Math, asked by ridhimachouhan, 5 months ago

In the given figure, find the area of the shaded portion.

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Answers

Answered by Anonymous
3

Step-by-step explanation:

this question's answer is 61.5 cm^2

Attachments:
Answered by Anonymous
4

AnswEr:-

Area of shaded portion = 61.5 cm².

ExplanatioN:-

GiveN:-

Side of square = 10 cm

Radius of circle = 3.5 cm

To FinD:-

The area of the shaded portion.

SolutioN:-

Analysis :

First we have to find the area of the whole square and then the area of the circle inscribed inside the square. After getting both the area of the square and the circle we have to subtract the area of square from the area of the circle. We will get out answer.

Solution :

Area of square :

We know that if we are given the measure of each side of square then our required formula to find the area of square is,

Area of square = (side)²

where,

  • Side = 10 cm

By using the formula for the area of square and substituting the values in the required formula,

⇒ Area of square = side × side

⇒ Area of square = 10 × 10

Area of square = 100 cm².

Area of circle :

We know that if we are given the radius of the circle then our required formula to find the area of circle is,

Area of circle = πr²

where,

  • π = 22/7 or 3.14
  • r = radius = 3.5 cm

By using the formula for the area of circle and substituting the values in the required formula,

⇒ Area of circle = πr²

⇒ Area of circle = 22/7 × 3.5 × 3.5

⇒ Area of circle = 22/7 × 12.25

⇒ Area of circle = 269.5/7

∴ Area of circle = 38.5 cm².

Area of the shaded portion :

We know that if we know the area of the square and area of circle then for finding the area of the shaded portion our formula is,

Area of shaded portion = Area of square - Area of circle

where,

  • Area of square = 100 cm²
  • Area of circle = 38.5 cm²

Substituting the values in the required formula,

⇒ Area of shaded portion = Area of square - Area of circle

⇒ Area of shaded portion = 100 - 38.5

∴ Area of shaded portion = 61.5 cm².

Area of the shaded portion is 61.5 cm² .

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