# NCERT Solutions for Class 8 Maths Chapter 11 Mensuration (EX 11.1) Exercise 11.1

## NCERT Solutions for Class 8 Maths Chapter 11 Mensuration (EX 11.1) Exercise 11.1 Free PDF download of NCERT Solutions for Class 8 Maths Chapter 11 Exercise 11.1 (EX 11.1) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 8 Maths Chapter 11 Mensuration Exercise 11.1 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.

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Exercise (11.1)

1. A square and a rectangular filled with measurements as given in the figure have the same perimeter. Which field has a larger area?

(Image Will Be Updated Soon)

Ans: Given that,

Side of the square $=60\text{ m}$

Length of the rectangle $=80\text{ m}$

Both the square and rectangle have same perimeter

To find,

Which field has larger area

Firstly to find the breadth of the rectangle

We know that,

Perimeter of the square = Perimeter of the rectangle

$4\left( \text{side of the square} \right)=2\left( length\times breadth \right)$

$4\left( 60 \right)=2l+2b$

$240=2\left( 80 \right)+2\left( b \right)$

$2b=240-160$

$2b=80$

$b=40\text{ m}$

Now area of the square$={{\left( side \right)}^{2}}$

$={{\left( 60 \right)}^{2}}$

$=3600\text{ }{{\text{m}}^{2}}$

Area of the rectangle $=length\times breadth$

$=80\times 40$

$=3200\text{ }{{\text{m}}^{2}}$

Area of the square > Area of the rectangle

$\therefore$ The area of the square has a comparatively larger area than the area of the rectangle.

2. 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 $Rs.55$ per ${{m}^{2}}$.

(Image Will Be Updated Soon)

Ans: Given that,

Side of the square plot $=25\text{ m}$

Length of the rectangular house $=20\text{ m}$

Breadth of the rectangular house $=15\text{ m}$

Cost of developing garden per ${{m}^{2}}=Rs.55$

To find,

The total cost of developing a garden around the house

Area of the square $={{\left( side \right)}^{2}}$

$={{\left( 25 \right)}^{2}}$

$=625\text{ }{{\text{m}}^{2}}$

Area of the rectangular house $=length\times breadth$

$=20\times 15$

$=300\text{ }{{\text{m}}^{2}}$

From the diagram, it can be noted that,

Area of the garden = Area of the square plot – Area of the rectangular house

$=625-300$

$=325\text{ }{{\text{m}}^{2}}$

Cost of developing garden per ${{m}^{2}}=Rs.55$

Cost of developing $325\text{ }{{\text{m}}^{2}}$ garden $=325\times 55$

$=Rs.17,875$

$\therefore$ The total cost for developing garden around the house will be $Rs.17,875$

3. 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 the garden Length of rectangle is $20-\left( 3.5+3.5 \right)\text{ meters}$

(Image Will Be Updated Soon)

Ans: Given that,

Total length of the garden $=20\text{ m}$

Length of the rectangular part $=20-\left( 3.5+3.5 \right)$

$=20-7$

$=13\text{ m}$

Breadth of the rectangular part $=7\text{ m}$

Diameter of semi-circular part$=7\text{ m}$

Radius of semi-circular part $=\dfrac{7}{2}$

$=3.5\text{ m}$

To find,

The perimeter of the garden and the area of the garden

Finding the perimeter of the garden

Perimeter of the garden = Perimeter of rectangular part + Perimeter of two semi-circular part

Perimeter of rectangular part $=2\left( l+b \right)$

$=2\left( 13+7 \right)$

$=2\left( 20 \right)$

$=40\text{ m}$

Perimeter of two semi-circular part $=2\left( \dfrac{2\pi r}{2} \right)$

$=2\pi r$

$=2\times \dfrac{22}{7}\times 3.5$

$=22\text{ m}$

Perimeter of the garden $=40+22$

$=62\text{ m}$

Area of the garden = Area of the rectangular part + Area of the two semi-circular part

Area of rectangular part $=l\times b$

$=13\times 7$

$=91\text{ }{{\text{m}}^{2}}$

Area of two semi-circular part $=2\left( \dfrac{\pi {{r}^{2}}}{2} \right)$

$=\pi {{r}^{2}}$

$=\dfrac{22}{7}\times 3.5\times 3.5$

$=38.5\text{ }{{\text{m}}^{2}}$

Area of the garden $=91+38.5$

$=129.5\text{ }{{\text{m}}^{2}}$

$\therefore$ The perimeter of the garden is $62\text{ m}$ and the area of the garden is $129.5\text{ }{{\text{m}}^{2}}$.

4. A flooring tile has the shape of a parallelogram whose base is $24\text{ cm}$ and the corresponding height is $10\text{ cm}$. How many such tiles are required to cover a floor of area $1080\text{ }{{\text{m}}^{2}}$?(If required you can split the tiles in whatever way you want to fill up the corners).

Ans: Given that,

Length of the parallelogram tile$=24\text{ cm}$

Breadth of the parallelogram tile$=10\text{ cm}$

Area of the floor with tiles $=1080\text{ }{{\text{m}}^{2}}$

Area of the a single parallelogram tile$= length\times breadth$

$=24\times 10$

$=240\text{ c}{{\text{m}}^{2}}$

Number of tiles required $=\dfrac{\text{Total area of the floor}}{Area\text{ of each tile}}$

We know that,

$1\text{ m}=100\text{ cm}$

$1\text{ }{{\text{m}}^{2}}=10000\text{ c}{{\text{m}}^{2}}$

$1080\text{ }{{\text{m}}^{2}}=1080\times 10000\text{ c}{{\text{m}}^{2}}$

$n=\dfrac{1080\times 10000}{240}$

$=45,000$ tiles

$\therefore 45,000$ tiles are required to make a floor.

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 $c=2\pi r$, where $r$ is the radius of the circle.

(Image Will Be Updated Soon)

Ans:

(a) Given that,

Diameter of the semicircle $=2.8\text{ cm}$

Radius of the semi-circle$=\dfrac{2.8}{2}$

$=1.4\text{ cm}$

Circumference of the semi-circle $=\dfrac{2\pi r}{2}$

$=\dfrac{22}{7}\times 1.4$

$=4.4\text{ cm}$

Total distance around the food piece (a) $\text{= Diameter+Perimeter of the semi-circle}$$= 2.8+4.4$

$=7.2\text{ cm}$

Total distance covered by ant in the food piece (a) is $7.2\text{ cm}$

(b) Given that,

Length of the given figure $=2.8\text{ cm}$

Breadth of the given figure $=1.5\text{ cm}$

Radius of semi-circular part $=\dfrac{2.8}{2}$

$=1.4\text{ cm}$

Perimeter of the semicircular part $=\dfrac{2\pi r}{2}$

$=\dfrac{22}{7}\times 1.4$

$=4.4\text{ cm}$

$=10.2\text{ cm}$

Total distance covered by ant in food piece (b) is $10.2\text{ cm}$

(c) Given that,

Two sides of the triangle are $2\text{ cm, 2 cm}$

Diameter of the semi-circular part $=2.8\text{ cm}$

Radius of the semi-circular part $=\dfrac{2.8}{2}$

$=1.4\text{ cm}$

Circumference of semi-circular part $=\dfrac{2\pi r}{2}$

$=\dfrac{22}{7}\times 1.4$

$=4.4\text{ cm}$

Total distance covered by ant in food piece (c)$=2+2+4.4$

$=8.4\text{ cm}$

$\therefore$ For the food piece (b), the ant have to take a longer time since the perimeter is comparatively larger for food piece (b)

## NCERT Solutions for Class 8 Maths Chapter 11 Mensuration (Ex 11.1) Exercise 11.1

Opting for the NCERT solutions for Ex 11.1 Class 8 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 11.1 Class 8 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 8 students who are thorough with all the concepts from the Subject Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 8 Maths Chapter 11 Exercise 11.1 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

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1. What does Class 8 Maths Chapter 11 of NCERT book depict?

Ans:Class 8 Maths Chapter 11 titled Mensuration deals with geometric figures and their parameters like length, volume, shape, surface area, lateral surface area, etc. Here, the questions/problems related to mensuration are explained thoroughly.

This chapter contains all the important mensuration formulas and properties of different geometric shapes and figures. The chapter also contains various problems related to perimeter and quadrilaterals which are solved by Vedantu’s top-notch subject matter experts.

2. How many questions are there in Class 8 Maths Chapter 11 Exercise 11.1 of NCERT textbook?

Ans: Class 8 Maths Chapter 11 Exercise 11.1 of NCERT textbook consists of five long answer type questions. Solutions to those problems have also been provided by Vedantu, India’s leading ed-tech portal. These solutions are created by the top-notch subject matter experts from the relevant industry.

3. What are the topics/ subtopics covered under the Class 8 Maths Chapter 11?

Ans: The topics/ subtopics covered under the Class 8 Maths Chapter 11 are as follows:

• 11.1 - Introduction

• 11.2 - Let us Recall

• 11.3 - Area of Trapezium

• 11.4 - Area of a General Quadrilateral

• 11.5 - Area of a Polygon

• 11.6 - Solid Shapes

• 11.7 - Surface Area of Cube, Cuboid and Cylinder

• 11.8 - Volume of Cube, Cuboid and Cylinder

• 11.9 - Volume and Capacity

4. Is it beneficial to refer to the NCERT Solutions Class 8 Maths Chapter 11 Exercise 11.1?

Ans: NCERT Solutions for Class 8 Maths Chapter 11 titled Mensuration contains all the answers for all the questions asked in the textbook. All the solutions are prepared by Vedantu’s subject matter experts as per the latest CBSE syllabus and guidelines.

NCERT Solutions Class 8 Maths Chapter 11 Exercise 11.1 helps all the students to prepare for the exams as per the syllabus and revise accordingly prior to the exam. These solutions will help the students to practise and learn the question and answer patterns as well. NCERT Solutions provided by Vedantu are of the best quality and help to score the highest possible marks in the final exam. SHARE TWEET SHARE SUBSCRIBE