Question

In the given figure triangle ABC is an equilateral triangle having each side equal to 10 cm and triangle PBC is right angled at P in which PB = 8 cm . Find the area of the shaded region...

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Answer 1

Answer:

Step-by-step explanation:

AREA OF THE EQUILATERAL TRIANGLE ABC

=√3÷4 a^2

=√3÷4 (10)^2. {100÷4=25}

=√3×25

=25√3

AREA OF THE RIGHT ANGLE TRIANGLE,

IN ∆PBC

(H)^2=P^2 + B^2

(10)2=(PC)^2+(PB)^2

100=(8)^2+ PC^2

100-64=PC^2

√36=PC

PC=6

AREA

=1÷2×PC×PB

=1÷2×8×6

=8×3

=24CM^2

NOW THE AREA OF SHADED PORTION

AREA OF ∆ABC-∆PBC

25√3-24

25×1.723 - 24

43.075-24

19.075