Verified answer

[tex]\mathbb\red{ \tiny A \scriptsize \: N \small \:S \large \: W \Large \:E \huge \: R}[/tex][tex] \rule{300pt}{0.1pt}[/tex]

Area of bigger circle [tex]\rm =\dfrac{22}{7} \times 14 \times 14=616 cm ^{2}[/tex]

Area of 2 small circles [tex]\rm =2 \times \pi r^{2}[/tex]

[tex] \rm=2 \times \dfrac{22}{7} \times 2.8\times 2.8 [/tex]

[tex] \rm = 49.28 \: \: {cm}^{2} [/tex]

[tex] \text{Area of rectangle = Length \( \times \) Breadth}[/tex]

[tex]\rm=4 \times 3=12 \ \ cm ^{2} [/tex]

Area of remaining sheet = Area of bigger circle - (Area of 2 small circles + Area of rectangle)

[tex]=616-(49.28+12)[/tex]

[tex] =616 - 61.28[/tex]

[tex] \color{olive}\ \boxed{\bf = 554.72 \: \: {cm}^{2} }[/tex]

Step-by-step explanation:

Given:-

From the circular card sheet of radius 14cm,two circles of radius 2.8 cm and a rectangle of length 4cm and breadth 3cm are removed

find the area of the remaining sheet. (shaded region){ Take π = 22/7 }

Solution:-

We know that,

• Area of circle si πr²
• Area of rectangle is l×b

Area of shaded region= Area of Sheat- Area of 2 circles+ Area of rectangle

[22/7×14²]-[2(22/7×2.8)+4×3]

616-61.28

554.72

Hence Area of shaded region is 554.72cm²

Hope this helps you dear

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