Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3155706

Let the three equal lengths be x. a = x+x+x+8+8 a = (16+3x) units. a is the diameter of the ascribed half circle. b = ½(a) b = ½(16+3x) units. b is the radius of the ascribed half circle. c² = x²+8² c = √(64+x²) units. It implies; Calculating x. √(64+x²) ~ (16+3x) x ~ √(64+x²) Cross Multiply. 16x+3x² = 64+x² 2x²+16x-64 = 0 x²+8x-32 = 0 (x+4)² = 32+(4²) (x+4)² = 48 x = -4±√(48) x = -4±4√(3) It implies; x ≠ -4-4√(3) units. x =4(√(3)-1) units. x = 2.9282032302...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3155704

Calculating r2, radius of the smaller inscribed circle. r2 will be; Let it be x. Therefore; (4+x)²=(8-x)²+(4-x)² 16+8x+x²=64-16x+x²+16-8x+x² x²-32x+64=0 (x-16)²=-64+256 x = 16±8√(3) x ≠ 16+8√(3) x = 16-8√(3) Therefore; x = r2 = 8(2-√(3)) units. Again, r2 is the radius of the smaller inscribed circle.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3155027

Kindly move the question left/right one time to review the analysis of the question (plane shape). Thank you nicely. Calculating R, radius of the quarter circle. a = 3+4 a = 7 units. b² = 2(4)² b = 4√(2) units. c = R-b c = (R-4√(2)) units. d = 180-45 d = 135° R is; Observing cosine rule. R² = 7²+(R-4√(2))²-2*7(R-4√(2))cos135 R² = 49+R²-8√(2)R+32+7√(2)(R-4√(2)) 8√(2)R = 81+7√(2)R-56 √(2)R = 25 R = ½(25√(2)) units. R = 17.6776695297 units. Or e² = 3²+4² e...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3154046

Calculating shaded blue area. The radius, r if the small circle will be; (5+r)²=2(5-r)² 25+10r+r²=50-20r+2r² Therefore r²-30r+25=0 (r-15)²=-25+225 r = 15±√(200) r = 15±10√(2) It implies; r ≠ 15+10√(2) r = 15-10√(2) units. Therefore area blue will be; ¼(100-25π) - (15-10√(2))²π = 3.05304958693 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3154043

Calculating shaded inscribed area. a = 6+6 a = 12 units. a is the side length of the ascribed square. b = a-4 b = 8 units. tanc = 8/6 c = atan(4/3)° d = 180-2c d = 2atan(3/4)° e = ½(d) e = atan(3/4)° tan(atan(3/4)) = f/6 3/4 = f/6 f = ½(9) units. g = a-f g = 12-½(9) g = ½(15) units. Therefore, shaded area is; ½*g*fsin(2atan(3/4)) = ½*½(15)*½(9)sin(2atan(3/4)) = ⅛*135*(24/25) = ⅕(81) square units. = 16.2 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3152716

Calculating area blue. a²+1² = 2² a = √(3) units. a is the diameter of the inscribed white circle. b = ½(a) b = ½√(3) units. b is the radius of the inscribed white circle. cos30 = 1/c ½√(3) = 1/c √(3)c = 2 c = ⅔√(3) units. c is the radius of the bigger ascribed circle. Or d = 180-30-30 d = 120° Therefore; 2² = 2c²-2c²cos120 4 = 3c² c² = 4/3 c = 2/√(3) c = ⅔√(3) units. Area Blue is; π(c)²-π(a)² = π(⅔√(3))²-π(½√(3))² = π((4/3)-(3/4)) = 7π/12 squa...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3152714

a² = 9²+12² a = √(81+144) a = √(225) a = 15 cm. It implies; Calculating r. (9*3r)+(12*r)+(15*r) = (9*12) 27r+12r+15r = 9*12 54r = 9*12 6r = 12 r = 2 cm.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3151652

Let XM be x. Let AZ be y. b = (8-x) units. b is MY. c = (4-y) units. c is AY. Therefore; ½(4x) = ½(8y) 2x = 4y x = 2y --- (1). Recall. b = 8-x And x = 2y b = (8-2y) units. It implies; Calculating y. 4 ~ (8-2y) 2y ~ (4-y) Cross Multiply. 8y-2y² = 8-2y 4y-y² = 4-y y²-5y+4 = 0 y²-4y-y+4 = 0 y(y-4)-1(y-4) = 0 (y-1)(y-4) = 0 Therefore; y ≠ 4 units. y = 1 unit. Again, y is AZ. Recall. At (1). x = 2y And y = 1 unit. x = 2*1 x = 2 units....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3151650

Let the radius of the quarter circle be 1 unit. 2a² = 1 a² = ½ a = ½√(2) units. a is the side length of the inscribed square. b = 1-a b = 1-½√(2) b = ½(2-√(2)) units. b is the side length of the green square G. Area green square is; b² = (½(2-√(2)))² = ½(3-2√(2)) square units. Calculating Area R, Red Area. (½√(2)+c)²+c² = 1² ½+√(2)c+2c² = 1 4c²+2√(2)c-1 = 0 (c+¼√(2))² = ¼ (c+¼√(2))² = ¼+(¼√(2))² c = -¼√(2)±√(6/16) c = -¼√(2)±¼√(6) Therefore; c ≠ -¼√(2)-¼√(6) c = -¼√(2)+¼√(6) c = ¼(√(6...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3150389

a = ½(10) a = 5 units. a is the radius of the ascribed half circle. cosb = 4/5 b = 36.8698976458° b is angle alpha. c²+8² = 10² c = √(100-64) c = 6 units. sin(2acos(4/5)) = d/10 d = 9.6 units. cosf = (0.5*9.6)/6 cosf = 4.8/6 f = acos(0.8)° tan(0.5acos(0.8)) = g/(0.5*9.6) g = 4.8tan(0.5acos(0.8)) g = 1.6 units. g is the radius of the inscribed circle. Area inscribed circle is; πg² = π(1.6)² = 2.56π square units.

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