Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2452469

Notice! Radius of the circle is 4 cm. a² = 4²+8² a² = 16+64 a = √(80) cm. a = 4√(5) cm. a is length AC. tanb = 2/4 b = atan(½)° c = 180-2b c = (180-2atan(½))° It implies; d² = 4²+4²-2*4*4cos(180-2atan(½)) d = 7.155417528 cm. d is length CB. Therefore, the required length, AB is; AC-BC = 4√(5)-7.155417528 = 1.788854382 cm. = ⅕(4√(5)) cm.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2452115

Calculating x. (2+x)² = (2x)²+x² 4+4x+x² = 5x² 4x²-4x-4 = 0 x²-x-1 = 0 (x-½)² = 1+¼ x = ½±½√(5) It implies; x ≠ ½(1-√(5)) x = ½(1+√(5)) units. x = 1.6180339887 units. Shaded Area is; ¼(x²)π = ¼(½(1+√(5)))²π = π(6+2√(5))/16 = ⅛(3+√(5))π square units. = 2.0561990865 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2452027

Sir Mike Ambrose is the radius of the question. Area R is; Area triangle with height 4 cm and base (4tan52.5) cm - Area sector with angle 52.5° and radius 4 cm. = ½*4(4tan52.5) - (52.5*4*4π)/360 = 8tan52.5 - ⅓(7π) = ⅓(24tan52.5-7π) = 3.09542012435 cm² ≈ 3.10 cm² to 2 decimal places.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2452006

Let the inscribed bigger quarter circle radius be a. Let the inscribed big quarter circle radius be b. c = 5 cm. c is the radius of the inscribed semi circle. It implies; a+b = 10 --- (1). a+5 =√(10²+5²) a = √(125)-5 a = 5(√(5)-1) cm. Again, a is the inscribed bigger quarter circle radius. And; a+b = 10 b = 10-a b = 10-5(√(5)-1) b = (15-5√(5)) cm. Again, b is the inscribed big quarter circle radius. Area Blue is; = (10*10)-½(5²)π-¼(5√(5)-5)²π-¼(15-5√(5))²π = 100-½(25π)-¼(125-50√(5)+25...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2451524

Let a be the side length of the square and the diameter of the biggest inscribed semi circle. Let b = diameter of the bigger inscribed semi circle. Let c = diameter of the big inscribed semi circle. It implies; (½(a+c))² = (½(a))²+(8+(½(c)))² ¼(a²+2ac+c²) = ¼(a²)+64+8c+¼(c²) ½(ac) = 64+8c ac = 128+16c --- (1). (½(b+c))² = (½(c))²+(6+(½(b)))² ¼(b²+2bc+c²) = ¼(c²)+36+6b+¼(b²) ½(bc) = 36+6b bc = 72+12b --- (2). 8+c = 6+b --- (3). a = 8+c --- (4). Substituting (4) in (1) to get c. (8+c)c...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2450969

a = (√(3)-1) units. b = (1-c) units. Calculating c. Observing similar plane shape (right-angled triangle) side length ratios. (1-c) - (√(3)-1) (√(3)-1) - 1 Cross Multiply. 1-c = (√(3)-1)² c = 1- (√(3)-1)² c = 1-(3-2√(3)+1) c = (2√(3)-3) units. Therefore, x, the required angle is; tanx = √(3)/c x = atan(√(3)/(2√(3)-3)) x = 75°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2450605

Let the square side length be 2 units. It implies; R = 1 unit. a = (2+2x) units. a is diameter of the bigger circle. b = ½(a) b = ½(2+2x) b = (1+x) units. b is the radius of the bigger circle. c = (1+x-2x) c = (1-x) units. d = (1+x-x) d = 1 unit. Calculating x. (1+x)² = (1-x)²+1² 1+2x+x² = 1-2x+x²+1 4x = 1 x = ¼ units. 2x is; = 2*¼ = ½ units. b = (1+x) And x = ¼ units. b = 1+¼ b = 1.25 units. Again, b is the radius of the bigger circle. a = (2+2x) And x = ¼ a = 2+2*¼ a = 2.5 units. a...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2450034

a = (x+17) cm. a is the length of the the ascribed rectangle. b = a-9 b = (x+17)-9 b = (x+8) cm. b is the width of the ascribed rectangle and also, the diameter of the inscribed semi circle and the radius of the inscribed quarter circle. c² = 2(8+x)² c² = 2(64+16x+x²) c = √(128+32x+2x²) cm. d = ½(b) d = ½(x+8) cm. d is the radius of the inscribed semi circle. Therefore, calculating x. Observing the side length ratios of similar plane shapes (right-angled triangle). x - ½...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2449884

a² = 2²+2²-2*2*2cos120 a = √(12) a = 2√(3) units. a is the side length of each of the congruent 4 inscribed regular hexagon. Calculating the side length of the ascribed regular hexagon. sin30 = c/a ½ = c/(2√(3)) c = √(2) units. b = a+2(2c+a) b = 2√(3)+2(2√(3)+2√(3)) b = 2√(3)+2(4√(3)) b = 10√(3) units. Again, b is the side length of the ascribed regular hexagon. (10√(3))² = 2d²-2d²cos120 300 = 3d² d² = 100 d = 10 units. d is the side length of the ascribed regular hexagon. sin60 = e/(2√(3...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2449817

a = 5-4 a = 1 cm. b² = 12²+1² b² = 145 b = √(145) cm. c = 4+5 c = 9 cm. Therefore, length x is; x²+9² = √(145)² x² = 145-81 x² = 64 x = √(64) x = 8 cm.

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