Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2134580

Area yellow is; Area circle with radius 4 cm - 3(area triangle with two side 4 cm each and angle 120°). = 16π - ¼(3*16√(3)) = 16π-12√(3) = 4(4π-3√(3)) cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2134561

Sir Mike Ambrose is the author of the question. Let the square side be 2 units. Area Square is; 2² = 4 square units. Calculating r, radius of the inscribed circle. √(2)r+½√(2)r+½√(10)r = ½√(2)*√(2) r = ¼(3√(2)-√(10)) units. Area Shaded is; Area triangle with height √(2) units and base ½√(2) unit - Area circle with radius ¼(3√(2)-√(10)) unit. = ½*√(2)*½√(2) - π(¼(3√(2)-√(10)))² = ½ - ¼*π(7-3√(5)) = ¼(2-7π+3√(5)π) square units. It implies; Area Shaded ÷ Area Square exactly in terms of π as...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2133254

Let x be 2 unit. Therefore; Area square is; 8² = 64 square unit. Calculating area rectangle. The length is; √(8²+6²) + 2cos(atan(4/3)) = 10+(6/5) = 56/5 unit. The width is; 2sin(atan(4/3)) = 8/5 unit. Area rectangle is; (56/5)*(8/5) = 448/25 square unit. It implies; Area rectangle : Area square is; (448/25) : 64 = 448 : (25*64) = 56 : (25*8) = 14 : (25*2) = 7 : 25

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2131968

Sir Mike Ambrose is the author of the question. Let the side of the inscribed square be x. Calculating x. (10√(29)-√(29)x)/(5sin(45+atan½)) = (√(2)x)/(sin(90-atan½-atan(2/5)) It implies; 3√(5)x = 16√(5) - ⅕(8√(5))x 23√(5)x = 80√(5) x = 80/23 units. Calculating Area Purple exactly. It is; Area triangle with height (80√(2)/23) units and base (30√(29)/23)sin(45+atan(2/5)) units - Area triangle with height and base (80/23) units respectively. = ½*(80√(2)/23)*(30√(29)/23)sin(45+atan(2/5)) -...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2130637

Sir Mike Ambrose is the author of the question. Let the side length of the inscribed equilateral triangle ABD be x. Calculating x. (12√(2))² = (½√(3)x)² + ((12√(2))-½(12√(2)-x))² It implies; x = 3(√(26-√(2)) cm. Therefore; Area Yellow exactly in cm² is; Area trapezoid with parallel sides (3√(26)-3√(2)) cm and (2√(26)-2√(2)) cm, and height ½(√(78)-√(6)) cm. = ½*½(√(78)-√(6))*((3√(26)-3√(2))+(2√(26)-2√(2))) = ¼(√(78)-√(6))(5√(26)-5√(2)) = 35√(3)-5√(39) = 5(7√(3)-√(39)) square...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2129899

Sir Mike Ambrose is the author of the question. Let the square side be x. Calculating x. 25 = (x²-9)+2x²-2√(2)x*√(x²-9)cos(45+y) Where y is an angle. Notice; cosy = √(x²-9)/x and siny = 3/x. It implies; 25 = (x²-9)+2x²-2√(2)x*√(x²-9)((√(2x²-18)-3√(2))/2x) 34 = x²+18+6√(x²-9) 16 = x²+6√(x²-9) x = √(10) units. Therefore; Area A exactly in square units is; Area sector with radius 5 units and angle asin(4/5) - Area triangle with height 2√(5) units and base sin(45+atan3) units - Area trian...

Mathematics Question and Solution

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

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...

We are smart and clever!

https://onlineedumath.com/update/we-are-smart-and-clever/2123614

Our certified, experienced, resilient, productive and committed Math Educators will be glad to be at your service. We make teaching and learning Mathematics fun for learners, we expose the simplicity and beauty of Mathematics to our learners, and mentor them to appreciate and love Mathematics. Communicate us to mentor/educate you Mathematics better online at affordable tuition.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2123263

Let the square side be 1 unit. Area Square is; 1² = 1 square unit. Calculating Area Blue. a = 108-60-36 a = 12° b = 180-12-108 b = 60° 1 = 2c²-2c²cos108 1 = 2.6180339887c² c = 0.6180339887 units. c is the side length of the regular pentagon. (0.6180339887/sin60) = (d/sin12) d = 0.148374968 units. Area Blue is; 0.5sin30 - 0.5*0.148374968*0.6180339887sin108 = 0.25-0.043606309 = 0.206393691 square units. It implies; Area Blue ÷ Area Square to 2 decimal places is; 0.206393691÷1 = 0.206...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2122090

Let the side length of the regular pentagon be 1 unit. a = ½(108)-25 a = 54-25 a = 29° b = 72-29 b = 43° (c/sin108) = (1/sin43) c = 1.3945143742 units. tan25 = d/1.3945143742 d = 0.6502727321 units. Area Yellow is; 0.5*1*0.6502727321sin25 = 0.0687042829 square units. Calculating Shaded Area. e = 54+25 e = 79° f² = 1²+1.3945143742²-2*1.3945143742cos79 f = 1.5532219952 units. (1.5532219952/sin79) = (1/sing) g = 39.1973995412° h = 39.1973995412+25 h = 64.1...

You Searched for "post"

2242 result found.

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

We are smart and clever!

Mathematics Question and Solution

Mathematics Question and Solution

OnlineEdumath

Socials

Useful Tags

Quick Links

Reach Us

Safety and Abuse Reporting

Thanks for being awesome!

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

Search

You Searched for "post"

2242 result found.

Subscribe to our mailing list for latest updates.

Safety and Abuse Reporting

Thanks for being awesome!

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