NCERT Solutions Class 8 Maths Chapter 11 Mensuration – Here are all the NCERT solutions for Class 8 Maths Chapter 11. This solution contains questions, answers, images, explanations of the complete chapter 11 titled Mensuration of Maths taught in class 8. If you are a student of class 8 who is using NCERT Textbook to study Maths, then you must come across chapter 11 Mensuration. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 8 Maths Chapter 11 Mensuration in one place.
NCERT Solutions Class 8 Maths Chapter 11 Mensuration
Here on AglaSem Schools, you can access to NCERT Book Solutions in free pdf for Maths for Class 8 so that you can refer them as and when required. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions.
For a better understanding of this chapter, you should also see summary of Chapter 11 Mensuration , Maths, Class 8.
Class | 8 |
Subject | Maths |
Book | Mathematics |
Chapter Number | 11 |
Chapter Name |
Mensuration |
NCERT Solutions Class 8 Maths chapter 11 Mensuration
Class 8, Maths chapter 11, Mensuration solutions are given below in PDF format. You can view them online or download PDF file for future use.
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Question & Answer
Q.1: Match the following figures with their respective areas in the box.
Ans : Missing
Q.2: Write the perimeter of each shape.
Ans : Missing
Q.3: A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area?
Ans : Perimeter of square = 4 (Side of the square) = 4 (60 m) = 240 m. Perimeter of rectangle = 2 (Length + Breadth) = 2 (80 m + Breadth) = 160 m + 2 x Breadth It is given that the perimeter of the square and the rectangle are the same. 160 m + 2 x Breadth = 240 m Breadth of the rectangle = \( \left(\frac{80}{2}\right) \mathrm{m} \) = 40 m Area of square = \( (\text { Side })^{2}=(60 \mathrm{m})^{2}=3600 \mathrm{m}^{2} \) Area of rectangle = Length x Breadth = (80 x 40) \( m^{2} \) = 3200 \( m^{2} \) Thus, the area of the square field is larger than the area of the rectangular field.
Q.4: Mrs. Kaushik has a square plot with the measurement as shown in the following 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 \( \mathrm{m}^{2} \)
Ans : Area of the square plot = \( (25 m)^{2}=625 m^{2} \) Area of the house = (15 m) x (20 m) = 300 \( m^{2} \) Area of the remaining portion = Area of square plot - Area of the house = 625 \( m^{2} \) - 300 \( m^{2} \) = 325 \( m^{2} \). The cost of developing the garden around the house is Rs 55 per \( m^{2} \). Total cost of developing the garden of area 325 \( m^{2} \) = Rs (55 x 325) = Rs 17,875
Q.5: The shape of a garden is rectangular in the middle and semi circular at the ends as shown in the diagram. Find the area and the perimeter of this garden [Length of rectangle is 20 – (3.5 + 3.5) metres].
Ans : Length of the rectangle = [20 - (3.5 + 3.5)] metres = 13 m Circumference of 1 semi-circular parts = nr \( =\left(\frac{22}{7} \times 3.5\right) \) = 11 m Circumference of both semi-circular parts = ( 2 x 11 ) m = 22 m Perimeter of the garden = AB + Length of both semi-circular regions BC and DA + CD = 13m + 22m + 13m = 48 m Area of the garden = Area of rectangle + 2 x Area of two semi-circular regions \( =\left[(13 \times 7)+2 \times \frac{1}{2} \times \frac{22}{7} \times(3.5)^{2}\right] \mathrm{m}^{2} \) \( =(91+38.5) \mathrm{m}^{2} \) \( =129.5 \mathrm{m}^{2} \)
NCERT / CBSE Book for Class 8 Maths
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