Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2488271

Sir Mike Ambrose is the author of the question. Exact Area R in its simplest from will be; Area semi circle with radius 2 unit - Area right-angled triangle with height 2√(2+√(2)) unit and width 2√(2-√(2)) unit - Area sector with radius 2 unit and angle 45° + Area isosceles triangle with two equal length, 2 unit each and angle 45° - Area quarter circle with radius 2 unit + Area isosceles right-angled triangle with height and width 2 unit respectively - Area sector with radius 2 unit and...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2486193

Let a be the radius of the quarter circle. It implies; a² = a²+4²-2*a*4cosb 8cosb = 16 acosb = 2 cosb = 2/a --- (1). 8² = a²+4²+2*a*4cosb 64 = a²+16+8acosb 48 = a²+8acosb --- (2). Substituting (1) in (2) to get a, radius of the quarter circle. 48 = a²+8a(2/a) 48 = a²+16 a² = 48-16 a² = 32 a = √(32) a = 4√(2) units. Again, a is the radius of the quarter circle. Therefore, area blue (area quarter circle) is; ¼(4√(2))²π = ¼*4*4*2π = 8π square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2479061

Let the square side be 2a cm. It implies; a is the radius of the inscribed semi circle. Let b equal the radius of the big inscribed quarter circle. Therefore; (a+b)² = a²+(2a)² a²+2ab+b² = 5a² 2ab+b² = 4a² --- (1). (2a-2)² = 2²+b² 4a²-8a+4 = 4+b² b² = 4a²-8a --- (2) From (2). b = √(4a²-8a) --- (3). Substituting (2) and (3) in (1) to get a. 2ab+b² = 4a² 2a√(4a²-8a)+4a²-8a = 4a² 2a√(4a²-8a) = 8a 4a²(4a²-8a) = 64a² 4a²-8a = 16 a²-2a-4 = 0 Resolving the abo...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2476573

Area Blue is; Area semi circle with radius √(2) units - Area quarter circle with radius 2 units + Area triangle with height and base 2 units respectively. = ½(√(2)²)π - ¼(2²)π + ½(2²) = π - π + 2 = 2 square units

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2473449

Calculating R, radius of the small inscribed quarter circle. a = 2*18 a = 36 units. a is the radius of the ascribed quarter circle. b = (18+R) units. It implies; R, radius of the small inscribed quarter circle is; 36² = R²+(18+R)² 1296 = R²+324+36R+R² 2R²+36R-972 = 0 R²+18R-486= 0 Resolving the above quadratic equation. R = (9√(7)-9) units. R = 14.8118 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2466950

Notice! A regular triangle (equilateral triangle) with side length 2 units is inscribed the quarter circle due to the apex vertices of the two equilateral triangles on point the circumference of the quarter circle. Therefore, radius of the quarter circle is 2 units. It implies; Area quarter circle is; ¼(2²)π = π square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2459663

r² = x²+(6+8)² --- (1). r² = (10+x)²+6² --- (2). Where r is the radius of the ascribed quarter circle. Equating (1) and (2) to get x. r² = r² x²+(6+8)² = (10+x)²+6² x²+196 = 100+20x+x²+36 196-136 = 20x x = 60/20 x = 3 units. Calculating r, radius of the ascribed quarter circle. At (1). r² = x²+(6+8)² And x = 3 units. r² = 3²+14² r² = 9+196 r = √(205) units.

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/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/2447638

a = (1+x) units. a is the diameter of the semi circle. b = ½(a) b = ½(1+x) units. b is the radius of the semi circle. c = x units. c is the radius of the quarter circle. Notice! A-B = π A = B+π B = A-π It implies; Calculating x. It is; Area semi circle with radius ½(1+x) units - Area B + Area A - ½(area quarter circle with radius x units) = ½(area quarter circle with radius x units) ½(½(1+x))²π-B+A-½*¼(x²)π = ½*¼(x²)π ⅛(1+2x+x²)π-(A-π)+A-⅛(x²)π = ⅛(x²)π ⅛π+¼(x)π+π = ⅛(x...

You Searched for "nft art"

393 result found.

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

OnlineEdumath

Socials

Facebook Page

Quick Links

Reach Us

Safety and Abuse Reporting

Thanks for being awesome!

Are you sure you want to report abuse against this website?

Search

You Searched for "nft art"

393 result found.

Subscribe to our mailing list for latest updates.

Safety and Abuse Reporting

Thanks for being awesome!

Are you sure you want to report abuse against this website?