Question

find the area of the shaded portion . All corners are right angled ​

Answer 1

$\sf\Large{\underline{\underline{Answer:-}}}$

Divided the shaded portion into rectangles. (in attachment)

And find there areas.

First, Finding area of whole figure.

Formula using here:-

$\begin{gathered}\\\;\sf{:\rightarrow\;\;area\;=\;\bf{{length \times breadth}\:}}\end{gathered}$

Substitute the Values,

$\begin{gathered}\\\;\sf{:\rightarrow\;\;area\;=\;\bf{{19.6 \times 15}\:}}\end{gathered}$

$\begin{gathered}\\\;\sf{:\rightarrow\;\;area\;=\;\bf{{294m^{2} }\:}}\end{gathered}$

${\bold\blue{⇝}}$ The area of whole figure is 294m²

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Now,

find the Area of all rectangles.

• ${\bold\green{Rectangle \: I}}$

$\begin{gathered}\\\;\sf{:\rightarrow\;\;area\;=\;\bf{{19.6 \times 2}\:}}\end{gathered}$

$\begin{gathered}\\\;\sf{:\rightarrow\;\;area\;=\;\bf{{39.2m{2} }\:}}\end{gathered}$

The area of Rectangle I is 39.2m²

Now,

• ${\bold\red{Rectangle \: II}}$

$\begin{gathered}\\\;\sf{:\rightarrow\;\;area\;=\;\bf{{19.6 \times 2}\:}}\end{gathered}$

$\begin{gathered}\\\;\sf{:\rightarrow\;\;area\;=\;\bf{{39.2m^{2} }\:}}\end{gathered}$

The area of Rectangle II is 39.2m².

Now

• ${\bold\blue{Rectangle \: III}}$

15m - (2m + 2m)

15m - 4m

Breadth of rectangle III is 11m.

$\begin{gathered}\\\;\sf{:\rightarrow\;\;area\;=\;\bf{{5.4\times 11}\:}}\end{gathered}$

$\begin{gathered}\\\;\sf{:\rightarrow\;\;area\;=\;\bf{{59.4m^{2} }\:}}\end{gathered}$

The area of Rectangle III is 59.4m².

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At last,

• To find shaded portion.

$\Longrightarrow$Shaded portion = Area of whole figure – (area of all rectangles).

$\Longrightarrow$Shaded portion = 294 – ( 39.2 + 39.2 + 59.4)

$\Longrightarrow$Shaded portion = 294 – 137.8

${\bold\purple{Hence,}}$

Area of Shaded portion = 156.2m².

Answer 2

$\sf\Large{\underline{\underline{Answer:-}}}$

Divided the shaded portion into rectangles. (in attachment)

And find there areas.

First, Finding area of whole figure.

Formula using here:-

$\begin{gathered}\\\;\sf{:\rightarrow\;\;area\;=\;\bf{{length \times breadth}\:}}\end{gathered}$

Substitute the Values,

$\begin{gathered}\\\;\sf{:\rightarrow\;\;area\;=\;\bf{{19.6 \times 15}\:}}\end{gathered}$

$\begin{gathered}\\\;\sf{:\rightarrow\;\;area\;=\;\bf{{294m^{2} }\:}}\end{gathered}$

${\bold\blue{⇝}}$ The area of whole figure is 294m²

─────────────────────

Now,

find the Area of all rectangles.

${\bold\green{Rectangle \: I}}$

$\begin{gathered}\\\;\sf{:\rightarrow\;\;area\;=\;\bf{{19.6 \times 2}\:}}\end{gathered}$

$\begin{gathered}\\\;\sf{:\rightarrow\;\;area\;=\;\bf{{39.2m{2} }\:}}\end{gathered}$

The area of Rectangle I is 39.2m²

Now,

${\bold\red{Rectangle \: II}}$

$\begin{gathered}\\\;\sf{:\rightarrow\;\;area\;=\;\bf{{19.6 \times 2}\:}}\end{gathered}$

$\begin{gathered}\\\;\sf{:\rightarrow\;\;area\;=\;\bf{{39.2m^{2} }\:}}\end{gathered}$

The area of Rectangle II is 39.2m².

Now

${\bold\blue{Rectangle \: III}}$

15m - (2m + 2m)

15m - 4m

Breadth of rectangle III is 11m.

$\begin{gathered}\\\;\sf{:\rightarrow\;\;area\;=\;\bf{{5.4\times 11}\:}}\end{gathered}$

$\begin{gathered}\\\;\sf{:\rightarrow\;\;area\;=\;\bf{{59.4m^{2} }\:}}\end{gathered}$

The area of Rectangle III is 59.4m².

─────────────────────

At last,

To find shaded portion.

$\Longrightarrow$Shaded portion = Area of whole figure – (area of all rectangles).

$\Longrightarrow$Shaded portion = 294 – ( 39.2 + 39.2 + 59.4)

$\Longrightarrow$Shaded portion = 294 – 137.8

${\bold\purple{Hence,}}$

Area of Shaded portion = 156.2m².