Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2796126

Calculating area of the inscribed triangle. a = 2r a = 2√(5) units. a is the side length of the ascribed square. tanb = √(5)/(2√(5)) b = atan(½)° c = 2b c = 2atan(½)° d = 90-2atan(½) d = 90-atan(½)-atan(½)° d = (atan(2)-atan(½))° Therefore, area inscribed triangle is; 0.5*2√(5)*2√(5)sin(atan(2)-atan(½)) = 10*0.6 = 6 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2795785

Sir Mike Ambrose is the author of the question. Let the side of the square be 2 units. Area square is; 2² = 4 square units. Area orange is; Area triangle with two side 0.70710678119 units and 1.2928932188 units, and angle (90+asin(⅓))° + Area triangle with two side 1.66764761969 units and 1.20710678117 units, and angle 66.436529825° = 0.5*0.70710678119*1.2928932188sin(90+asn(⅓)) + 0.5*1.66764761969*1.20710678117sin(66.436529825) = 0.43096440627 + 0.9225889843 = 1.3535533905...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2795727

Sir Mike Ambrose is the author of the question. Area red is; Area triangle with two side 4 cm and 8 cm, and angle 110°. = 0.5*4*8sin110 = 15.0350819326 cm² Area blue is; Area triangle with height 4.47952822702 cm and base (4.47952822702tan33.67018696) cm = 0.5*4.47952822702*4.47952822702tan33.67018696 = 6.6836970062 cm² Therefore; Area Blue + Area Red in cm² to 2 d. p. is; 6.6836970062 + 15.0350819326 = 21.7187789388 cm² ≈ 21.72 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2795137

a² = 2²+2²-2*2*2cos120 a = √(8+½(8)) a = √(12) a = 2√(3) units. a is the side length of each of the 4 inscribed small regular hexagon. sin30 = b/a b = asin30 b = ½*2√(3) b = √(3) units. c = 4b+3a c = 4(√(3))+3(2√(3)) c = 4√(3)+6√(3) c = 10√(3) units. c is the height of the big ascribed regular hexagon. d = ½(c) d = 5√(3) units. sin60 = d/e ½√(3)e = 5√(3) e = 10 units. e is the side length of the big ascribed regular hexagon. f² = 2(2√(3))²-2(2√(3))²cos120 f = 6...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2794675

Let a be the congruent side lengths each. b = ½(2) b = 1 unit. c²+1² = a² c = √(a²-1) units. d = ½(180-45) d = ½(135) d = 67.5° tan67.5 = e/1 e = 2.4142135624 units. e = (1+√(2)) units. f = 1+e f = (2+√(2)) units. f is the radius of the ascribed quarter circle. g = f-2 g = √(2) units. Therefore, blue area is; ½*(fg) = ½*(2+√(2))*√(2) = ½(2+2√(2)) = (1+√(2)) square units. = 2.4142135624 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2794161

a = ¼(60) a= 15 units. b² = 40²+15² b = 5√(73) units. b = 42.7200187266 units. c = ½(b) c = 2.5√(73) units. c = 21.3600093633 units. c is each of the congruent lengths. d = 60-15 d = 45 units. tane = 15/40 e = atan(3/8)° f = 90-e f = atan(8/3)° Therefore, the required length, ? is; Let it be g. g² = 21.3600093633²+45²-2*21.3600093633*45cos(atan(8/3)) g = 42.5 units. Again, g is the required length.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2793964

a = 180-54-30-18 a = 180-102 a = 78° a is angle ABC. b = 78+54 b = 132° b is angle BAD. c = 180-b-18 c = 180-150 c = 30° c is angle ADB. Let AD = 1 unit. (1/sin18) = (d/sin30) d = 1.6180339887 units. d is AB. (1.6180339887/sin54) = (e/sin(18+30)) e = 1.486289651 units. e is AC. f² = 1²+1.486289651²-2*1*1.486289651cos54 f = 1.2090569266 units. f is CD. Therefore, the required angle, ? is; Let it be g. (1.2090569266/sin54) = (1/sing) g = 42° Again, g i...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2793942

Calculating h. tan60 = h/a a = ⅓(√(3)h) units. tan30 = h/b b = √(3)h units. It implies; a + b = 100 ⅓(√(3)h)+√(3)h = 100 √(3)h+3√(3)h = 300 4√(3)h = 300 √(3)h = 75 h = ⅓(75√(3)) units. h = 25√(3) units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2793317

Let a be the radius of the half circle. a-x = 4 x = a-4 --- (1). b = (a-2x) units. It implies; a² = 4²+(a-2x)² a² = 16+a²-4ax+4x² 4ax = 16+4x² --- (2). Substituting (1) in (2) to get a, radius of the half circle. 4a(a-4) = 16+4(a-4)² 4a²-16a = 16+4(a²-8a+16) 4a²-16a = 16+4a²-32a+64 16a = 80 a = 5 units. Again, a is the radius of the half circle. Recall. a-x = 4 And a = 5 units. 5-x = 4 x = 1 unit. c = 2x c = 2 units. d = 8-c d = 6 units. It implie...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2792535

Radius of the quarter circle is (2+3) = 5 units. Let a be the side length of the inscribed square. b²+3² = a² b = √(a²-9) units. c = 3+b c = (3+√(a²-9)) units. Calculating a. (3+√(a²-9))²+3² = 5² 9+6√(a²-9)+a²-9+9 = 25 9+6√(a²-9)+a² = 25 6√(a²-9) = 16-a² 36(a²-9) = (16-a²)² 36a²-324 = 256-32a²+a⁴ a⁴-68a²+580 = 0 It implies; a = √(10) units a = 3.1622776602 units. Again, a is the side length of the inscribed square. Recall. b = √(a²-9) units And a = √(10)...

You Searched for "post"

2242 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

Useful Tags

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 "post"

2242 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?