Question

find the shaded area ​

Answer 1

$\huge\mathbb\colorbox{black}{\color{red}{Question}}$

• $\sf\red{To \: Find \: The \: Area \: Of \: The \: Shaded \: Portion}$

$\small\mathbb\colorbox{black}{\color{red}{Solution \: For \: The \: Correct \: Answer:-}}$

$\sf\red{First \: of \: all \: we \: see \: that \: the \: whole}$$\sf\red{area \: is \: a \: rectangle}$

$\sf\red{So \: we \: will \: first \: find \: the \: area \: of \: the}$$\sf\red{Rectangle}$

$\small\mathbb\colorbox{black}{\color{lime}{Formula \: To \: Find \: Area \: of \: The \: Rectangle}}$

• $\sf\green{Area \: = \: L \: × B \:}$

• $\sf\green{B=3m+6m \: = \: 18m}$

• $\sf\green{A \: = \: 5m×9m}$

• $\sf\green{Area \: = \: 45m²}$

$\sf\red{So \: now \: we \: have \: found \: the \: area \: of \: the }$

$\sf\red{Rectangle}$

Now we can notice two triangles as the shaded portion.So we will find the area of both the triangles and then add their area.

1st Triangle:-

• $\sf\red{Height \: = \: 5m}$
• $\sf\red{Base \: = \: 9m - 7m}$

$\small\mathbb\colorbox{black}{\color{lime}{Formula \: To \: Find \: Area \: of \: The \: Triangle}}$

$\sf\green{Area \: = \: 1/2 \: × \: Base \: × \: Height}$

• $\sf\green{A=1/2×5×2}$
• $\sf\green{A=1/2×10}$
• $\sf\green{Area \: = \: 5m²}$

2nd Triangle

• $\sf\red{Height \: = \: 5m}$
• $\sf\red{Base \: = \: 6m}$

$\sf\green{Area \: = \: 1/2 \: × \: Base \: × \: Height}$

• $\sf\green{A=1/2×5×6}$
• $\sf\green{A=1/2×30}$
• $\sf\green{Area \: = \: 15m²}$

Now we have got the area of both triangles so we will sum up both to get the area of the shaded portion.

• $\sf\red{triangle \: 1 \: + \: Triangle \: 2}$
• $\sf\red{5m² \: + \: 15m²}$

$\sf\green{:. \: Area \: of \: the \: shaded \: portion \: = \: 20m²}$

Answer 2

AREA OF SHADED PORTION IS FOUND BY FOLLOWING STEPS.

Step-by-step explanation:

• from the figure we can say that length=9m and breadth=5 m of the rectangle.
• area of shaded region = area of rectangle - area of unshaded region
• area of rectangle=length *breadth=a*b=9*5=45$m^{2}$
• unshaded region is in shape of trapezium
• area of unshaded trapezium =half of product of sum of parallel lines and distance between them=$\frac{(a+b)*h}{2}$=$\frac{(7+3)*5}{2} =\frac{10*5}{2}=25$$m^{2}$
• area of shaded region=45-25=20$m^{2}$