Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2565398

Calculating R. a²+3² = R² a = √(R²-9) units. b = 1+a b = (1+√(R²-9)) units. c = a-1 c = (√(R²-9)-1) units. d²+1² = R² d = √(R²-1) units. e = 3+d e = (3+√(R²-1)) units. f = d-3 f = (√(R²-1)-3) units. Therefore; (√(R²-1)-3)(3+√(R²-1)) = (1+√(R²-9))(√(R²-9)-1) √(R²-1)-9 = √(R²-9)-1 √(R²-1)-8 = √(R²-9) R²-1-16√(R²-1)+64 = R²-9 16√(R²-1) = 72 2√(R²-1) = 9 4(R²-1) = 81 4R² = 85 R = √(85/4) R = ½√(85) units. R = 4.6097722286 units. R is the radius of the ascr...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2565373

Let r be the radius of the circle. a = (r-1) units. b = (r-0.5) units. Calculating r. It implies; r² = (r-1)²+(r-0.5)² r² = r²-2r+1+r²-r+(¼) 0 = r²-3r+(5/4) 4r²-12r+5 = 0 4r²-2r-10r+5 = 0 2r(2r-1)-5(2r-1) = 0 Therefore; r ≠ ½ units. r = ½(5) units. r = 2.5 units. 2r is; 2*(2.5) = 5 units. 5 units is the diameter of the circle. c = 5-1 c = 4 units. c is the side length of the coloured square. Area Coloured Square is; c² = 4² = 16 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2564679

Let a be 1 units. a is the side length of the regular hexagon. Calculating b. c = 120-90 c = 30° 120° is the single interior angle of the regular hexagon. cos30 = 1/d d = 2/√(3) d = ⅓(2√(3)) units. d = 1.1547005384 units. e² = 2-2cos120 e = √(3) units. e = 1.7320508076 units. f = 90-(½(180-120)) f = 90-30 f = 60° Length b is; b² = 1.7320508076²+1.1547005384²-2*1.7320508076*1.1547005384cos60 b = 1.5275252317 units. Therefore; a/ b is; 1/1.5275252317 = 0.6546536707 exactly in decimal.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2563885

a = 6+4 a = 10 cm. b²+6² = 10² b = √(100-36) b = √(64) b = 8 cm. c = b+4 c = 12 cm. d = c-r d = (12-r) cm. r is the ascribed half circle radius. e = 6+d e = 6+(12-r) e = (18-r) cm. Calculating r, radius of the ascribed half circle. r² = (18-r)²+6² r² = 324-36r+r²+36 36r = 360 r = 10 cm. Area Painted is Area half circle with radius 10 cm - Area half circle with radius 6 cm - Area half circle with radius 4 cm. = ½*π*10²-½*π*6²-½*π*4² = ½(100π-36π-16π) =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2563864

Let the three equal lengths be 1 unit each. a² = 1²+1² a = √(2) units. b = 150-45 b = 105° c² = √(2)²+1²+2√(2)cos 105 c = 1.9318516526 units. (1.9318516526/sin105) = (1/sind) d = 30° Therefore, the required angle, x is; x = 45+d x = 75°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2563853

Let AD be 4 units. Therefore; AB will be; ½(AD)+½(½(AD)) AB = ½(4)+½(½(4)) AB = 2+1 AB = 3 units. Area ABCD is; 4*3 = 12 square units. Calculating Area Red (Area Trapezoid). 1 unit is the radius of the big inscribed quarter circle. It implies; a² = 2(1)² a = √(2) units. a is the height of the red area. 2 units is the radius of the bigger inscribed quarter circle. It implies; b² = 2²+2² b = √(8) b = 2√(2) units. b is the top parallel length of the red a...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2563815

9² = 16²+17²-2*16*17cosa 81 = 545-544cosa 544cosa = 545-81 cosa = 464/544 a = 31.4669762933° 13² = 16²+11²-2*16*11cosb 169 = 377-352cosb 352cosb = 377-169 cosb = (208/352) b = 53.7784533802° c = b-a c = 22.3114770869° It implies, the required length, x is; x² = 11²+17²-2*11*17cos22.3114770869 x = 8 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2563797

Let the side length of the red square be a units. b²+3² = a² b = √(a²-9) units. c²+6² = a² c = √(a²-36) units. Observing similar plane shape (right-angled) side length ratios. √(a²-9) - 6 3 - √(a²-36) Cross Multiply. √(a²-36)√(a²-9) = 18 (a²-36)(a²-9) = 18² (a²-36)(a²-9) = 324 a⁴-45a²+324 = 324 a⁴ = 45a² a² = 45 square units. Where a² is the red square area. Notice! a = 3√(5) units is the side length of the red square.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2563037

Notice; Diameter of the inscribed semi circle is ½ unit. Therefore radius of the inscribed semi circle is ¼ unit. Let the radius of the inscribed small circle be r. Calculating r. (1-√((¼+r)²-(¼-r)²))²+(1-r)²=(1+r)² (1-√(r))²+1-2r+r²=1+2r+r² 1-2√(r)+r=4r 1-3r=2√(r) (1-3r)²=4r 1-6r+9r²=4r Therefore; 9r²-10r+1=0 9r(r-1)-1(r-1)=0 (9r-1)(r-1)=0 Therefore; r ≠ 1 r = 1/9 unit. Radius of the small inscribed circle r is; 1/9 unit.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2562836

Radius of the inscribed quarter circle is 36 cm. Radius of the inscribed semi circle is; ½(½(36))=9 cm. Radius of the small inscribed circle is; ⅓(⅓(36))=4 cm. Therefore the side lengths of triangle blue will be; (36+9)=45 cm. (36+4)=40 cm. (9+4)=13 cm. Therefore area triangle blue is; 0.5x45x40sin(cos–¹((45²+40²-13²)÷(2x45x40))) = 252 cm²

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