Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2437851

Let the side length of the regular pentagon be 1 unit. a = ⅕(180(5-2)) a = 108° a is the single interior angle of the regular pentagon. b² = 2-2cos108 b = 1.6180339887 units. c = ½(180-108) c = 36° d = 108-36 d = 72° e = 180-72-30 e= 78° (1.6180339887/sin78) = (f/sin30) f = 0.8270909153 units. g = 180-78 g = 102° Therefore, the required angle alpha is; Let it be h. (1/sin102) = (0.8270909153/sinh) h = 54°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2437741

Notice! Radius of the ascribed semi circle is 6 cm. It implies; Radius, a of the inscribed bigger circle is; a = ½(6) a = 3 cm. Calculating b, radius of the smaller inscribed circle. c = (6-b) cm. c² = b²+e² (6-b)² = b²+e² e² = 36-12b e = √(36-12b) cm. f = (3-b) cm. g = (3+b) cm. Therefore; g² = f²+e² (3+b)² = (3-b)²+36-12b 9+6b+b² = 9-6b+b²+36-12b 12b = 36-12b 24b = 36 b = (36/24) b = (3/2) cm. Again, b is the radius of the smaller inscribed circle. Painted Area is; Area semi circle...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2437685

Calculating area of the inscribed blue square. Let the inscribed blue square side be x. a² = 2x² a = √(2)x units. a is the diagonal of the inscribed blue square. x² = 2b² b² = x²/2 b = ½√(2)x units. It implies; 1² = a²+b² 1 = 2x²+(x²/2) 1 = ½(5x²) 2 = 5x² x² = (2/5) square units. x² = 0.4 square units. Where x² is the area of the inscribed blue square.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2436610

Notice! AQ = 6 cm. a² = 6²+3²-2*6*3cos108 108° is the single interior angle of the regular pentagon. a = 7.4916361229 cm. a is AP. cosb = 6/7.4916361229 b = acos(6/7.4916361229)° b is angle PAQ. Therefore, area triangle PAQ (area blue) is; 0.5*6*7.4916361229sin(acos(6/7.4916361229)) = 13.4581390311 cm² ≈ 13.46 cm² to 2 decimal places.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2436205

(4+5+3)² = 2a² a² = 72 a = 6√(2) cm. a is the congruent adjacent sides of the right-angled triangle. b² = √(72)²+3²-2*√(72)*3cos45 b² = 81-36 b = √(45) b = 3√(5) cm. (3√(5)/sin45) = (3/sinc) c = 18.4349488229° d² = √(72)²+4²-2*√(72)*4cos45 d² = 88-48 d = √(40) d = 2√(10) cm. (2√(10)/sin45) = (4/sine) e = 26.5650511771° f = 90-c-e f = 90-18.4349488229-26.5650511771 f = 45° 2g² = 5² g² = 25/2 g = √(25/2) g = ½(5√(2)) cm. g = 3.5355339059 cm. It implies; Shaded Area is; Area quarter circl...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2435756

Let the base of the ascribed triangle be 1 unit. a = 180-20-20-10-10 a = 120° (1/sin120) = (b/sin20) b = 0.3949308436 units. c = 180-120-20 c = 40° (0.3949308436/sin40) = (d/sin120) d = 0.5320888862 units. e = 180-40-60 e = 80° (0.3949308436/sin80) = (f/sin60) f = 0.3472963553 units. g² = 0.3472963553²+0.5320888862²-2*0.3472963553*0.5320888862cos20 g = 0.2375646984 units. (0.2375646984/sin20) = (0.5320888862/sinh) h = 50° j = 180-h j = 180-50 j = 130° Therefore the required angle x,...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2435595

a = 180-20-50-40 a = 70° It implies the triangle is isosceles. Let the base of the isosceles triangle be 1 unit. 1² = 2b²-2b²cos40 1 = 0.4679111138b² b = √(1/0.4679111138) b = 1.4619022 units. b is the two equal lengths of the isosceles triangle. c = 180-40-20 c = 120° (1.4619022/sin120) = (d/sin20) d = 0.5773502692 units. (1.4619022/sin120) = (e/sin40) e = 1.0850635751 units. f = e-d f = 1.0850635751-0.5773502692 f = 0.5077133059 units. g = b-d g = 1.461902...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2435543

Notice! R is the radius of the ascribed semi circle. Calculating R. a² = 2*√(2)² a² = 4 a = 2 units. a is the diagonal of the inscribed red and white square. √(2)² = 2b² b² = 2/2 b = 1 unit. It implies; R² = 2²+1² R² = 5 R = √(5) units. Again, R is the radius of the ascribed semi circle. tanc = 2/(√(5)+1) c = atan(2/(√(5)+1))° c = 31.7174744115° tan(atan(2/(√(5)+1))) = d/√(5) d = 1.3819660112 units. e = d-b e = 1.3819660112-1 e = 0.3819660112 units. Therefore, white area is; Area tria...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2435037

a² = 2*5² a = √(50) a = 5√(2) cm. tanb = 10/5 b = atan(2)° c = b-45 c = (atan(2)-45)° c = 18.4349488229° d = 90-b d = (90-atan(2))° d = 26.5650511771° e = 180-2d e = 180-2*26.5650511771 e = 126.8698976458° f² = 5²+5²-2*5*5cos126.8698976458 f = 8.94427191 cm. Therefore, Blue Area is; Are triangle with height 5√(2) cm and base 8.94427191sin18.4349488229 cm. = 0.5*5√(2)*8.94427191sin18.4349488229 = 10 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2434402

Calculating x. Notice! AD = 100 cm. Therefore; Area triangle ABD + Area triangle ACD = Area triangle ABC 0.5*2x*100sin60+0.5*x*100sin60 = 0.5*2x*xsin120 100x+50x = x² x = 100+50 x = 150 cm. It implies; AB = 2*x AB = 2*150 AB = 300 cm. AC = x AC = 150 cm. Therefore, calculating BC. Let it be y. y² = 300²+150²-2*300*150cos120 y² = 157500 y = √(157500) y = 150√(7) cm. y = 396.8626966597 cm.

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