Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2569002

Let the single side length of the square be (x+10) unit. Calculating x. 10²=(5+x)²+5² 100-25-25=10x+x² Therefore; x²+10x-50=0 Resolving the above quadratic equation via completing the square approach to get x. (x+5)²=50+25 x =-5±√(75) Therefore; x ≠ -5-5√(3) x = (5√(3)-5) cm. Square length is; (x+10) And x = (5√(3)-5) cm. = 10+5√(3)-5 cm. = (5+5√(3)) units. Area of the square is; (5+5√(3))² = 100+50√(3) = 50(2+√(3)) cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2565814

Showing that quadrilateral ABCD is cyclic. Notice! For quadrilateral ABCD to be cyclic; Angle ADC will equal 2 times angle ABC. Angle ADC = 2(Angle ABC) It implies; Since Angle ABC = 60° Angle ADC must equal 2(60°) = 120° The working! a² = 10²+6²-2*10*6cos60 a = 2√(19) units. a = 8.7177978871 units. a is AC. Therefore, calculate angle ADC. Let it be b. (2√(19))² = 6²+4²-2*6*4cosb 76 = 52-48cosb 48cosb = 52-76 cosb = -24/48 b = acos(-1/2) b = 120° b is ang...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2563885

a = 6+4 a = 10 cm. b²+6² = 10² b = √(100-36) b = √(64) b = 8 cm. c = b+4 c = 12 cm. d = c-r d = (12-r) cm. r is the ascribed half circle radius. e = 6+d e = 6+(12-r) e = (18-r) cm. Calculating r, radius of the ascribed half circle. r² = (18-r)²+6² r² = 324-36r+r²+36 36r = 360 r = 10 cm. Area Painted is Area half circle with radius 10 cm - Area half circle with radius 6 cm - Area half circle with radius 4 cm. = ½*π*10²-½*π*6²-½*π*4² = ½(100π-36π-16π) =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2560327

Let the square side length be 3 units. tana = (3/2) a = atan(3/2)° tanb = 1.5/1 b = atan(3/2)° c = 180-45-a c = (135-atan(3/2))° It implies; (2/sin(135-atan(3/2))) = (d/sin(atan(3/2))) d = 1.6970562748 units. (2/sin(135-atan(3/2))) = (e/sin45) e = 1.4422205102 units. f² = 1.5²+1² f = 1.8027756377 units. g = (180-2atan(3/2))° Therefore; Area Blue is; 0.5*1.6970562748*2sin45 = ⅕(6) square units. Area Red is; 0.5*1.8027756377*1.4422205102sin(180-2atan(3/2)) = ⅕(6) square units. Noti...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2559056

5² = x²+a² a = √(25-x²) units. b = x-a b = (x-√(25-x²)) units. c² = 4²-x² c = √(16-x²) units. Calculating x, the ascribed square side length. Observing similar plane shape (right-angled) side length ratios. 3 - 4 (x-√(25-x²)) - √(16-x²) Cross Multiply. 3√(16-x²) = 4(x-√(25-x²)) 3√(16-x²) = 4x-4√(25-x²) 3²(16-x²) = (4x-4√(25-x²))² 144-9x² = 16x²-32x√(25-x²)+16(25-x²) 144-9x² = 16x²-32x√(25-x²)+400-16x² 144-9x² = 400-32x√(25-x²) 32x√(25-x²) = 256+9x² (32x)²(25-x²) = (...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2558853

Notice! Radius, a of the inscribed circle is 4 units. Radius, b of the inscribed semi circle is 7 units. 3²+4² = c² c = √(25) c = 5 units. d² = 5²+4² d = √(41) units. tane = 3/4 e = atan(3/4)° f = 90+e f = (90+atan(3/4))° g² = 2(4)²-2*4²cos(90+atan(3/4)) g = 7.155417528 units. h = ½(180-(90+atan(3/4))) h = 26.5650511771° j = 90-h j = 63.4349488229° k = 180-2j k = 53.1301023542° It implies; Length x is; tan53.1301023542 = 7/x x = 5¼ units. It i...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2558608

Let the inscribed square side be a. b² = 2a² b = √(2)a cm. b is the diagonal of the inscribed square. c = 90+45 c = 135° It implies, calculating a, the inscribed square side length. 10² = 5²+(√(2)a)²-2*5*√(2)acos135 100 = 25+2a²+10a 2a²+10a-75 = 0 Resolving the above quadratic equation via completing the square approach to get a, the side length of the inscribed square. (a+(5/2))² = (75/2)+(5/2)² (a+(5/2))² = ¼(150+25) (a+(5/2))² = ¼(175) a = -(5/2)±√(¼(175)) a = -(5...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2555576

Area red is; Area semi circle with radius 5 cm - Area sector with radius 5 cm and angle (180-2atan(2))° - Area kite with diagonal lengths √(10²+5²) cm and √(200-200cos(2atan(½))) cm + Area sector with radius 10 cm and angle 2atan(½)° - 2(area quarter circle with radius 5 cm - area triangle with height and base 5 cm respectively). = 12.5π - 11.591190225 - 50 + 46.3647608999 - 14.2699081699 = 9.77357067487 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2554584

0.5*4*5sina = 8 Where a is the supplementary angle of angle ABC. It implies; 10sina = 8 a = asin(8/10) a = asin(4/5)° or a = acos(3/5)° or a = atan(4/3)° a = 53.1301023542° Therefore; Angle ABC is; Let it be b. It implies; b+a = 180 b = 180-a And a = atan(4/3)° b = 180-atan(4/3) b = (90+atan(¾))° b = 126.8698976458°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2547705

Sir Mike Ambrose is the author of the question. Let the square side be 1 unit. Therefore; Area B is; Area triangle with height 1 unit and base sin(53.13010235416) unit - Area triangle with height 1 unit and base (0.71428571429sin(53.13010235416)) unit. = ½*1(sin(53.13010235416)) - ½*1*(0.71428571429sin(53.13010235416)) = 0.4 - 0.28571428572 = 0.11428571428 square unit. Area A is; Area triangle with height 1 unit and base 0.5 unit + Area triangle with height 0.5√(5) unit...

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