Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2805007

tana = 5/10 a = atan(½)° b = 2a b = 2atan(½)° c = b-45 c = 8.1301023542° d² =10²+10²-2*10*10cos8.1301023542 d = 1.4177804018 units. e = ½(180-8.1301023542) e = 85.9349488229° f² = 2*5² f = 5√(2) units. g = 10-f g = 2.9289321881 units. tanh = 10/5 h = atan(2)° j = 2h j = 2atan(2)° k = j-90 k = 36.8698976458° It implies, area red is; Area triangle with height 1.4177804018 units and base 2.9289321881sin85.9349488229 units + Area sector with radius 10 un...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2804990

Sir Mike Ambrose is the author of the question. Let length AB be 3 units. Therefore; AC = 20 units. It implies; Area large square is; 20² = 40 square units. Area orange is; Area triangle with two side 17 units and 32.0319709782 units, and angle (180-atan(10/7))° + Area triangle with two side 20 units and 44.0408623507 units, and angle 53.4269678346° - Area square with side 20 units + Area triangle with height 3 units and base 4.2857142857 units. = (0.5*32.0319709782...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2804295

Let EB be a. Let AE be b. It implies; a/b = √(2) --- (1). a+b = 1 --- (2). From (1). a = √(2)b --- (3). Substituting (3) in (2) to get b. √(2)b+b = 1 b(√(2)+1) = 1 b = (√(2)-1)/(2-1) b = √(2)-1 b = 0.4142135624 units. b is AE. At (2). a+b = 1 And b = 0.4142135624 units. a = 1-0.4142135624 a = 0.5857864376 units. a is EB. tanc = 1/0.4142135624 c = 67.5° c is angle AED. d =½(c) d = 33.75° Let x be the small inscribed circle radius. e = (0.4...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2803507

Sir Mike Ambrose is the author of the question. Let the side of the square be 2 units. Therefore; Area square is; 2² = 4 square units. Area S is; Area triangle with two side 0.98921899145 units and 1.02554900641 units, and angle 14.318078177° - Area sector with radius 2 units and angle 28.6361563539° + Area triangle with two side 2 units each, and angle 28.6361563539° = (0.5*0.98921899145*1.02554900641sin14.318078177) - (28.6361563539*2²π÷360) + (0.5*2²sin28.6361563539)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2803491

Let the square side be 2 units. tan30 = a/2 a = ⅓(2√(3)) unit. sin60 = 2/b b = ⅓(4√(3)) units. c = 2-a c = ⅓(6-2√(3)) units. It implies; Area Shaded is; 2(½*(⅓(6-2√(3)))²sin60) + 2*⅓(6-2√(3)) = ⅓(4√(3)) square units. Area of the composite shape is; Area Shaded + 2(½*(⅓(4√(3)))²sin60) = ⅓(4√(3)) + ⅓(8√(3)) = 4√(3) square units. Therefore; Fraction Shaded is; ⅓(4√(3)) ÷ 4√(3) = ⅓

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2802130

Sir Mike Ambrose is the author of the question. Let the side of the inscribed square be 2 units. Therefore; Area S is; Area triangle with two side 1.03528 units and 0.53589838486 units, and angle 45° = 0.5*1.03528*0.53589838486sin45 = 0.1961531464 space units. Area R is; Area triangle with two side (2√(2)-2) units and 0.65961372961 units, and angle 72.367664609° = 0.5*(2√(2)-2)*0.65961372961sin72.367664609 = 0.26038499708 space units. Area T is; Area sector with radi...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2802122

cos15 = 12/a a = 12.42331416492 cm. Where a is the side length of the inscribed regular triangle. Radius, r of the inscribed circle is; tan30 = r/(0.5* 12.42331416492) r = 3.58630188867 cm. Area Circle is; π(3.58630188867)² = 40.40578629489 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2800432

a = 90+45 a = 135° b² = 6²+(4√(2))²-2*6*4√(2)cos135 b = 10.7703296143 units. b = 2√(29) units. b is BC. (10.7703296143/sin135) = (6/sinc) c = 23.1985905136° c is angle ACD. tan23.1985905136 = d/(4√(2)) d = 2.4243661069 units. Therefore, area triangle ABD is; 0.5*6*2.4243661069sin45 = 5.1428571428 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2800406

Let the diameter of the blue semi circle be a. Calculating a. It implies; (½*a*6)+(½*10*16) = 140 6a+160 = 2*140 6a = 280-160 6a = 120 a = 20 cm. Again, a is the diameter of the blue semi circle. b = ½(a) b = 10 cm. b is is the radius of the blue semi circle. Area Blue Semi Circle is; ½*10*10π = 50π cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2799000

Let x be the side of the small inscribed square. Calculating x. Notice; The side length of the ascribed square is 14 cm. A small triangle with height ½ cm on the same vertical length with AC and extends from C, and hypotenuse 0.69047619048 cm is attached to the quadrilateral inscribed square ACDE, that have the small square inscribed to make the small square inscribed a right-angled triangle with adjacent sides 10.5 cm and 10 cm, and hypotenuse 14.5 cm. The interior angles of...

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