Answer:
1. D
To find its surface area we have to find the measure of its height first
h= √(s²-r²) by phytagorean theorem
= √(10²-5²)
= √(100-25)
= √75
h= 8.66025403784
now that we have the height, we can now solve for its surface area
SA= πr(r+√(r²+h²)
= (3.14×5)[5+√(5²+8.66025403784²)]
= (15.7)(5+10)
= 15.7×15
= 235.5
2. E
SA= 2πrh+2πr²
= (2×3.14×4×8)+(2×3.14×4²)
= (3.14×64)+(3.14×32)
= 200.96+100.48
= 301.44
3. A
SA= A+1/2(s+p)
A-Area of base
s- slant height
p-perimeter
s= √4²+3²
= √25
s= 5
p= l+l+w+w
= 4+4+2+2
= 12
A= lw
= 4×2
SA= 8+1/2(12×5)
= 8+1/2×60
= 8+30
= 38
4. B
SA= 4πr²
= 4×3.14×2²
= 4×3.14×4
= 16×3.14
= 50.24
5. F
SA= πr[r+√(r²+h²)]
h= √(s²-r²)
= √(1²-6²)
=10.39230484541
SA= (3.14×6)[6+√(6²+10.39230484541²)]
= 339.12
Pa brainliest po please thank you
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Answers & Comments
Answer:
1. D
To find its surface area we have to find the measure of its height first
h= √(s²-r²) by phytagorean theorem
= √(10²-5²)
= √(100-25)
= √75
h= 8.66025403784
now that we have the height, we can now solve for its surface area
SA= πr(r+√(r²+h²)
= (3.14×5)[5+√(5²+8.66025403784²)]
= (15.7)(5+10)
= 15.7×15
= 235.5
2. E
SA= 2πrh+2πr²
= (2×3.14×4×8)+(2×3.14×4²)
= (3.14×64)+(3.14×32)
= 200.96+100.48
= 301.44
3. A
SA= A+1/2(s+p)
A-Area of base
s- slant height
p-perimeter
s= √4²+3²
= √25
s= 5
p= l+l+w+w
= 4+4+2+2
= 12
A= lw
= 4×2
SA= 8+1/2(12×5)
= 8+1/2×60
= 8+30
= 38
4. B
SA= 4πr²
= 4×3.14×2²
= 4×3.14×4
= 16×3.14
= 50.24
5. F
SA= πr[r+√(r²+h²)]
h= √(s²-r²)
= √(1²-6²)
=10.39230484541
SA= (3.14×6)[6+√(6²+10.39230484541²)]
= 339.12
Pa brainliest po please thank you