N the given figure, diameter AB is 12 cm long. AB is trisected at points P and Q. Find the area of the shaded region.
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15
the question is incomplete with a figure
Answered by
64
AP=PQ=QB
AP+PQ+QB=12 cm__________________(Trisected)
AP=PQ=QB=4 cm___________________(Trisected)
3AP=3PQ=3QB=12 cm
Now,
AQ=2AP=8 cm
&
PB=2QB=8 cm
R=AQ/2=4 cm
Similarly,r=AP/2=2 cm
Therefore,
Area of shaded region= 2×(1/2πR²-1/2πr²)=37.7 cm²
AP+PQ+QB=12 cm__________________(Trisected)
AP=PQ=QB=4 cm___________________(Trisected)
3AP=3PQ=3QB=12 cm
Now,
AQ=2AP=8 cm
&
PB=2QB=8 cm
R=AQ/2=4 cm
Similarly,r=AP/2=2 cm
Therefore,
Area of shaded region= 2×(1/2πR²-1/2πr²)=37.7 cm²
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