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23 Aug 2019, 10:54
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87% (02:17) correct
13%
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From a rectangle ABCD of area 768 sq cm, a semicircular part with diameter AB and area 72π sq cm is removed. The perimeter of the leftover portion, in cm, is
A. 88 + 12π
B. 80 + 16π
C. 86 + 8π
D. 82 + 24π
E. 80 + 8π
A. 88 + 12π
B. 80 + 16π
C. 86 + 8π
D. 82 + 24π
E. 80 + 8π
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23 Aug 2019, 12:51
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\((\frac{1}{2})π*(AB/2)^2= 72π\)
\(AB^2=72*8\)
AB=24
BC= 768/24=32
The perimeter of the leftover portion= 2*BC+AB+π*(AB/2)= 2*32+24+12π= 88+12π
A
\(AB^2=72*8\)
AB=24
BC= 768/24=32
The perimeter of the leftover portion= 2*BC+AB+π*(AB/2)= 2*32+24+12π= 88+12π
A
AbdulMalikVT
From a rectangle ABCD of area 768 sq cm, a semicircular part with diameter AB and area 72π sq cm is removed. The perimeter of the leftover portion, in cm, is
A. 88 + 12π
B. 80 + 16π
C. 86 + 8π
D. 82 + 24π
E. 80 + 8π
A. 88 + 12π
B. 80 + 16π
C. 86 + 8π
D. 82 + 24π
E. 80 + 8π
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Untitled.png [ 7.29 KiB | Viewed 1415 times ]
27 Aug 2019, 01:35
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nick1816
\((\frac{1}{2})π*(AB/2)^2= 72π\)
\(AB^2=72*8\)
AB=24
BC= 768/24=32
The perimeter of the leftover portion= 2*BC+AB+π*(AB/2)= 2*32+24+12π= 88+12π
A
\(AB^2=72*8\)
AB=24
BC= 768/24=32
The perimeter of the leftover portion= 2*BC+AB+π*(AB/2)= 2*32+24+12π= 88+12π
A
AbdulMalikVT
From a rectangle ABCD of area 768 sq cm, a semicircular part with diameter AB and area 72π sq cm is removed. The perimeter of the leftover portion, in cm, is
A. 88 + 12π
B. 80 + 16π
C. 86 + 8π
D. 82 + 24π
E. 80 + 8π
A. 88 + 12π
B. 80 + 16π
C. 86 + 8π
D. 82 + 24π
E. 80 + 8π
But doesn't 2xBC+AB gives you perimeter of the entire rectangle ? (including the cut out part)
