You Searched for "100"

289 result found.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3214346

mathematics-question-and-solution_1755575144WaxcW2.jpeg
Notice. The ascribed quadrilateral is a square. Calculating x, the side length of the square. a²+x² = 8² a = √(64-x²) units. b²+x² = 10² b = √(100-x²) units. It implies;. c = (x-√(64-x²)) units. d = (x-√(100-x²)) units. Notice again. The inscribed triangle is a right-angle triangle. Therefore; √(x-√(100-x²) ~ √(64-x²) 6 ~ 8 Cross Multiply. 8(x-√(100-x²)) = 6√(64-x²) 4(x-√(100-x²)) = 3√(64-x²) 4x-4√(100-x²) = 3√(64-x²) 16x²-32x√(100-x²)+1600-16x² = 57...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3213003

mathematics-question-and-solution_1755473320rSTo95.jpeg
Calculating Shaded Blue Area. Notice. ½(10) = 5 cm is the radius of the inscribed circle. a² = 2(10)² a = 10√(2) cm. a is the diagonal of the ascribed square. b = ½(a) b = 5√(2) cm. It implies; 10² = 5²+(5√(2))²-2*5*5√(2)cosc 100 = 25+50-50√(2)cosc 50√(2)cosc = 75-100 cosc = -(25)/(50√(2)) c = acos(-¼*√(2)) c = 110.704811055° d = 2c d = 221.409622109° e = 360-d e = 138.590377891° (5/sinf) = (10/sin110.704811055) f = 27.8855668363° g = 2f g = 55.771133...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3207712

mathematics-question-and-solution_1755121744dIhfIT.jpeg
Sir Mike Ambrose is the author of the question. Calculating Area Blue. It is; Area triangle with height 25 units and base ⅓(100) units - Area square with single side length 5 units - Area triangle with height 30 units and base 30tan(atan(5/35)) units - Area triangle with height 5 units and base ⅓(20) units -  Area triangle with height 5 units and base 25sin(atan(3/4)) units. = ⅓(1250) -25 - 450tan(atan(5/35)) - ⅓(50) - ½(75) = ½(675) - 450tan(atan(5/35)) = ½(675 - 900tan(ata...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3205271

mathematics-question-and-solution_17549778935iCgYX.jpeg
Calculating Angle x. Let AC = 1 units. sin(10+10) = a/1 a = 0.34202014333 units. a is BC. tan40 = b/0.34202014333 b = 0.28698897612 units. c = 180-10-30-40 c = 100° (1/sin100) = (d/sin10) d = 0.17632698071 units. e = a-d e = 0.34202014333-0.17632698071 e = 0.16569316262 units. Therefore, the required angle x is; tanx = e/b x = atan(0.16569316262/0.28698897612) x = 30°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3201140

mathematics-question-and-solution_1754704753tmtFs2.jpeg
Calculating the circumference of the blue inscribed circle. Let x be the base of the inscribed right-angled triangle with height 10 units.  x²+10² = a² a = √(x²+100) units. b = ½(a) b = ½√(x²+100) units. Calculating x. 10² = 10²+(½√(x²+100))²-10√(x²+100)cosy 10√(x²+100)cosy = ¼(x²+100) --- (1). cosy = x/√(x²+100) --- (2). Substituting (2) in (1). 10√(x²+100)*x/√(x²+100) = ¼(x²+100) 10x = ¼(x²+100) 40x = x²+100 x²-40x+100 = 0 (x-20)² = -100+(-20)² (x-20)² = 300 x = 20±√(300) x =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3195459

mathematics-question-and-solution_1754381581L6tUJE.jpeg
Calculating x. Let a be the side length of the square. b² = 2a² b = √(2)a units. c = ½(b) c = ½√(2)a units. a ~ 1 0.5 ~ d Cross Multiply. d = 1/(2a) units. Calculating a. c+d = a ½√(2)a+(1/(2a)) = a √(2)a²+1 = 2a² (2-√(2))a² = 1 a = √(1/(2-√(2))) a = 1.30656296488 units. Again, a is the side length of the square. Recall. d = 1/(2a) units. And a = 1.30656296488 units. d = 1/2(1.30656296488) d = 0.38268343236 units. e² = d²+d² e² = 2(0.38268343236)² e...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3177715

mathematics-question-and-solution_1753260801CHCCVd.jpeg
Calculating area ABCD. Let x be the radius of the inscribed circle. a = ½(16) a = 8 units. x² = 8²+a² a = √(x²-64) units. b = ½(25) units. It implies; x ~ 8 ½(25) ~ x Cross Multiply. x² = 100 x = 10 units. Again, x is the radius of the inscribed circle. c = 2x c = 20 units. c is the diameter of the inscribed circle, and also, the height of the isosceles trapezoid. sind = 10/(0.5*25) d = asin(20/25) d = asin(4/5)° tan(asin(4/5)) = 20/f f = 15 units. g = x-√(10²-(0.5*16)²) g = 10-√(1...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3175040

mathematics-question-and-solution_17530876882vPbgp.jpeg
Calculating the area of the ascribed regular octagon. Let the equal longest length of the both inscribed yellow triangles be x. a = ⅛*180(8-2)) a = ⅛(180*6) a = ½(45*6) a = 135° a is the single interior angle of the ascribed regular octagon. sin(0.5*135) = (0.5x)/b b = 0.54119610015x units. b is the side length of the regular octagon. c = 135-(0.5*45)-½(360-2(135)) c = 135-22.5-45 c = 135-67.5 c = 67.5° c is the second bigger interior angle of inscribed triangle 21 square units. 2e² = (0....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3173559

mathematics-question-and-solution_1752985501Vy6OMU.jpeg
Calculating Yellow Area. Let a be the side length of the square. b = ½(a) units. c² = a²+(½(a))² c² = ¼(5a²) c = ½√(5)a units. c is the radius of the half circle. It implies; a ~ ½√(5)a d ~ a Cross Multiply. ½√(5)d = a d = 2a/√(5) d = ⅕(2√(5)a) units. e = ½(d) e = ⅕√(5)a units. f = c-e f = ½√(5)a-⅕√(5)a f = ⅒(3√(5))a units. Calculating a. ½*a*½(a)+½*⅒(3√(5))a*⅕(2√(5)a) = 15 ¼(a²)+⅒(3a²) = 15 5a²+6a² = 300 11a² = 300 a = 10√(3)/√(11) a = 10√(33)/11 units. g = 2f g = 2*⅒(3√(5))a g = ⅕(...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3170567

mathematics-question-and-solution_1752809768qk2VBK.jpeg
Calculating the area of the ascribed half circle. Let a be the radius of the circle. b² = a²+(3√(3))² b = √(a²+27) units. c = ½(3+b) c = ½(3+√(a²+27)) units. d = c-3 d = ½(3+√(a²+27))-3 d = ½(√(a²+27)-3) units. e² = (3√(3))²-(½(√(a²+27)-3))² e² = 27-¼(a²+27-6√(a²+27)+9) e² = ¼(108-a²-27+6√(a²+27)-9) e² = ¼(72-a²+6√(a²+27)). It implies; Calculating a. e²+c² = a² ¼(72-a²+6√(a²+27))+(½(3+√(a²+27)))² = a² ¼(72-a²+6√(a²+27))+¼(9+6√(a²+27)+a²+27) = a² 72-a²+6√(a²+27)+36+6√(a²+27)+a² = 4a²...
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