Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2720389

a² = 6²+9²-2*6*9cos60 a = 7.9372539332 units. (7.9372539332/sin60) = (6/sinb) b = 40.8933946491° (7.9372539332/sin120) = (6/sinc) c = 40.8933946491° d = c+b d = 81.7867892983° e = 180-d e = 98.2132107017° f = 180-120-40.8933946491 f = 19.1066053509° (g/sin19.1066053509) = (6/sin40.8933946491) g = 3 units. h = 180-60-98.2132107017 h = 21.7867892983° (3/sin21.7867892983) = (j/sin98.2132107017) j = 8 units. Therefore, area yellow is; 0.5*3*8sin60 = 6√(3) s...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2720381

Calculating area of the ascribed square and area of the inscribed blue triangle. a²+8² = 10² a = 6 units. 6 - 8 b - x Cross Multiply. 8b = 6x b = ¼(3x) units c²+x² = 8² c = √(64-x²) units. It implies; b+c = x ¼(3x)+√(64-x²) = x √(64-x²) = ¼(x) 64-x² = x²/16 (64*16)-16x² = x² 17x² = (64*16) x = √((64*16)/17) x = 32/√(17) x = (32√(17))/17 units. x = 7.7611400012 units. x is the side length of the ascribed square. Therefore, area of the ascribed square (x²)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2720039

Let AB be 1 unit. a = 180-18-36 a = 126° a is angle APB. (1/sin126) = (b/sin18) b = 0.3819660113 units b is BP. c = 180-(12+18)-(36+36) c = 180-102 c = 78° c is angle ACB. (1/sin78) = (d/sin30) d = 0.5111702974 units. d is BC. e² = 0.5111702974²+0.3819660113²-2*0.5111702974*0.3819660113cos36 e = 0.3021133709 units. e is CP. Therefore, the required angle, x is; (0.3021133709/sin36) = (0.3819660113/sinx) x = asin(0.7431448256) x = 48°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2720030

Notice. The complete plane shape is half a circle with radius 8 cm. Area green is; Area semi circle with radius 8cm - 2(area quarter circle with radius 8 cm - area triangle with height and base 8 cm respectively). = (½*8*8π)-2((¼*8*8π)-(½*8*8)) = 32π-2(16π-32) = 32π-32π+64 = 64 cm² Or (¼*8*8π)+(¼*8*8π)-(¼*8*8π)+(½*8*8)-(¼*8*8π)+(½*8*8) = 2(½*8*8) = 8*8 = 64 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2719425

Calculating area of the inscribed red circle. Let the bigger circle radius be a. Let the big circle radius be b. It implies; a+b = 6 --- (1). ½(b²)+½(a²)+14 = 6*a b²+a²+28 = 12a --- (2). At (1). b = 6-a --- (3). Therefore, substituting (3) in (2) to get a. (6-a)²+a²+28 = 12a 36-12a+a²+a²+28 = 12a 2a²-24a+64 = 0 a²-12a+32 = 0 (a-6)² = -32+(-6)² (a-6)² = 4 a = 6±2 Therefore; a = 6-2 a = 4 units. Again, a is the radius of the bigger circle. At (3)....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2718708

a = ⅐(360)° a is the interior angle of each of the congruent 7 sectors. Let the radius of the ascribed circle be 1 unit. Area ascribed circle is; π(1²) = π square units. c = (1-b) units. b is the radius of each of the 7 congruent inscribed half circles. Calculating b. sin(⅐(360)) = b/(1-b) 0.7818314825 = b/(1-b) 0.7818314825-0.7818314825b = b 1.7818314825b = 0.7818314825 b = 0.7818314825/1.7818314825 b = 0.4387796995 units. Again, b is the radius of each of the 7 co...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2718556

Calculating x. a²+4² = 5² a² = 25-16 a = 3 units. Therefore; 3*(x+5) = 4*(2x-5) 3x+15 = 8x-20 5x = 35 x = 7 units. It implies; b = 3+x+5 And x = 7 units. b = 3+7+5 b = 15 units. c = 4+2x-5 And x = 7 units. c = 4+2*7-5 c = 18-5 c = 13 units. d = ½(c) d = 6.5 units. e = ½(b) e = 7.5 units. f = e-3 f = 4.5 units. Calculating g, radius of the circle. g² = d²+f² g² = 6.5²+4.5² g = ½(5√(10)) square units. = 7.9056941504 units. Or g² = 7.5²+(6...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2718337

Calculating the required angle, x. a = ⅐*180(7-2) a = ⅐(900) a is the single interior angle of the regular heptagon. It implies, the required angle, x is; Notice. Length IC is parallel to length EK, meaning 'angle ICE is alternative to angle CEK'. Therefore; x = ½(180-a) x = ½(180-⅐(900)) x = ½*⅐(1260-900) x = ½*⅐(360) x = ⅐(180)° x = 25.7142857143°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2718331

Calculating shaded area. Let the equal lengths of the ascribed trapezoid be a. Calculating a. a² = 6²+(18-a)² 0 = 36+324-36a 36a = 360 a = 10 units. Therefore, shaded area is; ½(10+18)*6-½(6*18) = 84-54 = 30 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2718323

Calculating area of the bigger square. Let the bigger square side length be a. b² = 4 b = 2 cm. b is the side length of the big square. c = (a+2) cm. d = (a-2) cm. Calculating a², area of the bigger square. (a-2)²+(a+2)² = 10² a²-4a+4+a²+4a+4 = 100 2a²+8 = 100 2a² = 92 a² = ½(92) a² = 46 cm² Again, a² is the area of the bigger square.

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