10. From a circular card sheet of radius 14 cm, two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1cm are removed. (as shown in the adjoining figure). Find the area of the remaining sheet. (Take π = 22/7)
Answers
Answer:
1cm are removed. (as shown in the adjoining figure). Find the area of the remaining sheet. (Take π = \sf\dfrac{22}{7})
7
22
)
\bf{\underline{\underline{\orange{\red{Given:}}}}}
Given:
Radius of the circular card sheet = 14 cm
Radius of the two small circle = 3.5 cm
Length of the rectangle = 3 cm
Breadth of the rectangle = 1 cm
\bf{\underline{\underline{\pink{\purple{To\:Find:}}}}}
ToFind:
The area of remaining sheet.
\bf{\underline{\underline{\orange{\green{Solution:}}}}}
Solution:
\bf\underline{Remaining\:Area\:=\: Area\:of\: larger\: Circle}
RemainingArea=AreaoflargerCircle
{\implies}⟹ \sf{ – \:2\:×\:Area\:of\: smaller\: Circle}–2×AreaofsmallerCircle
{\implies}⟹ \sf{ – \:Area\:of\: Rectangle}–AreaofRectangle
\bf\underline{ Area\:of\: larger\: Circle}
AreaoflargerCircle
\sf{Radius\:of\: larger\: Circle\:=\:R\:=\:14 cm}RadiusoflargerCircle=R=14cm
\sf{Area\:of\: larger\: Circle\:=\:πr²}AreaoflargerCircle=πr
2
{\implies}⟹ = \sf\dfrac{22}{7}
7
22
\sf{×\:(14)²}×(14)
2
\begin{gathered}\\\end{gathered}
{\implies}⟹ = \sf\dfrac{22}{7}
7
22
\sf{×\:14\:×\:14}×14×14
\begin{gathered}\\\end{gathered}
{\implies}⟹ = \sf\dfrac{22}{7}
7
22
\sf{×\:2\:×\:14}×2×14
\begin{gathered}\\\end{gathered}
{\implies}⟹ \sf{=\: 44\:×\:14}=44×14
\begin{gathered}\\\end{gathered}
{\implies}⟹ = \sf{616\:Cm²}616Cm
2
\begin{gathered}\\\end{gathered}
\bf\underline{ Area\:of\: Smaller\: Circle}
AreaofSmallerCircle
\sf{Radius\:of\: larger\: Circle\:=\:R\:=\:3.5 cm}RadiusoflargerCircle=R=3.5cm
\sf{Area\:of\: Smaller\: Circle\:=\:πr²}AreaofSmallerCircle=πr
2
\begin{gathered}\\\end{gathered}
{\implies}⟹ = \sf\dfrac{22}{7}
7
22
\sf{×\:(3.5)²}×(3.5)
2
\begin{gathered}\\\end{gathered}
{\implies}⟹ = \sf\dfrac{22}{7}
7
22
\sf\bigg(\dfrac{35}{10}(
10
35
\sf\bigg)) ²
\begin{gathered}\\\end{gathered}
{\implies}⟹ = \sf\dfrac{22}{1}
1
22
\sf\bigg(\dfrac{7}{2}(
2
7
\sf\bigg)) ²
\begin{gathered}\\\end{gathered}
{\implies}⟹ \sf\dfrac{22}{7}
7
22
× \sf\dfrac{7}{2}
2
7
× \sf\dfrac{7}{2}
2
7
\begin{gathered}\\\end{gathered}
{\implies}⟹ \sf\dfrac{11}{7}
7
11
× \sf\dfrac{7}{1}
1
7
× \sf\dfrac{7}{2}
2
7
\begin{gathered}\\\end{gathered}
{\implies}⟹ \sf\dfrac{22}{7}
7
22
1 × 1 \sf\dfrac{7}{2}
2
7
× \sf\dfrac{77}{2}
2
77
\sf{ = \: 38.5\:cm²}=38.5cm
2
\begin{gathered}\\\end{gathered}
\bf\underline{ Area\:of\: Rectangle:}
AreaofRectangle:
\sf{Length\:of\: Rectangle\:=\:l\:=\:3cm}LengthofRectangle=l=3cm
\sf{Breadth\:of\: Rectangle\:=\:l\:×\:b}BreadthofRectangle=l×b
\sf{Area\:of\: Rectangle\:=\:l\:×\:b}AreaofRectangle=l×b
\sf{ =\:3\:×\:1}=3×1
\sf{ = \:3\:cm²}=3cm
2
\bf\underline{ Therefore:}
Therefore:
∴ Remaining Area = Area of larger Circle
– 2 × Area of Smaller Circle
– Area of Rectangle
{\implies}⟹ \sf{=\:616\:–\:2\:×}=616–2× \sf\bigg(\dfrac{77}{2}(
2
77
\sf\bigg)) – 3
\begin{gathered}\\\end{gathered}
{\implies}⟹ \sf{ =\:616\:–\:77\:–\:3}=616–77–3
\begin{gathered}\\\end{gathered}
{\implies}⟹ \sf{=\:616\:–\:80}=616–80
\begin{gathered}\\\end{gathered}
{\implies}⟹ \sf{=\:536\:=\:cm²}=536=cm
2
\begin{gathered}\\\end{gathered}
\bf{\underline{\underline{\red{\purple{∴\: Required\:Area\:is\:536\:Cm²}}}}}
∴RequiredAreais536Cm
2
Step-by-step explanation:
hope it helps.