Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2491932

Let AB = BC = 1 unit. tan30 = a/1 a = ⅓√(3) units. a = 0.5773502692 units. a is CD. cos30 = 1/b b = ⅓(2√(3)) units. b = 1.1547005384 units. b is BD. c = 90-30 c = 60° c is angle BDC. 3(90)+90+60+d = 180(5-2) 270+150+d = 540 540° is the sum of the interior angles of the irregular pentagon ABCDE. d = 540-270-150 d = 540-420 d = 120° d is angle BAE. e = 90-30 e = 60° e is angle ABD. f² = 1.1547005384²+1-2*1.1547005384cos60 f = 1.0856485596 units. f is AD. (...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2482706

a = (10-x) cm. 10 cm is the radius of the ascribed semi circle. 12² = a²+b² b² = 144-(100-20x+x²) b² = 44+20x-x² b = √(44+20x-x²) cm. 10² = x²+b² 100 = x²+44+20x-x² 56 = 20x 14 = 5x x = (14/5) cm. x = 2.8 cm. Let r be the radius of the inscribed orange circle. c = r-x c = (r-(14/5)) cm. d = (10-r) cm. It implies; d² = r²+c² (10-r)² = r²+(r-(14/5))² 100-20r+r² = r²+r²-(28/5)r+(196/25) 100-20r = r²-(28/5)r+(196/25) 2500-500r = 25r²-140r+196 25r²+360r-2304 = 0...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2479159

a = 120π*2*30÷360 a = ⅓(60)π a = 20π cm. a = 62.8318530718 cm. a is the length of arc of the sector and also, the base circumference of the cone. 20π = 2πb b = 10 cm. b is the radius of the base of the circular cone. 30² = 10²+c² c² = 900-100 c = √(800) c = 20√(2) cm. c is the height of the cone. Volume of the cone is; ⅓*10*10π*20√(2) = ⅓(2000√(2))π cm³ = 2961.9219587722 cm³ Calculating the volume outside the sphere. 2*30d+20d = 20*20√(2) 80d = 400√(2) d = 5√(2...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2478768

Let the side length of the regular pentagon be 1 unit. a² = 2-2cos108 108° is the single interior angle of the regular pentagon. a = 1.6180339887 units. a is BE. b = 108-60 b = 48° c = ½(180-48) c = 66° d = 180-60-c d = 54° e = 180-d-b e = 78° f = 180-e f = 102° g = 180-f-72 g = 6° g is angle BEG = angle BGE. h = 180-2g h = 168° h is angle EBG. j = h-72 j = 96° j is angle CBG. k = a-1 k = 1.6180339887-1 k = 0.6180339887 units.. k is BF l² = 0....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2473303

sin60 = 18/a a = 20.7846096908 units. tan60 = 18/b b = 10.3923048454 units. c+270+60+30+135 = 180(5-2) c = 540-495 c = 45° d² = 2(18²) d = 18√(2) units. d = 25.4558441227 units. e = a+d e = 20.7846096908+25.4558441227 e = 46.2404538135 units. f = (85+z) units. f is the required length x. Therefore: x = f = (85+z) units. g = (46.2404538135+z) units. tan60 = h/a h = 20.7846096908tan60 h = 36 units. j = (25.4558441227+z) units. j is the required length y. y...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2452006

Let the inscribed bigger quarter circle radius be a. Let the inscribed big quarter circle radius be b. c = 5 cm. c is the radius of the inscribed semi circle. It implies; a+b = 10 --- (1). a+5 =√(10²+5²) a = √(125)-5 a = 5(√(5)-1) cm. Again, a is the inscribed bigger quarter circle radius. And; a+b = 10 b = 10-a b = 10-5(√(5)-1) b = (15-5√(5)) cm. Again, b is the inscribed big quarter circle radius. Area Blue is; = (10*10)-½(5²)π-¼(5√(5)-5)²π-¼(15-5√(5))²π = 100-½(25π)-¼(125-50√(5)+25...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2449446

cosa = ½ a = acos(½)° a = 60° Calculating Red Perimeter. It is; 2(length of arc of a sector with radius 2 units and angle acos(½)°). = 2(acos(½)π*2*2/360) = 4.1887902048 units. Calculating Green Area. It is; 2(half perimeter red - area triangle with height 2 units and base 2sin(acos(½)) units.) = 2(½*4.1887902048-½*2*2sin(acos(0.5))) = 2(2.0943951024-1.7320508076) = 0.7246885897 square units. Therefore; Red Perimeter ÷ Green Area is; = 4.1887902048÷0.7246885897 = 5.7801244078 = 5.78...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2446155

a² = 3²+4² a = √(25) a = 5 units. 2b² = 5² b² = (25/2) b = √(25/2) b = ½(5√(2)) units. b is the radius of the semi circle. c = 90+45 c = 135° (½(5√(2))/sin135) = (3/sind) d = 36.8698976462° e = 180-135-36.8698976462 e = 8.1301023538° (f/sin8.1301023538) = (0.5*5√(2)/sin135) f = 0.7071067812 units. g² = 2r² g = √(2)r units. r is the radius of the inscribed circle. Calculating r. (0.5(5√(2))-r)² = (√(2)r)²+0.7071067812² 12.5-5√(2)r+r² = 2r²+0.5 r²+5√(2)r-12 = 0 Resolving the quadratic eq...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2444075

a = 10 units. a is the radius of the circle. b = (20-2x) units. c = ½(b) c = ½(20-2x) c = (10-x) units. It implies; Calculating x. 10² = 2c² 100 = 2(10-x)² 100 = 200-40x+2x² 2x²-40x+100 = 0 x²-20x+50 = 0 Resolving the above quadratic equation via completing the square approach to get x. ( x-10)² = -50+(-10)² ( x-10)² = 50 x-10 = ±√(50) x = 10±5√(2) It implies; x ≠ 10+5√(2) x = 10-5√(2) units. x = 2.9289321881 units. The length, y of the side of the square is;...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2440551

a² = 12²+6² a = 6√(5) cm. a = AE. b = atan(1/2)° b = angle DAE. c = 180-atan(2) c = 116.56505117708° c = angle AFC. d = 180-116.56505117708-atan(1/2) d = 36.86989764584° (6/sin36.86989764584) = (e/sin116.56505117708) e = 8.94427191 cm. f = 6√(5)-8.94427191 f = 4.472135955 cm. g = 180-2atan(2) g = 53.13010235416° g = angle AEB. cos53.13010235416 = h/4.472135955 h = 2.683281573 cm. It implies; Area S1 is; 0.5*2.683281573*4.472135955sin53.13010235416 = 4.8 cm² Calculating S2 Area. j =...

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