AB = BC = CD = AD = 8 cm

Area of ΔABC = Area of ΔADC = \( (\dfrac{1}{2})×8×8 = 32 cm^2\)

Area of quadrant AECB = Area of quadrant AFCD

= \(\dfrac{1}{4}\times\dfrac{22}{7}×8^2 \)

= \(\dfrac{352}{7}cm^2 \)

Area of shaded region = (Area of quadrant AECB – Area of ΔABC)

= (Area of quadrant AFCD – Area of ΔADC)

= \(( \dfrac{352}{7}-32)+(\dfrac{352}{7}-32)cm^2\)

= \( 2×(\dfrac{352}{7}-32)cm^2 \)

= \(\dfrac{256}{7}cm^2 \)

Answered by Abhisek | 1 year agoProve that the area of a circular path of uniform width hsurrounding a circular region of radius r is πh(2r + h).

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