Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2563037

Notice; Diameter of the inscribed semi circle is ½ unit. Therefore radius of the inscribed semi circle is ¼ unit. Let the radius of the inscribed small circle be r. Calculating r. (1-√((¼+r)²-(¼-r)²))²+(1-r)²=(1+r)² (1-√(r))²+1-2r+r²=1+2r+r² 1-2√(r)+r=4r 1-3r=2√(r) (1-3r)²=4r 1-6r+9r²=4r Therefore; 9r²-10r+1=0 9r(r-1)-1(r-1)=0 (9r-1)(r-1)=0 Therefore; r ≠ 1 r = 1/9 unit. Radius of the small inscribed circle r is; 1/9 unit.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2562836

Radius of the inscribed quarter circle is 36 cm. Radius of the inscribed semi circle is; ½(½(36))=9 cm. Radius of the small inscribed circle is; ⅓(⅓(36))=4 cm. Therefore the side lengths of triangle blue will be; (36+9)=45 cm. (36+4)=40 cm. (9+4)=13 cm. Therefore area triangle blue is; 0.5x45x40sin(cos–¹((45²+40²-13²)÷(2x45x40))) = 252 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2562365

Notice! The square side length is; 4+4+4 = 12 units. Calculating Area Red. Area quarter circle with radius 12 units - Area triangle with height and base 12 units respectively - Area quarter circle with radius 8 units + Area triangle with height and base 8 units respectively + Area quarter circle with radius 8 units - Area triangle with height and base 8 units respectively - Area quarter circle with radius 4 units + Area triangle with height and base 4 units respectively. = ¼(π*12²...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2562190

Let the diameter of the big inscribed half circle be a units. It's radius is ½(a) units. Let the diameter of the small inscribed half circle be b units. It's radius is ½(b) units. It implies; a+b = 2 --- (1). c = ½-½(b) c = ½(1-b) units. d = (a-½) units. Equation c and d. c = d ½(1-b) = a-½ a = 1+½(b) --- (2). Substituting (2) in (1). a+b = 2 And a = 1+½(b) 1+½(b)+b = 2 ½(b)+b = 1 1.5b = 1 ½(3b) = 1 3b = 2 b = ⅔ units. Again, b is the diameter of the...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2562085

Equation of the curve is; y = ½(x^(3/2)) It implies; PR = √(4²+6²) PR = 2√(13) units QR = (2√(13))/(cos(90-atan(⅔))) = 13 units. Therefore; Shaded Area is; (Area under the curve at x = 4 and x = 10) - Area trapezium with parallel sides 4 units and 13 units, and height 6 units. = 20√(10) - (32/5) - ½*6(4+13) = 20√(10) - (32/5) - 51 = (100√(10)-287)/5 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2561729

sin60 = 5/a a = 5/(√(3)/2) a = ⅓(10√(3)) units. a is the side length of the inscribed regular triangle. b = ½(a) b = ⅓(5√(3)) units. sin60 = c/⅓(5√(3)) √(3)/2 = 3c/(5√(3)) 6c = 15 c = 2.5 units. Let the inscribed square side length be d. e = (2.5+d) units. It implies; 5² = (2.5+d)²+(0.5d)² 25 = 6.25+5d+d²+0.25d² ¼(5d²)+5d-18.75 = 0 ¼(d²)+d-3.75 = 0 d²+4d-15 = 0 Resolving the above quadratic equation via completing the square approach to get d, the side length o...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2561674

Notice! The ascribed triangle is equilateral. Calculating the required angle x. a² = 1+3²-2*3cos60 a = √(7) units. (√(7)/sin60) = (1/sinb) b = 19.1066053509° c = 60-b c = 40.8933946491° d = 180-b-c d = 120° (3/sin120) = (e/sin40.8933946491) e = 2.2677868381 units. f² = 2.2677868381²+3²-2*2.2677868381*3cos40.8933946491 f = 1.9639610121 units. Therefore, the required angle x is; (3/sinx) = (1.9639610121/sin40.8933946491) x = 90°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2561633

Let CE be 3 units. Therefore: Area Orange is; Area triangle with height √(19) units and base 6sin(60-acos(4/√(19))) units - Area triangle with height (12√(19)/19) units and base (36√(57)/133) units. = ½*√(19)*6sin(60-acos(4/√(19))) - ½*(12√(19)/19)*(36√(57)/133) = ½(9√(3)) - (216√(3))/133 = (765√(3)/266) square units. Area ACEB is; Area trapezium with parallel sides 3 units and 6 units, and height ½(3√(3)) units. = ½*½(3√(3))*(3+6) = ¼*9*3√(3) = ¼(27√(3)) square unit...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2561621

Let x be the single side length of the square. Therefore; x²=28-4√(x²-16) via similar triangle. 16(x²-16)=(28-x²)² 16x²-256=784-56x²+x⁴ x⁴-72x²+1040=0 Let x²(blue area) = P Therefore; P²-72P+1040=0 P²-20P-52P+1040=0 P(P-20)-52(P-20)=0 (P-20)(P-52)=0 Therefore; P ≠ 52 P = 20 And P = x² (blue area) Therefore blue area (area inscribed square) is; x² = 20 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2560976

Let a be the radius of the half circle. b² = (3√(3))²+a² b = √(27+a²) units. cosc = 3√(3)/√(27+a²) --- (1). a² = 3²+(3√(3))²-2*3*3√(3)cosc a² = 36+18√(3)cosc --- (2). Substituting (1) in (2). a² = 36+18√(3)(3√(3)/√(27+a²)) a² = 36+((162/√(27+a²)) a²-36 = (162/√(27+a²)) Therefore; a = 3√(6) units. Again, a is the radius of the semi circle. It implies, area semi circle is; ½*πa² = ½*π(3√(6))² = ½(54)π = 27π square units.

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