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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1969883

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Let the radius of the three congruent inscribed circles be r. Calculating r. (1+r)² = (1-r)²+(1-3r)² 1+2r+r² = 1-2r+r²+1-6r+9r² 1+2r+r² = 10r²-8r+2 9r²-10r+1 = 0 Therefore, solving the quadratic equation. r ≠ 1 unit. r = ⅑ units.

We are smart and clever!

https://onlineedumath.com/update/we-are-smart-and-clever/1969696

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Our certified, experienced, resilient, productive and committed Math Educators will be glad to be at your service. We make teaching and learning Mathematics fun for learners, we expose the simplicity and beauty of Mathematics to our learners, and mentor them to appreciate and love Mathematics. Communicate us to mentor/educate you Mathematics better online at affordable tuition. We are smart and clever.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1969678

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(AC)² = 2a² AC = √(2)a units. b² = 3²+4² b = 5 units. c = √(16-a²) units. d = a-c d = (a-√(16-a²)) units. Therefore, considering similar triangle ratios. a/4 = d/3. And d is (a-√(16-a²)). Therefore; a/4 = (a-√(16-a²))/3 Cross Multiply. 3a = 4a-4√(16-a²) 4√(16-a²) = a 16(16-a²) = a² 256-16a² = a² 256 = 17a² a² = 256/17 a = 16/√(17) a = (16√(17))/17 units. a = 3.88057000058 units Where a is a side length of the square ABCD.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1968584

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4 = 18-18cosa a = acos(7/9) a = 38.94244126898° a is angle BOC. b = 90-38.94244126898 b = 51.05755873102°  b is angle COD. d = 0.5(180-38.94244126898) d = 70.52877936551° d is angle OCB = angle OBC. e = 90-70.52877936551 e = 19.47122063449° e is angle OCD. f = 19.47122063449+51.05755873102 f = 70.52877936551° f is angle ADC. g = 0.5*38.94244126898+51.05755873102 g = 70.52877936551° sin70.52877936551 = 1/h h = 1.06066017178 units. h is OD. tan(0.5*70.528779365...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1967906

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Calculating r, radius of the circle. (r/tan40)+(r/tan30) = 10 r = 3.42020143326 cm. (a/sin60) = (10/sin40) a = 13.47296355334 cm. tan40 = 3.42020143326/b b = 4.07603734548  (4.07603734548/sin50) = (c/sin80) c = 5.24005260459 cm. d = 13.47296355334-4.07603734548 d = 9.39692620786 cm. e² = 9.39692620786²+5.24005260459²-2*9.39692620786*5.24005260459cos130 e = 13.38142726359 cm. (13.38142726359/sin130) = (5.24005260459/sinf) f = 17.45618925878° g = 100-17.45618925878  g = 82.54381074122°...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1967879

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Let the three congruent lengths be 1 unit. a² = 2-2cos96 a = 1.48628965095 units. (1/sin12) = (1.48628965095/sinb) b = 18° It implies; Alpha = b = 18°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1967262

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AC = √(4²+4²) AC = √(32) AC = 4√(2) Units. (4√(2))² = (4+r)²+r² 32 = 16+8r+r²+r² 2r²+8r-16 = 0 r²+4r-8 = 0 (r+2)² = 12 r = -2±2√(3) r ≠ -2-2√(3) units. r = (2√(3)-2) units. a = r+4 a = (2+2√(3)) units. tanb = (2√(3)-2)/(2+2√(3)) b = 15° c = 45-b c = 30° Therefore, Red Shaded Area is; 0.5(AC)²sin30 0.5*(4√(2))²sin30 = 0.25*32 = ¼(32) = 8 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1966974

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r, radius of the inscribed circle is; 2(r/tan22.5) = 12√(2) r = 6√(12)tan22.5 r = 3.51471862576 cm. a = 12-3.51471862576 a = 8.48528137424 cm. b = (8.48528137424-c) units. c is the radius of the inscribed half circle. c² = (8.48528137424-c)²+3.51471862576²) c² = 72-16.97056274848c+c²+12.35324701826  16.97056274848c = 84.35324701826  c = 4.97056274848 cm. d = 12-4.97056274848-3.51471862576 d= 3.51471862576 cm. Therefore, AB is; √(3.51471862576²+8.48528137424²) AB = 9.18440237676 cm.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1966632

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a² = 5²+2² a = √(29) cm. b² = 3²+2² b = √(13) cm c = atan(2/5)° d = atan(1/3)° e = atan(2/3)° f = 90-c-d f = 90-atan(2/5)-atan(1/3) f = 49.76364169073° g = atan(3)° h = g-e h = atan(3)-atan(2/3) h = 37.8749836511° cos49.76364169073 = j1/√(29) j1= 3.47850542618 cm sin49.76364169073 = j2/√(29) j2 = 4.11096095822 cm. cos37.8749836511 = k1/√(13) k1 = 2.84604989415 cm. sin37.8749836511 = k²/√(13) k2 = 2.21359436212 cm. Therefore, Blue Area is; 0.5*j1*j2 + 0.5*k1*k2 0.5*3.47850542618*4...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1966150

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Let the height of the quadrilateral with parallel sides 7 cm and 9 cm be a. b =√(a²+4) cm. c = ½√(a²+4) cm. d = √((½√(a²+4))²-1) d = √(¼(a²+4)-1) d = √(¼(a²)) d = ½(a) cm. Considering similar triangle ratios. ½(a) = 8       1 = e, cross multiply. e = (16/a) cm. f = d-e f = ½(a)-(16/a) f = (a²-32)/2a cm. Calculating a. ½(a)+8(½(a)+(a²-32)/2a) = 2*26 ½(a)+4a+(4a²-128)/a = 52 ½(a²)+4a²+4a²-128 = 52a 17a²-256 = 104a 17a²-104a-256 = 0 Therefore; a = 8 cm. b =...
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