Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2410652

Let c3 = a = 6 units. Let c2 = b = 8 units. Therefore; c1 = c = √(6²+8²) c = √(100) c = 10 units. Calculating Area S3. It is; Area equilateral triangle with side 6 units - 2(area equilateral triangle with side 3 units) - Area sector with radius 3 units and angle 60° = ½*6*6sin60-2(½*3*3sin60)-(60π*3*3/360) = 9√(3)-½(9√(3))-½(3π) = ½(18√(3)-9√(3)-3π) = ½(9√(3)-3π) square units. Calculating Area S2. It is; Area equilateral triangle with side 8 units - 2(area equilateral triangle with side...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2401648

Let the side length of the green square be x. Notice, radius of the two congruent circles is 5 units each. Therefore; a = (5-x) units. b = (5-½(x)) units. Observing Pythagoras Rule to get x, side length of the green square. 5² = b²+a² 5² = (5-½(x))²+(5-x)² 25 = 25-5x+(x²/4)+25-10x+x² 0 = 25-15x+(5x²/4) 5x²-60+100 = 0 x²-12x+20 = 0 Resolving the above quadratic equation via completing the square approach. (x-6)² = -20+(-6)² (x-6)² = 16 x = 6±√(16) x = 6±4 It...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2400035

Let AB = 1 unit. Calculating CD. a = 180-40-60 a = 80° a is angle ADB. Observing Sine Rule. (1/sin80) = (b/sin40) b = 0.6527036447 units. b is AD. c = 180-80 c = 100° c is angle ADC. d = 180-100-30 d = 50° d is angle CAD. Therefore, Observing Sine. (0.6527036447/sin30) = (e/sin50) e = 1 unit. e is CD. It implies; AB = CD Proved!

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2393496

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)² = 12+16 (CE)² = 28 CE = √(28) 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/2387055

Let the two equal lengths be 1 unit. a = 100-60-20 a = 100° (b/sin20) = (1/sin100) b = 0.3472963553 units. c = 360-120+100 c = 360+220 c = 140° It implies; d² = 1+0.3472963553²-2*0.3472963553cos140 d = 1.2855752193 units. Therefore, the required angle x, is; (1.2855752193/sin140) = (0.3472963553/sinx) x = 10°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2385722

Calculating area red. a = 4+6 = 10 units. a is the side length of the two congruent regular composite plane shape. b = ½(10) b = 5 units. c = 5-1 c = 4 units. tans = 4/6 d = atan(2/3)° e² = 6²+4² e = √(36+16) e = √(52) e = 2√(13) units. f = 90-d f = 90-atan(2/3) f = atan(3/2)° tanf = g/4 tan(atan(3/2)) = g/4 3/2 = g/4 g = 6 units. It implies, h = e h = 2√(13) units. i = 10-6-1 i = 3 units. Therefore, area red is; Area trapezoid with parallel sides 3 units and 5 units respectively,...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2380531

Notice. Radius of the ascribed circle is 7 cm. Calculating a, radius of the big inscribed circle. (7-a)² = 2a² 49-14a+a² = 2a² a²+14a-49 = 0 Resolving the above quadratic equation via completing the square approach. (a+7)² = 49+(7²) (a+7)² = 98 a = -7±√(98) a = -7±7√(2) It implies; a ≠ -7-7√(2) a = -7+7√(2) a = 2.8994949366 cm. b = 2(7)-a b = 14-2.8994949366 b = 11.1005050634 cm. b is the diameter of the orange circle. c = ½(b) c = ½(11.1005050634) c = 5.5502525317 cm. c is the radius...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2377561

Let the radius of the small yellow inscribed circle be a. Let the radius of the big green inscribed circle be b. Notice! PQ = a+b Calculating radius of the ascribed semi circle. c² = 12²+16² c = √(144+256) c = √(400) c = 20 units. c is the diameter of the ascribed semi circle. d = ½(c) d = ½(20) d = 10 units. d is the radius of the ascribed semi circle. tane = 16/12 e = atan(4/3)° e is angle BAC. sin(atan(4/3)) = f/12 4/5 =f/12 f =48/5 f = 9.6 units. cos(atan(4/3)) = g/12 3/5 = g/12 g =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2373947

Calculating the interior angles. It the interior angles be a, b and c identifying them clockwise. Notice. The plane shape consist of two composite regular nonagon. It implies; d = ⅑(180(9-2)) d = 140° d is a single interior angle of the composite regular nonagon. Calculating a. a = 180-40 a = 40° Let the side length of the composite regular nonagon be 1 unit. e² = 1²+1²-2*1*1cos100 e = 1.5320888862 units. f² = 1.5320888862²+1²-2*1.5320888862*1cos100 f = 1.969615506 units. Calculating...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2373159

a+a+b = 20 2a+b = 20 --- (1). 20² = c²+10² c² = 400-100 c = = √(300) c = 10√(3) units. It implies; a+b = c a+b = 10√(3) --- (2). Solve (1) and (2) simultaneously, Calculating a and b. dSubtracting (2) from (1). 2a+b = 20 -- (1) a+b = 10√(3) -- (2) a = 20-10√(3) a = 2.6794919243 units. It implies; From (2). a+b = 10√(3) b = 10√(3)-a b = 10√(3)-2.6794919243 b = 14.6410161514 units. It implies; Area Blue is; Area square with diagonal 14.6410161514 uni...

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