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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1945925

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Let the side of the regular pentagon be 1 unit. Notice; The regular pentagon side is equal the regular heptagon side. a = ⅐(180*5) a = ⅐(900)°, the single interior angle of the regular heptagon. b = 2a b = ⅐(450)° tan(⅐(450)) = c/0.5 c = 1.03826069829 units. cos(⅐(450)) = 0.5/d d = 1.15238243548 units. Where d is the radius of the ascribed circle. e = 180-⅐(900) e = ⅐(360)° tan18 = 0.5/f f = 1.53884176859 units. g² = 1.15238243548²-0.5² g = 1.03826069828 units. h = f-g h = 0.5005810...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1945920

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Let the side of the regular pentagon be 1 unit. a² = 2-2cos108 a = 1.61803398875 units. sin30 = b/0.5 b = 0.25 unit. c = 2b c = 0.5 unit. Where c is the inscribed square side. Area Square is; (0.5)² = 0.25 square units. Inscribed circle radius d is; d = ½(b) d = 0.125 unit. e = 0.5-0.125 e = 0.375 unit. f = 2atan(0.125/0.375) f = 36.86989764584° tan36.86989764584 = g/0.5 g = 0.375 unit. Shaded Area is; 0.5*0.375*0.5-π*0.125*0.125 = 0.04466261479 square units. (a) Area Shaded ÷...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1944792

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a² =2*12² a = 12√(2) cm. a = AC. 12b+12b+12√(2)b = 12*12 b = 144/(24+12√(2)) b = 12-6√(2) cm. b = 3.51471862576 cm. b = radius of the inscribed circle. BF = ½(12√(2)) BF = 6√(2) cm. c² = (6√(2))²+3.51471862576²-12√(2)*3.51471862576cos45 c = 6.49435320175 cm. c = EF. (6.49435320175/sin45) = (6√(2)/sind) d = 112.5° d = Angle BEF. e = 2(180-112.5-45) e = 45° f = 112.5-90 f = 22.5° cos22.5 = 12/g g = 12.98870640351 cm. g = EG. FG = 12.98870640351-6.49435320175 FG = 6.49435320176 cm. tan22...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1937325

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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 = 12+16 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/1916258

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Calculating R. (5+R)² = (10-R)²+(5-R)² 25+10R+R² = 100-20R+R²+25-10R+R² 20R = 100-20R+R² R²-40R+100 = 0 Therefore; R ≠ 20+10√(3) units. R = 20-10√(3) units. R = 2.67949192431 units. Calculating r. (5-r)² = r²+a² 25-10r+r² = r²+a² a = √(25-10r) units. Therefore; (5+r)² = (5-r)²+ (5+√(25-10r))² 25+10r+r² = 25-10r+r²+25+10√(25-10r)+25-10r 20r = 25+10√(25-10r)+25-10r 30r-50 = 10√(25-10r) (3r-5)² = 25-10r 9r²-30r+25 = 25-10r 9r² = 20r 9r = 20 r = ⅑(20) units. r = 2.22 units. It implies; r ÷...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1904531

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Radius of the circle r, is; πr² = 25π r = 5 cm. Therefore diameter is; 2r = 10 cm. Let the side of the small yellow Inscribed square be x. a² = 2x² a = √(2)x cm. Where a is a diagonal of the small yellow Inscribed square. Extending a. We now have a rectangle inscribed the circle (cyclic quadrilateral) with diameter 10 cm. Where 6 cm is the width and (6+√(2)x) cm is the length. Calculating x (small yellow Inscribed square length) 10² = 6²+(6+√(2)x)² 100-36 = (6+√(2)x)² 8 = 6+√(2)x 2 = √...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1872693

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Calculating angle ABD. Let it be a. Using cosine rule. 25 = 36+9-2*18cosa 36cosa = 45-25 a = acos(20/36)° a = acos(5/9)° Calculating x. Using cosine rule. x² = 36+144-2*6*12cos(acos(5/9)) x² = 180-144(5/9) x² = 180-(5*16) x² = 100 x = 10 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1870080

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Notice; Diameter of the ascribed semi circle is 20 cm. Radius, r will be 10 cm. Calculating Area Shaded. It will be; Area half circle with radius 10 cm - 2(area semi circle with radius 5 cm) - Area sector with radius 5 cm and angle 60° + 2(area sector with radius 5 cm and angle 60°) - 2(area triangle with two equal lengths 5 cm each and the angle they form, 60°) = ½(100π)-2(½*25π)-⅙(25π)+2(⅙*25π)-2(½*25*(½√(3))) = 50π-25π-⅙(25π)+⅓(25π)-½(25√(3)) = 25π-⅙(25π)+⅓(25π)-½(25√(3...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1868088

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Calculating shaded area considering semi circle PQ. Diameter of semi circle PQ will be 10cm, therefore radius is 5cm. Shaded area will be; Area semi circle PQ of radius 5cm and angle 180° - 2(area sector of radius 5cm and 60° - area equilateral triangle of equal side lengths 5cm each and angle 60° each angle) - 2(area sector of radius 5cm and 60° - area equilateral triangle of equal side lengths 5cm each and angle 60° each angle). =(5²π)/2 - (2(5²π*60)/360 - (5²sin60)/2) - (2(5²π*6...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1855059

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Calculating BC. (BC/sin30) = (12/sin75) BC = 6.21165708246 cm. Calculating EF. (6.21165708246/sin75) = (EF/sin30) EF = 3.21539030917 cm. Angle FEG is; 360-2(75)-60 = 360-210 = 150° It implies; Area triangle EFG is; 0.5*3.21539030917*3.21539030917sin150 = 2.58468371008 cm²
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