hi friends you please follow question for me
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AOB IS AN EQUILATERAL Δ (OB=OA=10CM ⇒∠A=∠B=60°)
AREA OF SHADED REGION= AREA OF Δ-AREA OF SECTOR
= √3/4*10² - 60/360*π*8²
=√3/4*100 - 60/360*22/7*64
=25√3 - 33.52
=9.73 cm²
there you go!
AREA OF SHADED REGION= AREA OF Δ-AREA OF SECTOR
= √3/4*10² - 60/360*π*8²
=√3/4*100 - 60/360*22/7*64
=25√3 - 33.52
=9.73 cm²
there you go!
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hope u get the solution....and i m not sure for ur answer...i may be close to ur answer...its just an approximate....it actually depends o the values of Pie and Root3...
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