Answers
solution given:
For Cuboid
length[l]=11mm
breadth [b]=9mm
height[h]=6mm
For semi cylinder
height[H]=11mm
radius[r]=[tex] \frac{9}{2}=4.5mm[/tex]
Now
Totalsurface area=2(lb+bh+lh)+½(2πr(r+H))-l*b[/tex]
:2(11*9+9*6+11*6)+22/7*4.5(4.5+11)-11*9
:438+219.2-99
:558.2mm²
Here area of base is subtracted as it is not included.
Total surface area of composite figure is :558.2mm².
Answer:
[tex]\displaystyle SA_{Total} = \frac{279 \pi}{4} + 339 \ mm^2[/tex]
General Formulas and Concepts:
Pre-Algebra
Order of Operations: BPEMDAS
BracketsParenthesisExponentsMultiplicationDivisionAdditionSubtractionLeft to RightAlgebra I
FactoringGeometry
Shapes
Congruency
Congruent sides and lengthsRadius Formula: [tex]\displaystyle r = \frac{d}{2}[/tex]
d is diameterSurface Area of a Rectangular Prism Formula: SA = 2(wl + hl + hw)
w is widthl is lengthh is heightSurface Area of a Cylinder Formula: SA = 2πrh + 2πr²
r is radiush is heightStep-by-step explanation:
Step 1: Define
Identify
[Rectangular Prism] w = 9 mm
[Rectangular Prism] l = 11 mm
[Rectangular Prism] h = 6 mm
[Cylinder] d = 9 mm
[Cylinder] h = 11 mm
Step 2: Derive
Modify Surface Area equations and combine
[Surface Area of a Cylinder Formula] Factor: [tex]\displaystyle SA = 2(\pi rh + \pi r^2)[/tex][Surface Area of a Cylinder Formula] Divide by 2 [Semi-Cylinder]: [tex]\displaystyle SA = \pi rh + \pi r^2[/tex][Surface Area of a Semi-Cylinder] Substitute in r [Radius Formula]: [tex]\displaystyle SA = \pi (\frac{d}{2})h + \pi (\frac{d}{2})^2[/tex][Surface Area of a Semi-Cylinder] Evaluate exponents: [tex]\displaystyle SA = \pi (\frac{d}{2})h + \pi (\frac{d^2}{4})[/tex][Surface Area of a Semi-Cylinder] Multiply: [tex]\displaystyle SA = \frac{\pi dh}{2} + \frac{\pi d^2}{4}[/tex][Surface Area of a Rectangular Prism] Remove top: [tex]\displaystyle SA = 2(wh + lh) + lw[/tex]Combine Surface Area equations: [tex]\displaystyle SA_{Total} = \frac{\pi dh}{2} + \frac{\pi d^2}{4} + 2(wh + lh) + lw[/tex]Step 3: Find Surface Area
Substitute in variables [Combined Surface Area equation]: [tex]\displaystyle SA_{Total} = \frac{\pi (9 \ mm)(11 \ mm)}{2} + \frac{\pi (9 \ mm)^2}{4} + 2[(9 \ mm)(6 \ mm) + (11 \ mm)(6 \ mm)] + (11 \ mm)(9 \ mm)[/tex]Evaluate exponents: [tex]\displaystyle SA_{Total} = \frac{\pi (9 \ mm)(11 \ mm)}{2} + \frac{\pi (81 \ mm^2)}{4} + 2[(9 \ mm)(6 \ mm) + (11 \ mm)(6 \ mm)] + (11 \ mm)(9 \ mm)[/tex]Multiply: [tex]\displaystyle SA_{Total} = \frac{99\pi \ mm^2}{2} + \frac{81\pi \ mm^2}{4} + 2[54 \ mm^2 + 66 \ mm^2] + 99 \ mm^2[/tex][Brackets] Add: [tex]\displaystyle SA_{Total} = \frac{99\pi \ mm^2}{2} + \frac{81\pi \ mm^2}{4} + 2[120 \ mm^2] + 99 \ mm^2[/tex]Multiply: [tex]\displaystyle SA_{Total} = \frac{99\pi \ mm^2}{2} + \frac{81\pi \ mm^2}{4} + 240 \ mm^2 + 99 \ mm^2[/tex]Add: [tex]\displaystyle SA_{Total} = \frac{279 \pi}{4} + 339 \ mm^2[/tex]Related Questions
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A. 1.60
B. 1.65
C. 1.70
D. 1.75
Answers
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I dont know
Step-by-step explanation:
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—————
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Answers
Answer:
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Step-by-step explanation:
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Answers
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obtuse isosceles
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Answers
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Answers
A.
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Answers
Answer:
2
Step-by-step explanation:
Answer:
[tex]2[/tex]
Step-by-step explanation:
[tex]8+14-5-7+3-7-4[/tex]
We'll Follow the PEMDAS order of operations:-
Add and subtract:-
8+14=22
[tex]22-5-7+3-7-4[/tex]
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[tex]17-7+3-7-4[/tex]
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[tex]10+3-7-4[/tex]
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[tex]13-7-4[/tex]
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[tex]6-4[/tex]
[tex]2[/tex]
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Answers
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Answers
Answer:
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[tex]\huge{\textbf{\textsf{{\color{navy}{An}}{\purple{sw}}{\pink{er}} {\color{pink}{:}}}}}[/tex]
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Answers
Answer:
Step-by-step explanation:
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Answer:
Step-by-step explanation:
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Answer:
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335
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151
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104
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Answers
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Step-by-step explanation:
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Substitute the values in the above equation and you will get
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Step-by-step explanation:
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Answers
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Answers
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Answers
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multiply both sides
swap the sides
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Answers
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8.
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Answers
Answer:
D: y = 4x^2
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Answers
Answer:
$20.64
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if you cut 61.92 in to thirds
61.92/ 3= 20.64
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Answers
Answer:
C
Step-by-step explanation:
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Answers
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Answers
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Answers
Answer:
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Step-by-step explanation:
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Answers
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12
Step-by-step explanation:
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Answers
Answer:
40 waves
Step-by-step explanation:
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Answers
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Answers
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Answers
Answer:
160 and 20 degrees
Step-by-step explanation: