You Searched for "post"

2242 result found.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2395906

mathematics-question-and-solution_1709706989PWgs4o.jpeg
Let a be the radius of the ascribed semi circle. b = a-½(7) b = (a-3.5) units. 15² = b²+c² c² = 225+(a-3.5)² c² = 225-(a²-7a+12.25) c = √(225-a²+7a-12.25) c = √(212.75+7a-a²) units. Therefore; a² = 3.5²+c² a² = 12.25+212.75+7a-a² 2a² = 225+7a 2a²-7a-225 = 0 Resolving the above quadratic equation via completing the square approach to get a, radius of the ascribed semi circle. (a-(7/4))² = (225/2)+(-7/4)² (a-(7/4))² = (1800+49)/16 a = (7/4)±√(1849/16) a = ¼(7±43) It implies; a ≠ ¼(7-43) a...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2395874

mathematics-question-and-solution_1709705012IgsPYE.jpeg
a = 1+2+1 a = 4 units. a is the side length of the square. tanb = 4/2 b = atan(2)° tanc = 4/1 c = atan(4)° d = 180-b-c d = (180-atan(2)-atan(4))° Observing Sine Rule. (1/sin(180-atan(2)-atan(4))) = (e/sin(atan(4))) e = 1.490711985 units. f = 180-b f = (180-atan(2))° Observing Cosine Rule. g² = 1.490711985²+2²-2*2*1.490711985cos(180-atan(2)) g = 2.98142397 units. Therefore, the required angle, x is; Observing Sine Rule. (2.98142397/sin(180-atan(2))) = (2/sinx) sinx = (3/5) x = 36.869897...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2394828

mathematics-question-and-solution_1709651716IXERQ9.jpeg
Notice! Radius of the circle, r is; r = √(50) r = 5√(2) units. a² = 2²+6² a² = 4+36 a = √(40)  a = 2√(10) units. tanb = 6/2 b = atan(3)° c = ½(a) c = ½(2√(10)) c = √(10) units. cosd = c/r cosd = √(10)/(5√(2)) d = acos(⅕√(5))° e = b-d e = (atan(3)-acos(⅕√(5)))° Therefore, the required length, x is; x² = 2²+(5√(2))²-2*2*5√(2)cos(atan(3)-acos(⅕√(5))) x = 5.0990195136 units. x = 5.1 units to 2 decimal places.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2393496

mathematics-question-and-solution_170958588166apaf.jpeg
AB = BC = AC = 6 cm. cos60 = 2/BE BE = 4 cm. (DE)² = 4²-2² DE = 2√(3) cm. (CE)² = (2√(3))²+4² (CE)² = 12+16 (CE)² = 28 CE = √(28) CE = 2√(7) cm. EF = (2√(7)-4) cm. EF = 1.29150262213 cm. AE = 6-4 AE = 2 cm. Angle AEF = 180-(atan(2/2√(2√(3)))+atan(4/2√(3))) Angle AEF = 100.89339464913° Therefore; Area Shaded, Triangle AEF is; 0.5*2*1.29150262213sin100.89339464913 = 1.26823039846 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2393162

mathematics-question-and-solution_1709572399OBKYKs.jpeg
Calculating angle x (angle CBD). Let the two equal lengths of the plane shape be 1 unit. a = 180-30-24 a = 180-54 a = 126° a is angle ABC. It implies; Observing sine rule. (1/sin30) = (b/sin24) b = 0.8134732862 units. b is AB. Observing cosine rule. c² = 0.8134732862²+1-2*0.8134732862*1cos30 c = 0.5027541398 units. c is BD. Observing sine rule. (0.5027541398/sin30) = 1/sind) d = 84° d is angle ABD. It implies, the required angle x (angle CBD) is; x+d = a x = a-d x = 126-84 x = 42°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2392703

mathematics-question-and-solution_1709554248Lfpn0j.jpeg
Radius, r of the six congruent inscribed circles each is; 3r = 7 r = (7/3) units. Radius, R of the ascribed circle is; R = ½(6r) And r = (7/3) units. R = ½*6*(7/3) R = 7 units. Therefore, Area Orange is; Area ascribed circle with radius 7 units - 6(area circle with radius (7/3) units) = 7²π-6(7/3)²π = 49π-⅓(98π) = ⅓(147π-98π) = ⅓(49π) square units. = 51.3126800086 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2392377

mathematics-question-and-solution_1709540887oYB5TC.jpeg
a = ½(10+6) a = 8 units. a is the radius of the ascribed half circle. b = a-6 b = 8-6 b = 2 units. b is OB. Observing similar triangles side lengths ratio. b - (10-c) (6-c) - b Cross Multiply. b² = (6-c)(10-c) b² = 60-16c+c² b = √(60-16c+c²) units. d = (8-c) units. It implies; 8² = d²+b² 64 = (8-c)²+√(60-16c+c²)² 64 = 64-16c+c²+60-16c+c² 0 = 60-32c+2c² c²-16c+30 = 0 Resolving the above quadratic equation via completing the square approach to get c. (c-8)² = -30+(8)² (c-8)² = 34 c = 8±...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2391371

mathematics-question-and-solution_1709485420TaC13j.jpeg
Shaded Area is; 4(area quarter circle with radius 2 units - area triangle with height and base 2 units each) + 2(area square with side 2 √(2) units - area semi circle with radius 2 units) = 4(¼*4π-½*4)+2((2√(2))²-½*4π) = 4π-8+16-4π = -8+16  = 8 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2390872

mathematics-question-and-solution_1709466044rBNjex.jpeg
Let r be the radius of the ascribed semi circle. It implies; r² = 14²+a² a² = r²-196 a = √(r²-196) cm. a is OA. b = a+r b = (√(r²-196)+r) cm. b is AB. Therefore, calculating r. ½*14*(√(r²-196)+r) = 147 7(√(r²-196)+r) = 147 (√(r²-196)+r) = 21 √(r²-196) = 21-r r²-196 = (21-r)² r²-196 = 441-42r+r² 42r = 441+196 42r = 637 6r = 91 r = 15⅙ cm. r = 15.167 cm. AB = √(r²-196)+r AB = √((91/6)²-196)+(91/6) AB = (35/6)+(91/6) AB = 126/6 AB = 21 cm. Therefore, green area is; Area semi circle with ra...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2390822

mathematics-question-and-solution_1709463974kjtFf0.jpeg
Notice, the triangle is isosceles. It implies; ½(12x+22)+(6y-5) = 90 6x+11+6y-5 = 90 6x+6y = 84 x+y = 14 --- (1). ½(12x+22)+(10y-41) = 90 6x+11+10y-41 = 90 6x+10y = 120 3x+5y = 60 --- (2). Solving (1) and (2) simultaneously observing substitution approach to get x and y. From (1). x = 14-y --- (3). Substituting (3) in (2). 3(14-y)+5y = 60 42-3y+5y = 60 2y = 18 y = 9 It implies, x will be; At (1). x = 14-y  And y = 9 Therefore; x = 14-9 x = 6
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