Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2776225

½(xy)=24 xy = 48 ------ (1) x²+y²=10² ------- (2) And; x²+y²=(x+y)²-2xy Therefore; (x+y)²-2xy=100 And; xy = 48 (x+y)²-2(48)=100 (x+y)²=196 x+y=14 ----- (3) x = 14-y ------ (4) Substitute (4) in (1) y(14-y)=48 y²-14y+48=0 (y-7)²=-48+49 y = 7±1 Therefore; y = 8 units or 6 units. It implies; For y = 8 units, x = 14-y x = 14-8 x = 6 units. Calculating r, radius of the inscribed circle. ½(6r)+½(8r)+½(10r)=½(8*6) 3r+4r+5r=24 12r = 24...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2775410

Let the radius, r of the circle be 7 units. Therefore perimeter of the circle is; 2πr = 44 units. Calculating the length of the square. 14² = (14-2x)²+(14-4x)² 196 = 196-56x+4x²+196-112x+16x² 20x²-168x+196=0 5x²-42x+49=0 (x-(21/5))²= (-49/5)+(-21/5)² (x-(21/5))²= (196/25) x = (21/5)±(14/5) x = (21±14)/5 x ≠ (35/5) ≠ 7 x = (7/5) units. It implies; The square length is; 14-2x, and x = (7/5) units. = 14-2(7/5) = (70-14)/5 = 56/5 units. Therefore perimeter of...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2775407

Let r be the radius of the semi circle. Therefore; A right-angled triangle with hypotenuse 6 units, and adjacent height and base, (r-7) units and √(r²-49) units respectively is derived. It implies; 6²=(r-7)²+√(r²-49)² 36=r²-14r+49+r²-49 2r²-14r-36=0 r²-7r-18=0 (r-(7/2))²=18+(7/2)² (r-(7/2))²=(121/4) r = (7/2)±(11/2) r = ½(7±11) Therefore; r ≠ ½(-4) ≠ -2 r = ½(18) = 9 unit.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2774057

Let the side length of the regular pentagon be 1 unit. tan36 = a/0.5 a = 0.363271264 units. b² = 0.363271264²+0.5² b = 0.61803398875 units. tan72 = c/0.5 c = 1.53884176859 units. Area Green is; 0.5sin108+0.5*1.53884176859*0.5-0.5*0.5*0.363271264-0.5*0.61803398875*0.61803398875sin36 = 0.65716389015 square units. Calculating Area Blue. Let d be the radius of the inscribed blue circle. Calculating d. e = 0.5(90+36) e =63° (d/tan63) + (d/tan45) = a d = 0.3632...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2774044

Let r be the radius of the circle. (6√(2))²=r²+(6+r)² 72=r²+36+12r+r² 2r²+12r-36=0 r²+6r-18=0 (r+3)²=18+9 r = -3±3√(3) Therefore; r ≠ - 3 - 3√(3) r = (3√(3)-3) units. Therefore; Area triangle AOB is; Area triangle with two side 6√(2) unit each, and angle (45-atan((3√(3)-3)/(3√(3)+3)))° = ½(6√(2))²sin(45-atan((3√(3)-3)/(3√(3)+3))) = ½*72*½ = ¼(72) = 18 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2774023

Let the two equal lengths be 1 unit each. a² = 2(1²) a = √(2) units. b = 180-60-45 b = 75° (√(2)/sin45) = (c/sin75) c = 1.9318516526 units. d = c-a d = 0.5176380902 units. e = 180-45 e = 135° It implies; f² = 0.5176380902²+1.9318516526-2*0.5176380902*1.9318516526cos135 f = √(2) units . Calculating the required angle, x. (√(2)/sin135) = (1/sinx) x = 30°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2774015

Sir Mike Ambrose is the author of the question. Please, move the above question left/right one time to review the solution. Thank you. Let the side length of the inscribed square be 1 unit. Therefore; The required angle, theta is; 45.896821403° ≈ 45.9° to 1 d. p.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2772866

Sir Mike Ambrose is the author of the question. Calculating Area Purple. Let the side length of the square be x. Calculating x. 2(one-sixth x²) = Area Orange 2(1/6)x² = 96 x² = 288 x = 12√(2) cm. Therefore; Area purple is; Area triangle with height 12√(2) cm and base 6√(2) cm - Area triangle with two side 6√(2) cm and 10.5409255339 cm, and angle atan(0.5)° = (½*12√(2)*6√(2)) - (½*6√(2)*10.5409255339sin(atan(0.5)) = 72 - 20 = 52 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2772527

(0.5*(8+3)(8+4)sina)-(0.5*3*8sina) = 36 Calculating angle a. 66sina-12sina = 36 11sina-2sina = 6 9sina = 6 sina = 2/3 a = asin(2/3) a = 41.8103148958° Therefore, the required area is; 0.5*3*8sin(asin(2/3)) = 0.5*3*8*(2/3) = 8 square units. Or It is; 0.5*12*11sin(asin(2/3))-36 = 66*(2/3)-36 = 44-36 = 8 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2772524

Let the side length of the ascribed regular pentagon be 2 units. a = ⅕*180(5-2) a = 108° a is the single interior angle of the ascribed regular pentagon. b² = 2²+1²-2*1*2cos108 b = 2.497212041 units. (2.497212041/sin108) = (1/sinc) c = 22.3861775592° d = 180-108-c d = 49.6138224408° e = 180-d e = 130.3861775592° f = d-c f = 27.2276448816° g = 180-2d g = 80.7723551184° h = 180-g-f h = 72° (2.497212041/sin72) = (j/sin27.2276448816) j = 1.2013397144 units....

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