. d² = 1²+1² d = √(2) units. e = 90-15 e...
By OnlineEdumath   |  27th August, 2024
Let the length of the two congruent rectangle be 1 units. Let the width be x. Therefore; 1 ~ (1+x) x ~ 1 It implies; 1/x = (x+1)/1 x²+x-1 = 0 (x+½)²=1+¼ x = -½±½√(5) Therefore...
By OnlineEdumath   |  27th August, 2024
Let the side length of the blue inscribed equilateral triangle be 1 unit. Area blue equilateral triangle is, 0.5*1²*sin60 = ¼√(3) square units. = 0.4330127019 square units. It implies;...
By OnlineEdumath   |  27th August, 2024
Sir Mike Ambrose is the author of the question. The equation of the curve is; Using the points; (4, 4), (2, 0), (12, 0) And, y =ax²+bx+c Therefore; 4=16a+4b+c 0=4a+2b+c 0=144a+12b...
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