Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2147406

Sir Mike Ambrose is the author of the question. Let the side of the regular hexagon be 2 units. Area Regular hexagon is; ½(12)*(2/(2tan(180/6))) = 6/tan30 = 6√(3) square units. Area Shaded is; Area triangle with height and base 2 units respectively - Are triangle with height (2-√(3)) unit and base 0.28144275258sin45 - Area sector with radius 1 unit and angle 120° + Area triangle with height 1 unit and base sin120 units. ½(2*2) - ½(2-√(3))*0.28144275258sin45 - (120π÷360) + ½(sin120) = 2...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2147400

Sir Mike Ambrose is the author of the question. Let the side of the inscribed regular pentagon be 2 units. Therefore; Length Red (AD) is; 2+2(2cos72) = 2+4cos72 = 3.2360679775 units. Calculating Length Blue. Let the radius of the inscribed circle be r. r is; tan27 = r/(3.2360679775-2sin72-r) tan27=r/(1.33395494491-r) 0.67968399291=1.50952544949r r = 0.45026335471 unit. Length Blue is; 1.1163084098tan54 + r 1.1163084098tan54+0.45026335471 = 1.98673006763 units....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2145374

Notice; Quadrilateral KMHE and Quadrilateral HMLF are similar plane shapes. Therefore; Let MH be x. Calculate x. x = 4 9 = x, Cross Multiply. It implies; x² = 9*4 x² = 36 x = 6 cm. Therefore; MH = 6 cm.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2145356

Sir Mike Ambrose is the author of the question. Shaded Area exactly in square cm is; Area rectangle with length 6 cm and 2√(3) cm - Area triangle with height (10-4√(3)) cm and base (10-4√(3))sin60. = 6*2√(3) - ½(10-4√(3))²sin60 = 12√(3) - ¼(148√(3)-240) = 12√(3) - (37√(3) - 60) = 12√(3) - 37√(3) + 60 = 60 - 25√(3) = 5(12-5√(3)) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2144034

Sir Mike Ambrose is the author of the question. Let the side length of the square be 10 unit. Area square is; 10² = 100 square unit. Therefore; Area R is; Area semi circle with radius 5 unit - Area sector with radius 5 unit and angle (180-2atan(2))° - Area kite with diagonal lengths √(10²+5²) unit and √(200-200cos(2atan(½))) unit + Area sector with radius 10 unit and angle 2atan(½)° - 2(area quarter circle with radius 5 unit - area triangle with height and base 5 unit respectively) - area s...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2143050

Sir Mike Ambrose is the author of the question. Let CD = 1 unit. Therefore; AB = 2 units. (a) Calculating Alpha, let it be y to 2 decimal places. (0.84523652348/siny) = (1/sin115) siny = 0.84523652348sin115 y = asin(0.84523652348sin115) y = 49.99999999989° ≈ 50° (b) Calculating Area Hexagon ÷ Area Green to 2 decimal places. Let the side of the regular hexagon be x. Calculating x. 2x = 2.09302817926 Where 2.09302817926 units is the vertical length from A to the apex (vertical...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2141249

Sir Mike Ambrose is the author of the question. Let the side of the square be 2 units. (a) Calculating theta, let it be x. Angle x is; x+65+90 = 180 x = 180-155 x = 25° (b) Calculating Area Circle ÷ Area Square to 3 decimal places. Notice; Square side is 2 units. Area square is; 2² = 4 square units. Radius of the inscribed circle, r is; (r/tan35) + (r/tan22.5) = 2 It implies; r = (2tan22.5*tan35)/(tan22.5+tan35) r = 0.52051322189 unit. Area inscribed circle is;...

Mathematics Question and Solution

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

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/2137983

Sir Mike Ambrose is the author of the question. Let the square side be 1 unit. Calculating Area Yellow. Length CE is; (CE/sin105)=(1/sin60) CE = sin105/sin60 CE = (√(6)+√(2))/2√(3) units. Area Yellow is; Area triangle with height (√(6)+√(2))/2√(3)) units and base (√(6)+√(2))/2√(3))sin60 ½*((√(6)+√(2))/2√(3))²sin60 = ¼((√(3)+2)/√(3)) square units. Area Orange is; Area triangle with height ½(√(6)-√(2)) units and base ⅙(3√(2)-√(6)) units. = ½*½(√(6)-√(2))*⅙(3√(2)-√(6)) = ⅙(2√(3)-3) square u...

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

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