Question

Please answer this question ✨✨✍️​

Answer 1

Answer:

Radius of the circle = 32 cm

Draw a median AD of the triangle passing through the centre of the circle.

⇒ BD = AB/2

Since, AD is the median of the triangle

∴ AO = Radius of the circle = 2/3 AD

⇒ 2/3 AD = 32 cm

⇒ AD = 48 cm

In ΔADB,

By Pythagoras theorem,

AB2 = AD2  + BD2 

⇒ AB2 = 482  + (AB/2)2

⇒ AB2 = 2304  + AB2/4

⇒ 3/4 (AB2) = 2304

⇒ AB2 = 3072

⇒ AB = 32√3 cm

Area of ΔADB = √3/4 × (32√3)2 cm2 = 768√3 cm2

Area of circle = π R2 = 22/7 × 32 × 32 = 22528/7 cm2

Area of the design = Area of circle - Area of ΔADB

                              = (22528/7 - 768√3) cm2

Answer 2

❤️answer❤️

let ABC be the eq. and let O be in the centre of the circle if r = 32cm

area of circle =πr^2

= [ 22/7 × 32 × 32 ] cm^2

= 22528 / 7 cm^2

draw OM | BC

NOW ,

BOM = 1/ 2 ×120 degree = 60°

so , from / BOM , We have

OM | OB = COS 60° 1/ 2

OM = 16 CM

Also , BM / OB = COS 60° [1/2]

BM = 16 √3 CM

BC = 2BM = 32√3CM

HENCE ,

AREA OF BOC = 1/ 2 BC.OM

=1/3 ×32 √3×16

768 √3cm ^2

area of design = are of o - area of ABC

=(22528/7 - 768√3)cm ^2............

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