Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2735481

Blue area is; 8(area quarter circle with radius 4 cm - area triangle with height and base 4 cm each) - 8(area quarter circle with radius 2 cm - area triangle with height and base 2 cm each) = 8(4π-8) - 8(π-2) = 32π-64-8π+16 = 24π-48 = 24(π-2) cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2735475

a = ½(12) a = 6 cm. a is the radius of the inscribed half circle. c = (6+b) cm. b is the radius of the inscribed circle. d = (12-b) cm. Calculating b. (6+b)²+6² = (12-b)² 36+12b+b²+36 = 144-24b+b² 72+12b = 144-24b 36b = 72 b = 2 cm. Again, b is the radius of the inscribed circle. e²+6² = 12² e² = 144-36 e = √(108) e = 6√(3) cm. Painted Area is; 2(area sector with radius 12 cm and angle 60°) - Area triangle with height 6√(3) cm and base 12 cm + Area half circle...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2735470

tan60 = a/2 a = 2√(3) units. a is the height of the inscribed red regular triangle. b = 2+a b = (2+2√(3)) units. Area green is; (½*4*(2+2√(3)))-(½*2√(3)*4) = 4+4√(2)-4√(3) = 4 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2733670

Notice. OE = AE = ½(6) OE = AE = 3 cm. a²+3² = 6² a² = 36-9 a = √(27) a = 3√(3) cm. tanb = a/3 b = atan(√(3)) a = 60° Shaded Area is; (0.5*3*3√(3))+(60π*6*6/360)-(0.5*6*6sin60 = ½(9√(3))+6π-9√(3) = 6π-½(9√(3)) = ½(12π-9√(3)) cm² Or (60π*6*6/360)-(0.5*3*3√(3) = 6π-½(9√(3)) = ½(12π-9√(3)) cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2733657

a = (2+x) units. a is the radius of the ascribed sector. b = 2x units. x is the radius of the inscribed quarter circle. Calculating x. (2+x)² = x²+(2x)² 4+4x+x² = 5x² x²-x-1 = 0 (x-½)² = 1+(-½)² (x-½)² = (5/4) x = ½±√(5/4) x = ½±½√(5) It implies; x = ½(1+√(5)) units. x = 1.6180339887 units. Therefore, area inscribed quarter circle is; ¼(x²)π = ¼*½(1+√(5))*½(1+√(5))π = π(1+√(5))²/16 = 2.0561990865 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2733648

Calculating x. 2*2^(x)-9^(x) = 0 2*2^(x) = 9^(x) Dividing through by 2^(x). 2 = (4.5)^(x) Taking log base 10 at both sides of the equation. log2 = log(4.5)^(x) x = log2/log(4.5) x = 0.4608454206 Checking Accuracy. 2^(1+0.4608454206) + 3^(2*0.4608454206) = 0

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2731949

Area red is; ½(area square with single side length 24 units - 4(area circle with radius 6 units)) = ½(24²-4(36π)) = ½(576-144π) = 288-72π = 72(4-π) square units. Or 2(area square with side length 12 units - area circle with radius 6 units). = 2(12²-π6²) = 2(144-36π) = 72(4-π) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2731938

Sir Mike Ambrose is the author of the question. Let the single side length of the equilateral triangle be 2 unit. Therefore; Area Green is; ⅓√(3) * ⅔ = 2√(3)/9 square units. Area Yellow is; Area triangle with height (⅔) units and base 0.44905020937 units + Area triangle with two side length 0.8037975709 units and 0.6546536707 units, and angle 15.143370384°. = ½(⅔*0.44905020937) + ½(0.8037975709*0.6546536707)sin15.143370384 = 0.14968340312 + 0.06873217491 = 0.21841557803 sq...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2731933

Let the single side length of the blue area/square be x. Therefore; 5² = x² + (x+½(5-x))² 5² = x² + (½(5+x))² 25 = x² + ¼(25+10x+x²) 100 = 4x²+25+10x+x² 5x²+10x-75=0 x²+2x-15=0 (x+1)²=15+1 x = -1±√(16) x = -1±4 Therefore; x ≠ -5 x = 3 cm Therefore blue area/area square is; x² = 3² = 9 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2731257

a = 15+5 a = 20 units. a is the radius of the circle. d² = b²+c² --- (1). Where b is the width of the inscribed blue rectangle area and c is the length. 15²+20² = d² d = 25 units. d is the diagonal of the inscribed blue rectangle. Substituting d is (1) making c the subject formula. d² = b²+c² 25² = b²+c² c = √(625-b²) units. e = d+b e = (25+b) units. f = 2a f = 40 units. Calculating b. 40² = √(625-b²)²+(25+b)² 1600 = 625-b²+625+50b+b² 1600 = 1250+50b 160 =...

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