Question

please answer this question now ....

Answer 1

Hope it works..........

Answer 2

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step1 :
Find area of equilateral triangle ABC
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since , the triangle ABC is an equilateral triangle with length of each side = 10 cm

we know that ,

area of equilateral triangle

= √3/4( side )^2

=√3 /4( 10 )^2 = 100√3 / 4=25√3 cm^2

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step2 : Find area of right angle ∆ BOC which is inscribed inside the equilateral triangle :
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in ∆ BOC,

by Pythagoras theorem , we get

( BC )^2 = ( BO )^2 + (OC )^2

since , OC = 8 cm and BC = 10 cm , we get

( 10 )^2 = ( BO )^2 + (8 )^2

( BO)^2 = 100 - 64

(BO )^2 = 36 => BO = √36 = 6 cm

area of right angle ∆ BOC

= 1 /2 × BO × OC= 1/ 2 × 6 × 8 = 24cm^2

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Step 3 Find Area of shaded region :
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area of shaded region

= area of ∆ ABC - area of ∆ BOC

= 25√3 - 24 , { put value of √3 = 1.732 }

therefore , area of shaded region

=25 (1.732) - 24 = 43.3 - 24 =19.3 cm^2


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Your Answer:area = 19.3 cm^2 (approx.)
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