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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2668416

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Calculating r, radius of the circle. a = 6+4 a = 10 units. a is AD = AC. b²+6² = 10² b = √(100-36) b = √(64) b = 8 units. c = b-r c = (8-r) units. Therefore, r, radius of the circle is; r² = (8-r)²+4² r² = 64-16r+r²+16 16r = 80 r = 5 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2668401

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Calculating R, radius of the inscribed quarter circle (side length of the square). a = (R-3) units. b = (R+1) units. Therefore; (R+1)² = R²+c² c² = R²+2R+1-R² c = √(2R+1) units. d = c+3 d = (3+√(2R+1)) units. e = a+1 e = (R-3)+1 e = (R-2) units. R is; (3+√(2R+1))² = R²+(R-2)² 9+6√(2R+1)+2R+1 = R²+R²-4R+4 6√(2R+1) = 2R²-6R-6 3√(2R+1) = R²-3R-3 18R+9 = R⁴-3R³-3R²-3R³+9R²+9R-3R²+9R+9 18R+9 = R⁴-6R³+3R²+18R+9 0 = R⁴-6R³+3R² R²-6R+3 = 0 (R-3)² = -3+(-3)² (R-3)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2668392

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Calculating x. Notice. x is the radius of the big inscribed half circle. Let y be the radius of the small inscribed half circle. It implies; 2x+2y = 16 2y = 16-2x y = (8-x) units. Again, y is the radius of the small inscribed circle. Therefore, x, radius of the big inscribed half circle is; Area half circle with radius x units + Area half circle with radius (8-x) units + 15π = Area half circle with radius 8 units. ½(x²)π+½(8-x)²π+15π = ½(8²)π x²+(8-x)²+30 = 64 x²+64...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2666676

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a = 12+1 a = 13 units. a is the radius of the quarter circle. 13² = 12²+b² b² = 169-144 b = √(25) b = 5 units. R is the radius of the inscribed circle. c = b+R c = (5+R) units. d = (a-R) d = (13-R) units. Therefore, R is; (13-R)² = R²+(5+R)² 169-26R+R² = R²+25+10R+R² 169-26R = 25+10R+R² R²+36R-144 = 0 Resolving the above quadratic equation via completing the square approach to get R, radius of the inscribed circle. (R+18)² = 144+(18)² (R+18)² = 468 R = -18±...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2666606

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Sir Mike Ambrose is the author of the question. The equation of the curve is; y = ½(3x²)+4 Point P is; P(4, 28) Calculating OQ (x) y = ½(3x²)+4 dy/dx = 3x, and x = 4. Therefore; dy/dx (gradient) = 3 * 4  dy/dx = 12 Therefore; (28-0)/(4-x) = 12 12x = 48-28 x = 20/12 x (OQ) = 5/3 units. QXp = 4 - x = 4 - (5/3) QXp = (12-5)/3 QXp = 7/3 units. Area Orange is; Area under the curve at points (0, 4) and (4, 28) - Area triangle with height 28 unit and base (7/3)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2666595

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Sir Mike Ambrose is the author of the question. Let the two equal lengths of the isosceles triangle be 2 units. Therefore; Area triangle ABC is = 2sin36 square units. = 1.17557050458 square units. Area green is; Area triangle with lengths 0.93931804148 units and 0.51440724841 units, and angle 63°, the angle they form. = ½(0.93931804148*0.51440724841)sin63 = 0.21526361627 square units. It implies; Area Green ÷ Area Isosceles Triangle ABC to 2 d.p. is; 0.2152636162...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2666568

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Sir Mike Ambrose is the author of the question. Let the single side length of the regular heptagon be 1 unit. Therefore; Area Regular Heptagon is; 7/(4tan(180/7)) = 3.633912444 square units unit. Area Blue is; Area triangle with two equal lengths 2.24697960372 units, each and angle (180/7)° - Area triangle with lengths 1 unit and 1.08208834613 units, and angle (270/7)° - Area triangle with lengths 0.80193773581 units and 1 unit, and angle (180/7)° = ½(2.24697960372)²sin(18...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2665242

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Calculating a. b² = a²+1 b = √(a²+1) units. b is AC. Therefore; a - 2a 1 - √(a²+1) Cross Multiply. a√(a²+1) = 2a √(a²+1) = 2 a²+1 = 4 a² = 3 a = √(3) units. a is BC. Therefore; CD = 2a And a = √(3) units. CD = 2√(3) units. Recall. b = √(a²+1) And a = √(3) units. b = √(√(3)²+1) b = √(3+1) b = √(4) b = 2 units. Again, b is AC. Therefore, the required length, x is; x² = (CD)²+(AC)² x² = (2√(3))²+2² x² = 12+4 x² = 16 x = √(16) x = 4 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2664624

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πr² = 9π r² = 9 r = 3 units. r is the radius of the inscribed circle. cosa = 6/7 a = acos(6/7)° a = 31° b = 90-a b = (90-acos(6/7))° b = 59° tan(0.5*59) = r/c c = 3/tan(0.5*59) c = 5.3024820487 units. d = r+c d = 3+5.3024820487 d = 8.3024820487 units. tan59 = e/d e = 13.8176505263 units. f = e-7 f = 6.8176505263 units. sin31 = g/7 g = 3.6052665244 units. h = 180-59 h = 121° It implies; j² = 3.6052665244²+6.8176505263²-2*6.8176505263*3.6052665244cos1...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2664473

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Sir Mike Ambrose is the author of the question. Let the radius of the semi circle be 4 units. Area triangle ABC is; ½(2√(2)*4√(2)) = 8 square units. Area red is; ½((5, 2) (4, 4) (2, 2) (4, (4/3)) (5, 2)) ½(12+(-16/3)+(4/3)) ½(12+(-12/3)) ½(12-4) = ½(8) = 4 square units. Area Red ÷ Area Triangle ABC is; 4 ÷ 8 = ½
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