Question

if ABCD is a rectangle, find the area of shaded portion..​

Answer 1

${\blue{\underline{\underline{\purple{\textsf{\textbf{Answer : }}}}}}}$

The area of the shaded portion is 45.84m²

${\blue{\underline{\underline{\purple{\textsf{\textbf{Step-by-step explanation: }}}}}}}$

Step 1 :

• Find the area of the inner circle

Area of the circle is given by : πr²

we have,

r (radius) = 1.4m

taking π as 22/7

Area of the inner circle :

→ πr²

→ 22/7 × (1.4)²

→ 6.16m²

Step 2 :

• Find the area of the inner rectangle

Area of rectangle is given by : l × b

We have,

l (length) = 2m

b (breadth) = 1m

Area of the rectangle :

→ l × b

→ (2 × 1)m²

→ 2m²

Step 3 :

Finding the area of the rectangle ABCD :

$\begin{gathered} { \textsf{ \textbf{Here}}} \begin{cases} \sf{l(length) = 9m} \\ \sf{b(breadth) = 6m} \\ \end{cases} \end{gathered}$

Area of the rectangle :

→ l × b

→ (9 × 6)m²

→ 54m²

Step 4 :

Area of the shaded portion :

$\sf \to Area \: of \: ABCD\: - (area \: of \: the \: inner \: circle + area \: of \: the \: inner \: rectangle)$

$\begin{gathered} { \textsf{ \textbf{Here}}} \begin{cases} \sf{Area \: of abcd = 54 {m}^{2} } \\ \sf{Area \: of \: inner \: circle = 6.16 {m}^{2} } \\ \sf{Area \: of \: the \: inner \:rectangle = {2m}^{2} } \\ \end{cases} \end{gathered}$

Therefore :

➪ Area of the shaded portion :

→ 54m² - (6.16m² + 2m²)

→ 54m² - 8.16m²

→ 45.84m²

${ \underline{\therefore{ \textsf{ \textbf{The \: area \:of \: the \: shaded \: region \: is \: 45.84sq.m. }}}}}$

Answer 2

Answer:

$\huge \bold \red{answer}$

Step 1 :

Find the area of the inner circle

Area of the circle is given by : πr²

we have,

r (radius) = 1.4m

taking π as 22/7

Area of the inner circle :

→ πr²

→ 22/7 × (1.4)²

→ 6.16m²

Step 2 :

Find the area of the inner rectangle

Area of rectangle is given by : l × b

We have,

l (length) = 2m

b (breadth) = 1m

Area of the rectangle :

→ l × b

→ (2 × 1)m²

→ 2m²

Step 3 :

Finding the area of the rectangle ABCD :

l(length)=9m

b(breadth)=6m

Area of the rectangle :

→ l × b

→ (9 × 6)m²

→ 54m²

Step 4 :

Area of the shaded portion :

Areaofabcd=54m^2

Areaofinnercircle=6.16m^2

Areaoftheinnerrectangle=2m^2

Therefore :

➪ Area of the shaded portion :

→ 54m² - (6.16m² + 2m²)

→ 54m² - 8.16m²

→ 45.84m²

∴ The area of the shaded region is 45.84sq.m.