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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2413790

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Let r be the radius of the inscribed circle. Calculating r. √(5)² = (2r)²+r² 5 = 5r² r = 1 unit. Notice! The square side is 4r. Let a be the square side. a = 4r  And r = 1 unit. a = 4(1) a = 4 units. Therefore, total area of the composite plane shape is; Area square with side 4 units. = a² = 4² = 16 square units.

Mathematics Question and Solution

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sin30 = a/4 Where a is the red length. a = 4sin30 a = 2 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2413264

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Calculating PQ. Let r be the radius of the smaller circle. 7 units is the radius of the bigger circle. sin60 = a/7 √(3)/2 = a/7 a = ½(7√(3)) units. a = 6.0621778265 units. a is the radius, r of the smaller circle. b = 7+6.0621778265 b = 13.0621778265 units. Therefore, PQ is; PQ = b+½(7) PQ = 13.0621778265+½(7) PQ = 13.0621778265+3.5 PQ = 16.5621778265 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2412794

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Let r be the radius of the two equal inscribed circles. tana = 21/28 a = atan(3/4)° b = 90-a b = 90-atan(3/4)° b = atan(4/3)° c² = 28²+21² c = √(28²+21²) c = 35 units. tan(0.5atan(3/4)) = r/d ⅓ = r/d d = 3r units. tan(0.5atan(4/3)) = r/e ½ = r/e e = 2r units. Calculating r. It implies; 2r+d+e = c 2r+3r+2r = 35 7r = 35 r = 5 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2411675

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Observing cosine rule. a² = 12²+4²-2*12*4cos60 a = 10.5830052443 units. Observing cosine rule. b² = 3²+12²-2*3*12cos120 b = 13.7477270849 units. Observing cosine rule. 10.5830052443² = 13.7477270849²+7²-2*7*13.7477270849cosc 192.4681791886cosc = 126 c = acos(126/192.4681791886) c = 49.106605351° d = 2c d = 98.213210702° e = ½(10.5830052443) e = 5.2915026222 units. Therefore, r, radius of the circle is; sin49.106605351 = 5.2915026222/r r = (5.2915026222/sin49.106605351) r = 7 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2410652

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Let c3 = a = 6 units. Let c2 = b = 8 units. Therefore; c1 = c = √(6²+8²) c = √(100) c = 10 units. Calculating Area S3. It is; Area equilateral triangle with side 6 units - 2(area equilateral triangle with side 3 units) - Area sector with radius 3 units and angle 60° = ½*6*6sin60-2(½*3*3sin60)-(60π*3*3/360) = 9√(3)-½(9√(3))-½(3π) = ½(18√(3)-9√(3)-3π) = ½(9√(3)-3π) square units. Calculating Area S2. It is; Area equilateral triangle with side 8 units - 2(area equilateral triangle with side...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2410318

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Calculating area blue. 6 - a b - 2 Cross Multiply. ab = 12 cm² c = ½(ab) c = ½(12) c = 6 cm² Therefore, calculating a. 6+½(2a) = ½(2+b)*a 6+a = ½*2a*ab And ab = 12 cm² 6+a = ½*2a*12 6+a = 12a 6 = 12a-a 11a = 6 a = (6/11) cm. And ab = 12 b = (12/a) And a = (6/11) b = 12÷(6/11) b = 12*(11/6) b = 22 cm. Therefore, blue area is; ½*(b+2)*a = ½(22+2)*(6/11) = ½*24*(6/11) = 12*(6/11) = (72/11) cm²

We are smart and clever!

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Online Edumath Educators and Learners are Super Smart and Amazingly, Very Clever. Communicate us to mentor/teach/educate your child/children Mathematics online at affordable tuition, helping them become a lover of Mathematics, and also, a Mathematician. Call/WhatsApp: +2349076614992 or +2348152827445 Email: [email protected] [email protected] [email protected] Facebook Page: Online Edumath We are super smart and amazingly, very clever!

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2409919

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cos60 = a/4 ½ = a/4 a = 2 units. b = a+4 b = 2+4 b = 6 units. tan60 = c/6 c = 6√(3) units. d = b+4 d = 6+4 d = 10 units. It implies, area red is; ½(cd) = ½*6√(3)*10 = 30√(3) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2409591

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Let r be the radius of the circle. It implies, calculating r. r² = (r-1)²+(r-0.5)² r² = r²-2r+1+r²-r+(¼) 0 = r²-3r+(5/4) 4r²-12r+5 = 0 Resolving the above quadratic equation via completing the square approach to get r, radius of the circle. r²-3r = (-5/4) (r-(3/2))² = (-5/4)+(-3/2)² (r-(3/2))² = (-5/4)+(9/4) (r-(3/2))² = 1 r = (3/2)±1 It implies; r ≠ (3/2)-1 r = (3/2)+1 r = 2.5 cm. Notice! The blue area is a square. Therefore, a, side length of the square is;...
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