
Calculate the perimeter and area of a quadrant of a circle whose radii are
i) 98
ii) 70
iii) 42
iv) 28
Answer
600k+ views
Hint: Perimeter can be defined as the total length of the boundary of a geometrical figure. Area can be defined as the space occupied by a flat shape or the surface of an object. So, using this definition we can easily solve our problem.
Complete step-by-step answer:
If a circle is drawn from the origin, then the quadrant of a circle can be defined as a portion of the circle which lies between the positive x-axis and positive y-axis.
In general, the perimeter of the quadrant of circle $=\dfrac{\pi r}{2}+2r$.
The area of quadrant of a circle $=\dfrac{1}{4}\pi {{r}^{2}}$.
i) Perimeter of quadrant of a circle of radius 98 cm $=\dfrac{\pi (98)}{2}+2\times 98=350cm$
Area of quadrant of circle of radius 98 cm$=\dfrac{1}{4}\times \pi \times {{(98)}^{2}}=7546c{{m}^{2}}$
ii) Perimeter of quadrant of a circle of radius 70 cm $=\dfrac{\pi (70)}{2}+2\times 70=250cm$
Area of quadrant of circle of radius 70 cm$=\dfrac{1}{4}\times \pi \times {{(70)}^{2}}=3850c{{m}^{2}}$
iii) Perimeter of quadrant of a circle of radius 42 cm $=\dfrac{\pi (42)}{2}+2\times 42=150cm$
Area of quadrant of circle of radius 42 cm $=\dfrac{1}{4}\times \pi \times {{(42)}^{2}}=1386c{{m}^{2}}$
iv) Perimeter of quadrant of a circle of radius 28 cm $=\dfrac{\pi (28)}{2}+2\times 28=100cm$
Area of quadrant of circle of radius 28 cm $=\dfrac{1}{4}\times \pi \times {{(28)}^{2}}=616c{{m}^{2}}$
Note: The key step for solving this problem is the knowledge of area and perimeter of a geometrical figure. In this case, the given figure is a quadrant of a circle. So, on constructing the figure we obtain the area and parameter in generalized form. After putting values in the formula, we get the desired result.
Complete step-by-step answer:
If a circle is drawn from the origin, then the quadrant of a circle can be defined as a portion of the circle which lies between the positive x-axis and positive y-axis.
In general, the perimeter of the quadrant of circle $=\dfrac{\pi r}{2}+2r$.
The area of quadrant of a circle $=\dfrac{1}{4}\pi {{r}^{2}}$.
i) Perimeter of quadrant of a circle of radius 98 cm $=\dfrac{\pi (98)}{2}+2\times 98=350cm$
Area of quadrant of circle of radius 98 cm$=\dfrac{1}{4}\times \pi \times {{(98)}^{2}}=7546c{{m}^{2}}$
ii) Perimeter of quadrant of a circle of radius 70 cm $=\dfrac{\pi (70)}{2}+2\times 70=250cm$
Area of quadrant of circle of radius 70 cm$=\dfrac{1}{4}\times \pi \times {{(70)}^{2}}=3850c{{m}^{2}}$
iii) Perimeter of quadrant of a circle of radius 42 cm $=\dfrac{\pi (42)}{2}+2\times 42=150cm$
Area of quadrant of circle of radius 42 cm $=\dfrac{1}{4}\times \pi \times {{(42)}^{2}}=1386c{{m}^{2}}$
iv) Perimeter of quadrant of a circle of radius 28 cm $=\dfrac{\pi (28)}{2}+2\times 28=100cm$
Area of quadrant of circle of radius 28 cm $=\dfrac{1}{4}\times \pi \times {{(28)}^{2}}=616c{{m}^{2}}$
Note: The key step for solving this problem is the knowledge of area and perimeter of a geometrical figure. In this case, the given figure is a quadrant of a circle. So, on constructing the figure we obtain the area and parameter in generalized form. After putting values in the formula, we get the desired result.
Recently Updated Pages
Why is there a time difference of about 5 hours between class 10 social science CBSE

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Trending doubts
Who was the first woman to receive Bharat Ratna?

State and prove the Pythagoras theorem-class-10-maths-CBSE

State BPT theorem and prove it class 10 maths CBSE

What is the minimum age for fighting the election in class 10 social science CBSE

Describe the Salient Features of Indian Economy

The slogan Jai Hind was given by A Lal Bahadur Shastri class 10 social science CBSE

