sove q.16..........plz
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A(ABC) = (√3 /4)a^2 ....(as it is an eqilateral triangle)
so, A(ABC) = 25√3 sq.cm.
Now in PBC,
BY Pythagoras' theorem,
BC^2 = PB^2 + PC^2
100 = 64 - PC ^2
PC = √36 = 6cm.
A(PBC) = √s(s-a)(s-b)(s-c) where s= (a+b+c)/2 =12cm.
A(PBC) = √12(2)(4)(6) = 24sq.cm.
Now, area of shaded region = A(ABC) -A(PBC) = 25√3 - 24
Ans) Area of shaded region is -24+25√3 sq.cm.
:)
so, A(ABC) = 25√3 sq.cm.
Now in PBC,
BY Pythagoras' theorem,
BC^2 = PB^2 + PC^2
100 = 64 - PC ^2
PC = √36 = 6cm.
A(PBC) = √s(s-a)(s-b)(s-c) where s= (a+b+c)/2 =12cm.
A(PBC) = √12(2)(4)(6) = 24sq.cm.
Now, area of shaded region = A(ABC) -A(PBC) = 25√3 - 24
Ans) Area of shaded region is -24+25√3 sq.cm.
:)
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