By OnlineEdumath   |  27th July, 2024
Calculating angle x. Let the hypotenuse of the inscribed right-angled triangle of the ascribed quadrilateral be 1 unit. 2a² = 1² a = √(1/2) units. a = 0.7071067812 units. c = (b+0.707106...
By OnlineEdumath   |  27th July, 2024
Let a be the white inscribed circle radius. a² = (0.5*8)²+b² b = √(a²-16) units. Let c be the radius of the ascribed. c² = (0.5*8+4)²+b² c² = 64+√(a²-16)² c² = 64-16+a² c = √(a²+48) un...
By OnlineEdumath   |  27th July, 2024
a²+8² = 12² a = √(144-64) a = √(80) a = 4√(5) units. tanb = 8/(4√(5)) b = atan(2/√(5))° c = 90-b c = atan(√(5)/2)° cos(atan(√(5)/2)) = 4√(5)/d  d = 13.416407865 units. Therefore,...
By OnlineEdumath   |  27th July, 2024
Let a be the radius of the half circle. (2a)² = 8²+3² 4a² = 73 a = ½√(73) units. b = 90-atan(3/8) b = atan(8/3)° Therefore, area blue triangle is; 0.5*½√(73)*(8+3)sin(atan(8/3)) = 0...
By OnlineEdumath   |  26th July, 2024
a = ½(180-45) a = 67.5° sin67.5 = (0.5*18)/b b = 9.7415298026 units. b is the radius of the half circle and the width of the rectangle. 2c² = b² 2c² = 9.7415298026² c = 6.8883017826 unit...
By OnlineEdumath   |  26th July, 2024
a*6 = 12*8 a = 16 cm. a is DM. b² = 16²+12² b = 20 cm. b is BD. c = ½(16+6) c = 11 cm. d = ½(8+12) d = 10 cm. e = d-8 e = 2 cm. f² = 11²+2² f = 5√(5) cm. f is OD, radius of...
By OnlineEdumath   |  26th July, 2024
2a² = 2² a² = 2 a = √(2) units. a is the radius of the inscribed blue quarter circle. Calculating R, radius of the inscribed circle. b = (√(2)+R) units. c = (√(2)-R) units. It implie...
By OnlineEdumath   |  26th July, 2024
Observing the given yellow scalene triangle, an inscribed smaller scalene triangle that is similar the ascribed scalene triangle is identified. Please, move the above question left/right one time...
By OnlineEdumath   |  25th July, 2024
a = 180-60 a = 120° b² = 2²+3²-2*2*3cos60 b² = 13-6 b = √(7) units. (√(7)/sin120) = (2/sinc) c = 40.8933946491° d = 60-c d = 19.1066053509° (√(7)/sin60) = (2/sine) e = 40.893394649...
By OnlineEdumath   |  25th July, 2024
Calculating x, angle ABD a = ½(180-x)° a is angle BAD. It implies; x, angle ABD is, x+x+(a+39) = 180 (sum of the interior angles of the ascribed triangle) 2x+½(180-x) = 180-39 2x+½(180-x)...
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