NCERT Solutions For Class 8 Maths Chapter 11 Mensuration Ex 11.1
Here, Below you all know about NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.1 Question Answer. I know many of you confuse about finding this Chapter 11 Mensuration Ex 11.1 Of Class 8 NCERT Solutions. So, Read the full post below and get your solutions.
| Textbook | NCERT |
| Board | CBSE |
| Category | NCERT Solutions |
| Class | Class 8 |
| Subject | Maths |
| Chapter | Chapter 11 |
| Exercise | Class 8 Chapter 11 Mensuration 11.1 |
| Number of Questions Solved | 5 |
NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.1
NCERT TEXTBOOK EXERCISES
Question 1. A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area?
Solution.
Area of the square field = a x a
= 60 m x 60 m = 3,600 m2
Perimeter of the square field = 4a
= 4 x 60 m = 240 m
∴ Perimeter of rectangular field = 240 m
⇒ 2(l + b) = 240
⇒ 2(80 + b) = 240
where b m is the breadth of the rectangular field
⇒ 80 + b = 240/2 ⇒ 80 + bx = 120
⇒ b = 120 – 80 = 40
∴ Breadth = 40 m
∴ Area of rectangular field
= l x b = 80 m x 40 m = 3,200 m2
So, the square field (a) has a larger area
Question 2. Mrs. Kaushik has a square plot ‘ with the measurement as shown in the figure. She wants to construct a house in the middle of the plot. A garden is developed around the house. Find the total cost of developing a: garden around the house at the rate of ₹ 55 per m2.
Solution.
Area of the square plot = a x a
= 25 x 25 m2 = 625 m2
Area of the house = a x b
= 20 x 15 m2 = 300 m2
∴ Area of the garden
= Area of the square plot – Area of the house
= 625 m2 – 300 m2
= 325 m2
∵ The cost of developing the garden per square metre = ₹ 55.
∴ Total cost of developing the garden
= ₹ 325 x 55
= ₹ 17,875.
Question 3. The shape of a garden is rectangular in the middle and semicircular at the ends as shown in the diagram. Find the area and the perimeter of this garden.
Solution.
Question 4. A flooring tile has the shape of a parallelogram whose base is 24 cm and the cor-responding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m2? (If required you can split the tiles in whatever way you want to fill up the corners).
Solution.
Area of a flooring tile = bh
= 24 x 10 cm2
= 240 cm2
Area of the floor
= 1080 m2
= 1080 x 100 x 100 cm2
∵ m2 = 100 x 100 cm2
∴ Number of tiles required to cover the floor
= $\frac { Area\quad of\quad the\quad floor\quad }{ Area\quad of\quad a\quad flooring\quad tile }$
= $\frac { 1080\times 100\times 100\quad }{ 240 }$
= 45000.
Question 5. An ant is moving around a few food pieces of different shapes scattered on the floor. For which food-piece would the ant have to take a longer round? Remember, the circumference of a circle can be obtained by using the expression c = 2πr, where r is the radius of the circle.
Solution.
NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.1 Exercise 11.1 PDF
For NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.1, you may click on the link below and get your NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Exercise pdf file.
Finally, You all know about NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.1. If you have any questions, then comment below and share this post with others.
