You Searched for "100"

289 result found.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2139607

mathematics-question-and-solution_1695550452XExW1u.jpeg
Let O be the centre of the circle. Since AE = EB, therefore E is the midpoint of length AB. AB = 10cm, therefore AE = EB = 10/2 = 5cm. r will be; Triangle EBC Area+Triangle EOC Area+Triangle AOE Area+Triangle AOD Area+Triangle DOC Area = Square ABCD Area. The triangles height and base. EBC, height=5cm, base=10cm. EOC, height=rcm, base=√(5²+10²)=√(125)=5√(5)cm. AOE, height=(10-r)cm, base=5cm. AOD, height=rcm, base=10cm. DOC, height=rcm, base=10cm. Therefore r is; (5x10)/2+(5√(5)r)/2+(5(1...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2126302

mathematics-question-and-solution_1695050034bsD78p.jpeg
100 = 49+64-2*7*8cosA 112cosA = 113-100 A = acos(13/112) A = 83.3345727438° (7/cosB) = (10/cos83.3345727438) B = 44.0486256741° (8/cosC) = (10/cos83.3345727438) C = 52.6168015821° a²= 2.5²+10²-20*2.5cos52.6168015821 a= 8.7116506555 cm. (8.7116506555/sin52.6168015821) = (2.5/sinb) b = 13.1808743912° (b1)² = 2.5²+10²-20*2.5cos44.0486256741 b1 = 8.3852549156 cm. (8.3852549156/sin44.0486256741) = (2.5/sinc) c = 11.9635307444° d = 180-11.9635307444-13.1808743912 d = 1...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2121779

mathematics-question-and-solution_1694864171ofgjNB.jpeg
Let the regular hexagon side length be a. Calculating a. 0.5a²sin120 = 4√(3) √(3)a² = 16√(3) a = 4 units. b² = 4²+4²-2*4*4cos120 b = 4√(3) units. c = ½(b) c = 2√(3) units. d² = 2²+4²-2*2*4cos120 d = 2√(7) units. (2√(7)/sin120) = (2/sine) e = 19.1066053509° f = 120-e f = 100.8933946491° tan60 = 2√(3)/g g = 2 units. h² = 6²+28-24√(7)cos100.8933946491 h = 8.7177978871 units. (8.7177978871/sin100.8933946491) = (2√(7)/sinj) j = 36.5867755536° k = 120-36.5867755536 k = 83.4132244464° l =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2116290

mathematics-question-and-solution_16944957958OV77C.jpeg
Sir Mike Ambrose is the author of the question. Let R be radius of the inscribed circle. Calculating r. √(100-25) = r+√(25+r²) r = ⅓(5√(3)) cm. Therefore; Area Shaded as a single fraction in terms of π is; Area sector with radius 10cm and angle 30° - Area triangle with height 7.40536184888 cm and base ⅓(10√(3)sin30) cm. = (30*π*100÷360) - (0.5*7.40536184888*⅓(10√(3)sin30)) = ⅓(25π) - (889749425√(3)/144179222) = ( 3604480550π- 2669248275√(3))/ 432537666 cm²  = 15.4912196...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2101410

mathematics-question-and-solution_169389210217GXcb.jpeg
Let AB be 4 units. Let BC be 3 units. Area ABCD is; 4*3 = 12 square units. Calculating Area CEF. tana = 3/2 a = atan(3/2)° a is angle AMD. b = atan(2/3)° b is angle DEM. c = (180-atan(3/2))° c is angle CME. sin(atan(2/3)) = d/2 d = 4√(13)/13 units. d is EM. e² = (4√(13)/13)²+2²-4*(4√(13)/3)sin(180-atan(3/2)) e = 10√(13)/13 units. e is side length of the inscribed regular triangle. Area CEF is; 0.5*(10√(13)/13)²*sin60 = 0.5*(100/13)*(√(3)/2) = 25√(3)/13 square units. Area Triangle CEF...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2100314

mathematics-question-and-solution_1693843654Gizm3m.jpeg
Let AB be 4 units. BC = 6 units. Let a be the side length of the inscribed square. Calculating a. b² = 4²-a² b = √(16-a²) It implies; a = √(16-a²) 6 = 4 Cross Multiply. 4a = 6√(16-a²) 2a = 3√(16-a²) 4a² = 9(16-a²) 4a² = 144-9a² 13a² = 144 a = 12√(13)/13 units. a = 3.32820117735 units. c² = 4²-(12/√(13))² c² = 16-(144/13) c = √(64/13) c = 8√(13)/13 units. It implies, area red is; 0.5*(8√(13)/13)(12√(13)/13) = 48/13 square units. sind = (8√(13)/13)/4 d = asin(2√(13)/13) e = (90-asin(2...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2092520

mathematics-question-and-solution_16935193757IXI1P.jpeg
tan30 = 6/a a = 6√(3) cm. a = 10.39230484541 cm. sin30 = 6/b b = 12 cm. cos15 = c/12 c = (3√(6)+3√(2)) cm. c = 11.59110991547 cm. d = 11.59110991547²+10.39230484541²-2*11.59110991547*10.39230484541cos45 d = 8.48528137424 cm. (8.48528137424sin45) = (11.59110991547/sine) e = 75° f = √(100) f = 10 cm. g = 75-atan(3/4) g = 38.13010235416° cos38.13010235416 = h/10 h = 7.86610731855 cm. j² = 7.86610731855²+8²-2*7.86610731855*8cos75 j = 9.65925826289 cm. (9.65925826289/sin75) = (7.8661073185...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2088640

mathematics-question-and-solution_1693365997etlQnv.jpeg
Let the green and purple color regular triangle side be a. b = √(4²+2²) b = 2√(5) units. c = 0.5b c = √(5) units. d = √(8²+4²) d = 4√(5) units. e = √(a²+(a/2)²) e = √(5a²/4) e = ½(√(5)a) Calculating a, side length of the green and purple regular triangle. d-c = e 4√(5)-√(5) = ½(√(5)a) 3√(5) = ½(√(5)a) 3 = ½(a) a = 6 units. sin60 = f/3 f = ½(3√(3)) units. cos60 = g/3 g = (3/2) units. h = 6-(3/2) h = (9/2) units. j = 12+(9/2) j = (33/2) units. tank = (3√(3)/2)/(33/2) k = atan(√(3)/11)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2084922

mathematics-question-and-solution_1693220330m17SVo.jpeg
Let the side length of the regular nonagon be 1 unit. a = ⅑(180(9-2)) a = ⅑(180*7) a = 140° Area Nonagon is; 0.5*9*0.5tan(70) = ¼(9tan70) = 6.18182419377 square units. b = 540-3*140 b = 120° c = ½(b) c = 60° d = 140-60 d = 80° (e/sin30) = (1/sin50) e = 0.65270364467 units. sin80 = f/1 f = 0.98480775301 units. (1/sin20) = (g/sin80) g = 2.87938524157 units. sin80 = h/2.87938524157  h = 2.83564090981 units. j = 180-40-80 j = 60° (1/sin60) = (k/sin40) k = 0.74222719897 units. (0.74222...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2076443

mathematics-question-and-solution_1692877465xeWxDS.jpeg
Calculating equation of the curve. y-6√(2) = a(x-6)² --- (1). Substituting x = 0, y = 0 to get a. It implies; -6√(2) = 36a a = -(1/(3√(2)) --- (2). Substituting (2) in (1). y-6√(2) = -(1/(3√(2))(x-6)² y = -(1/(3√(2))(x²-12x+36)+6√(2) y = -(x²/(3√(2))+2√(2)x-6√(2)+6√(2) y = -(x²/(3√(2))+2√(2)x --- (3). Where (3) is the equation of the curve. Calculating x, when y is ⅓(10√(2)). ⅓(10√(2)) = -(x²/(3√(2))+2√(2)x  (x²/(3√(2))-2√(2)x+⅓(10√(2)) = 0 x²-12x+20 = 0 Resolving the quadratic equation. (...
WhatsApp Google Map
✕

Safety and Abuse Reporting

Thanks for being awesome!

We appreciate you contacting us. Our support will get back in touch with you soon!

Have a great day!

Are you sure you want to report abuse against this website?

Please note that your query will be processed only if we find it relevant. Rest all requests will be ignored. If you need help with the website, please login to your dashboard and connect to support

Cancel Confirm