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# NCERT Solutions for Class 8 Maths Chapter 9 - Mensuration Exercise 9.1

Last updated date: 09th Sep 2024
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## NCERT Solutions for Maths Class 8 Chapter 9 Mensuration Exercise 9.1 - FREE PDF Download

Access NCERT Solutions for Class 8 Maths Chapter 9, Exercise 9.1 - Mensuration, are crafted to simplify the learning of mensuration concepts for students. These solutions are carefully prepared by Vedantu’s experts, ensuring they follow the CBSE syllabus and guidelines. Each problem is solved with detailed explanations to make understanding easier for students.

Table of Content
1. NCERT Solutions for Maths Class 8 Chapter 9 Mensuration Exercise 9.1 - FREE PDF Download
2. Glance on Class 8 Maths Chapter 9 Exercise 9.1 Class 8 | Vedantu
3. Formulas Used in Class 8 Chapter 9 Exercise 9.1
4. Access NCERT Solutions for Maths Class 8 Chapter 9 Mensuration Exercise 9.1
5. Class 8 Maths Chapter 9: Exercises Breakdown
6. CBSE Class 8 Maths Chapter 9 Other Study Materials
7. Chapter-Specific NCERT Solutions for Class 8 Maths
FAQs

Class 8 Ch 9 Maths Ex 9.1 includes various mensuration problems that aid in thorough revision and practice. By using these class 8 maths chapter 9 exercise 9.1 solutions, students can improve their problem-solving abilities and build confidence for their exams. Download the free PDF of these solutions to enhance your exam preparation and excel in your studies.

## Glance on Class 8 Maths Chapter 9 Exercise 9.1 Class 8 | Vedantu

• Mensuration involves understanding the measurement of shapes and figures, forming the foundation for geometry.

• The area of a polygon is the measure of the space enclosed within its sides, calculated using specific formulas depending on the type of polygon.

• For irregular polygons, the area is typically calculated by dividing the shape into simpler figures like triangles and rectangles, calculating the area of each, and then summing these areas.

• Polygons are closed two-dimensional shapes with straight sides.

• The area is the total space inside the polygon's sides.

• Polygons are defined by the number of sides, such as a triangle having 3 sides and a quadrilateral having 4 sides.

• Regular polygons have equal sides and angles, while irregular polygons do not.

• Convex polygons have all interior angles less than 180O, while concave polygons have at least one interior angle greater than 180O.

• There are 11 questions in class 8 maths chapter 9 exercise 9.1 solutions  which are fully solved by experts at Vedantu.

## Formulas Used in Class 8 Chapter 9 Exercise 9.1

• Formulas for Regular Polygons: $\dfrac{1}{2}\times Perimeter\times Apothem$

• The sum of interior angles of an n-sided polygon is $\left ( n-2 \right )\times 180^{\circ}$

Competitive Exams after 12th Science

## Access NCERT Solutions for Maths Class 8 Chapter 9 Mensuration Exercise 9.1

### Exercise 9.1

1. The shape of the top surface of a table is a trapezium. Find the area of its parallel sides, that is, 1 m and $1.2$ m and the perpendicular distance between them is $0.8$ m.

Ans: The area of a trapezium is the half the product of sum of parallel sides and distance between the parallel sides.

Here, the sides are 1 m and $1.2$ m. The height of the table is $0.8$.

$A = \dfrac{1}{2} \times \left( {1 + 1.2} \right) \times 0.8$

$A = \dfrac{1}{2} \times 1.2 \times 0.8$

$A = 0.88\,{{\text{m}}^2}$

Hence, the area is $0.88\,{{\text{m}}^2}$.

2. The Trapezium Is 34 square cm and the length of the parallel sides is 10cm and it's height is 4cm. Find the length of the other parallel side.

Ans: The area of a trapezium is the half the product of sum of parallel sides and distance between the parallel sides. Mathematically, $A = \dfrac{1}{2} \times \left( {a + b} \right) \times h$, where $a$ and $b$ are the parallel sides, and $h$ is the height of the trapezium.

Substitute the given values

$34 = \dfrac{1}{2} \times \left( {10 + a} \right) \times 4$

$34 = \left( {10 + a} \right) \times 2$

Divide both sides by 2.

$\dfrac{{34}}{2} = 10 + a$

$17 = 10 + a$

Subtract 10 from both sides.

$17 - 10 = a$

$7 = a$

Thus, the length of the other parallel side is 7 cm.

3. The length of the fence of a trapezium-shaped field ${\text{ABCD}}$ is 120 m. If ${\text{BC}} = 48\,{\text{m}}$, ${\text{CD}} = {\text{17}}\,{\text{m}}$ and ${\text{AD}} = 40\,\,{\text{m}}$, find the area of this field. Side $AB$ is perpendicular to the parallel sides ${\text{AD}}$ and ${\text{BC}}$.

Ans: Here, the length of side ${\text{AB}}$ is not given. So, using the length of the fence of the trapezium we will find the length of the side ${\text{AB}}$.

$120 = {\text{AB}} + 48 + 17 + 40$

$120 = {\text{AB}} + 105$

Subtract 105 from both sides.

$120 - 105 = {\text{AB}}$

$15\,{\text{m}} = {\text{AB}}$

Thus, the length of side ${\text{AB}}$ is 15 m.

Now, use the formula of area of trapezium to find the area of ${\text{ABCD}}$.

${\text{Area of ABCD}} = \dfrac{1}{2} \times \left( {{\text{AD + BC}}} \right) \times {\text{AB}}$

Substitute the value of sides.

${\text{Area of ABCD}} = \dfrac{1}{2} \times \left( {{\text{40}} + 48} \right) \times {\text{15}}$

${\text{Area of ABCD}} = \dfrac{1}{2} \times 88 \times {\text{15}}$

${\text{Area of ABCD}} = 44 \times {\text{15}}$

${\text{Area of ABCD}} = 660\,{{\text{m}}^2}$

Thus, the area of ${\text{ABCD}}$is $660\,{{\text{m}}^2}$.

4. The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite sides are 8 m and 13 m. find the area of the field.

Ans: The given quadrilateral has two triangles.

The area of the triangle is the half of the product of base and height.

We will find the area of both the triangles and add them to get the required area.

$A = \dfrac{1}{2} \times 24 \times 13 + \dfrac{1}{2} \times 24 \times 8$

Take $\dfrac{1}{2}$ and 24 common.

$A = \dfrac{1}{2} \times 24 \times \left( {13 + 8} \right)$

$A = \dfrac{1}{2} \times 24 \times \left( {21} \right)$

$A = 252\,{{\text{m}}^2}$

Hence, the area of the given field is $252\,{{\text{m}}^2}$.

5. The diagonals of a rhombus are $7.5$ cm and 12 cm. Find its area.

Ans: The area of rhombus is half of the product of its diagonals.

$A = \dfrac{1}{2} \times \left( {{\text{product of diagonals}}} \right)$

Substitute the diagonals as $7.5$ and 12.

$A = \dfrac{1}{2} \times 7.5 \times 12$

$A = 45\,\,{\text{c}}{{\text{m}}^2}$

Hence, the area of the rhombus is $45\,\,{\text{c}}{{\text{m}}^2}$.

6. Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm. If one of the diagonals is 8 cm long, find the length of other diagonal.

Ans: Since a rhombus is also a kind of a parallelogram,

The area of a rhombus = $Base \times Altitude$ ……(1)

Substitute the values of Base = 5 cm and Altitude = 4.8cm in  (1)

Area of rhombus = $5 \times 4.8$

$= 24$

The area of a rhombus is $24\,c{m^2}$

Also, the area of rhombus in terms of diagonal is $= \,\left( {\frac{1}{2}} \right) \times {d_1}{d_2}$$Substituting the values of {d_1}\, = \,8 and area of rhombus = 24, we get 24\, = \,\left( {\frac{1}{2}} \right) \times 8 \times {d_2}$$$

$24\, = \,4{d_2}$

${d_2}\, = \,6$

Hence, the length of the other diagonal is $6\,cm$.

6. Find the area of the rhombus whose side is 6 cm and whose altitude is 4 cm. if one of its diagonals is 8 cm long, find the length of the other diagonal.

Ans: Let the length  of the other diagonal be $x$. Since a rhombus is a special parallelogram, then, use the formula  of area of parallelogram to find the area of the given rhombus.

$A = 6 \times 4$

$A = 24\,{\text{c}}{{\text{m}}^2}$

Now, the area of rhombus is half of the product of its diagonals.

$A = \dfrac{1}{2} \times \left( {{\text{product of diagonals}}} \right)$

Substitute $A$ as 24, and one of the diagonals as 8.

$24 = \dfrac{1}{2} \times \left( {\text{8}} \right) \times x$

$24 = 4 \times x$

Divide both sides by 4.

$\dfrac{{24}}{4} = x$

$6 = x$

Hence, the length of the other diagonal is 6 cm.

7. The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per ${{\text{m}}^2}$ is Rs 4.

Ans: The area of rhombus is half of the product of its diagonals.

$A = \dfrac{1}{2} \times \left( {{\text{product of diagonals}}} \right)$

On substituting the value of diagonals in the formula, we get,

$A = \dfrac{1}{2} \times \left( {{\text{45}} \times {\text{30}}} \right)$

$A = 675\,\,{\text{c}}{{\text{m}}^2}$

Now, we will find the area of 3000 tiles.

${A_{3000}} = 675 \times 3000\,\,{\text{c}}{{\text{m}}^2}$

${A_{3000}} = 2025000\,\,{\text{c}}{{\text{m}}^2}$

Convert it into meters by dividing it by 10000.

${A_{3000}} = 202.5\,\,{{\text{m}}^2}$

Now, we will find the total cost of polishing the floor by multiplying the cost of polishing per square metre by area of 3000 tiles.

${\text{cost}} = {\text{Rs}}\,\left( {4 \times 202.5} \right)$

${\text{cost}} = {\text{Rs}}\,810$

Hence, the cost of polishing the floor of the building is Rs. 810.

8. Mohan wants to buy a trapezium shaped field. Its side along the river is parallel to and twice the side along the road. If the area of this field is 10500 ${{\text{m}}^{\text{2}}}$ and the perpendicular distance between the two parallel sides is 100m, find the length of the side along the river.

Ans: Let the length of the field along the road be $l$ m. Thus, the length of the field along the river is $2l$ m.

The area of a trapezium is the half the product of sum of parallel sides and distance between the parallel sides. Mathematically, $A = \dfrac{1}{2} \times \left( {a + b} \right) \times h$, where $a$ and $b$ are the parallel sides, and $h$ is the height of the trapezium.

Substitute $A$ as 10500 ${{\text{m}}^{\text{2}}}$, and $a$ as $l$, $b$ as $2l$, and $h$ as 100 in the formula.

$10500 = \dfrac{1}{2} \times \left( {l + 2l} \right) \times 100$

$10500 = \dfrac{1}{2} \times \left( {3l} \right) \times 100$

$10500 = \left( {3l} \right) \times 50$

$10500 = 150l$

Divide both sides by 150.

$\dfrac{{10500}}{{150}} = l$

$70 = l$

Thus, the length of the field along the river is twice the length of $l$.

$2l = 2 \times 70$

$2l = 140\,{\text{m}}$

Hence, the length of the side along the river is 140 m.

9. Top surface of a raised platform is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface.

Ans:Label the octagon as ${\text{ABCDEFGH}}$.

We can see that the octagon is divided into three parts, two trapeziums and one rectangle.

The area of trapezium ${\text{ABCH}}$ is equal to the area of trapezium ${\text{DEFG}}$.

The area of trapezium ${\text{ABCH}}$ is $A = \dfrac{1}{2} \times \left( {{\text{AB + HC}}} \right) \times {\text{H}}$.

${A_{{\text{ABCH}}}} = \dfrac{1}{2} \times \left( {{\text{11 + 5}}} \right) \times 4$

${A_{{\text{ABCH}}}} = \dfrac{1}{2} \times \left( {{\text{16}}} \right) \times 4$

${A_{{\text{ABCH}}}} = 32\,\,\,{{\text{m}}^2}$

Also, ${A_{{\text{DEFG}}}} = 32\,{{\text{m}}^2}$.

Now, we will find the area of the rectangle ${\text{HGDC}}$.

${A_{{\text{HGDC}}}} = 11 \times 5$

${A_{{\text{HGDC}}}} = 55\,{{\text{m}}^2}$

Now, the area of octagon is the sum of area of trapezium, ${\text{ABCH}}$, ${\text{DEFG}}$, and rectangle, ${\text{HGDC}}$.

$A = 32 + 32 + 55\,{{\text{m}}^2}$

$A = 119\,{{\text{m}}^2}$

Hence, the area is $119\,{{\text{m}}^2}$.

10. There is a pentagonal shaped park as shown in the figure. To find its area Jyoti and Kavita divided it in two different ways.

Find the area of this park using both ways. Can you suggest some other ways for finding its area?

Ans: Jyoti can find the area of the part in the following ways.

The first way to find the area is dividing the given pentagon into two trapeziums.

The area of this pentagon is twice the area of trapezium ${\text{ABCF}}$.

${A_{{\text{ABCF}}}} = 2 \times \dfrac{1}{2} \times \left( {15 + 30} \right) \times \dfrac{{15}}{2}\,\,{{\text{m}}^2}$

${A_{{\text{ABCF}}}} = 337.5\,\,{{\text{m}}^2}$

In the second way, she can divide the given pentagon into a square and a triangle.

${A_{{\text{ABCDE}}}} = \left[ {\dfrac{1}{2} \times 15 \times \left( {30 - 15} \right) + {{\left( {15} \right)}^2}} \right]\,{{\text{m}}^2}$

${A_{{\text{ABCDE}}}} = \left[ {\dfrac{1}{2} \times 15 \times \left( {15} \right) + {{\left( {15} \right)}^2}} \right]\,{{\text{m}}^2}$

${A_{{\text{ABCDE}}}} = 112.5 + 225\,{{\text{m}}^2}$

${A_{{\text{ABCDE}}}} = 337.5\,\,{{\text{m}}^2}$

11. Diagram Of The Adjacent Picture frame has outer Dimensions=24cm×28cm and inner dimensions 16cm×20cm. Find The area of each section Of The Frame, if the width of each section is same.

Ans:Divide and label the frame as follows.

Now, the width of each section is the same.

$IB = BJ = CK = CL = DM = DN = AO = AP$

Observe that $IL$ is equal to the sum of $IB$, $BC$, and $CL$.

$IL = IB + BC + CL$

Substitute $IL$ as 28, $BC$ as 20 cm.

$28 = IB + 20 + CL$

Subtract 20 from both sides.

$28 - 20 = IB + CL$

$8 = IB + CL$

Since, $IB = CL$ then, $IB = 4\,{\text{cm}}$.

Hence, $IB = BJ = CK = CL = DM = DN = AO = AP = 4\,{\text{cm}}$.

Now, the area of section ${\text{BEFC}}$ is equal to the area of section ${\text{DGHA}}$.

${A_{{\text{BEFC}}}} = \dfrac{1}{2} \times \left( {20 + 28} \right) \times 4\,{\text{c}}{{\text{m}}^2}$

${A_{{\text{BEFC}}}} = 96\,{\text{c}}{{\text{m}}^2}$

Hence, the area of each section is $96\,{\text{c}}{{\text{m}}^2}$.

## Conclusion

The NCERT Solutions for maths class 8 chapter 9 exercise 9.1, for Class 8 by Vedantu are crucial for understanding mensuration concepts. This exercise helps students learn to calculate the area and perimeter of various geometric shapes, a fundamental skill in geometry.

Important points to focus on class 8 math ex 9.1 includes mastering the formulas for area and perimeter and practising the step-by-step solutions provided. Regular practice with these solutions will improve problem-solving skills and build confidence in geometry. Vedantu's expert-curated solutions provide clear and detailed explanations, making exam preparation more effective. Download the solutions PDF to strengthen your understanding and perform well in your exams.

## Class 8 Maths Chapter 9: Exercises Breakdown

 Exercise Number of Questions Exercise 9.2 10 Questions with Solutions Exercise 9.3 8 Questions with Solutions

## CBSE Class 8 Maths Chapter 9 Other Study Materials

 S. No Important Links for Chapter 9 Mensuration 1 Class 8 Mensuration Important Questions 2 Class 8 Mensuration Revision Notes 3 Class 8 Mensuration Important Formulas 4 Class 8 Mensuration NCERT Exemplar Solution 5 Class 8 Mensuration RD Sharma Solutions

## Chapter-Specific NCERT Solutions for Class 8 Maths

Given below are the chapter-wise NCERT Solutions for Class 8 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 8 Maths Chapter 9 - Mensuration Exercise 9.1

1. Can I avail NCERT Solutions for Class 8 Maths Chapter 9Mensuration Exercise 9.1 online?

Yes, you can definitely avail NCERT Solutions for class 8 maths chapter Mensuration exercise 9.1 as well as other exercises online. It is available on Vedantu, India’s top online learning platform. It provides the free PDF of exercise-wise NCERT Solutions for 8 Maths Chapter 9 Mensuration. These solutions are really helpful while preparing for exams. These solutions are prepared by subject experts who have years of teaching experience and vast subject knowledge. Without any further delay, avail NCERT Solutions for Class 8 chapter 9 maths exercise 9.1.

2. What will students learn from class 8 maths Mensuration exercise 9.1?

Students will learn how to calculate areas of different shapes including a Trapezium, a general quadrilateral and a polygon.  These are two-dimensional figures. Students will be taught how to solve problems based on such figures in class 8 math ex 9.1. For having a clear understanding of the various topics associated with the exercise, students must download the free PDF of maths class 8 chapter 9 exercise 9.1 available on Vedantu.

3. How can NCERT Solutions maths chapter Mensuration ex 9.1 class 8 ensure optimum revision?

NCERT Solutions for maths chapter Mensuration ex 9.1 class 8 is a great material for revision. The free PDF of NCERT Solutions for class 8 ch 9 maths ex 9.1 can be utilized to learn the chapter effectively at first place and then for revision during the exam. As it includes stepwise explanations for the exercise problems, it allows students to have a deeper understanding, which is important during exams. Vedantu’s NCERT Solutions for class 8 maths chapter Mensuration exercise 9.1 are easily available over the internet. Mensuration is an important chapter from the exam perspective. In order to revise and score well, students must download these solutions available on Vedantu.

4. What is required to find the area of a trapezium?

We need to know the length of the parallel sides and the perpendicular distance between these two parallel sides to find the area of a trapezium. The area of trapezium is given by half the product of the sum of the lengths of parallel sides and the perpendicular distance between them.

5. What is the total number of problems in Exercise 9.1 of Chapter 9 of Class 8 Maths?

Exercise 9.1 of Chapter 9 Class 8 Maths contains 11 questions. These questions are based on the topics named “Area of a General Quadrilateral”, “Area of Special Quadrilaterals” and “Area of Polygon”. By solving these questions, students will get a brief introduction about the topics, and it will help in understanding them. The answers to the questions are available free of cost on the Vedantu website and the Vedantu app, in the form of a PDF file. They can take help from these solutions to clarify their doubts.

6. What is trapezium discussed in Exercise 9.1 of Chapter 9 of Class 8 Maths?

The topic “Trapezium” is discussed in Exercise 9.1 of Chapter 9 “Mensuration” of Class 8 Maths.

The quadrilateral whose two opposite sides are parallel and the other two sides are non-parallel is termed trapezium. This figure can be split into other two figures, i.e. triangle and rectangle. This splitting of the trapezium is helpful in finding its area.

The formula for finding the area of trapezium is:

Area of trapezium = ½ (a + b) * h

7. Note down some properties of trapezium which will be helpful in solving questions of Exercise 9.1 of Chapter 9 Class 8 Maths.

Some of the properties of a trapezium are discussed below:

• There are three types of the trapezium. These are-

1. Isosceles Trapezium

2. Scalene Trapezium

3. Right Trapezium

• The opposite sides or the bases are parallel to each other.

• The diagonals are of equal length in the trapezium.

• The intersecting property is shown by the diagonals of the trapezium.

• The sum of the adjacent interior angles of the trapezium is 180 degrees.

• The interior angles of the trapezium sum up to 360 degrees.

8. Where can I find the NCERT Solutions of Exercise 9.1 of Chapter 9 Class 8 Maths?

The NCERT Solutions of Exercise 9.1 of Chapter 9 Class 8 Maths are easily available on Vedantu. You just have to follow some steps to get them:

• Visit the page NCERT Solutions for Exercise 9.1 of Chapter 9 of Class 8 Maths.

• On the screen of your device, you will find that there are the NCERT Solutions of Exercise 9.1 of Chapter 9 Class 8 Maths PDF format.

• Above this PDF file, visit the option of “Download PDF” to get the PDF file.

9. How can I attain satisfactory marks in Exercise 9.1 of Chapter 9 Class 8 Maths?

Chapter 9 “Mensuration” of Class 8 Maths requires lots of calculation. Students need to practice more in order to score well in Exercise 9.1 of Chapter 9 of Class 8 Maths. Firstly, read the theory and learn the formulas used in Exercise9.1. After that, solve each example and the questions of the exercise to have a better understanding of the concepts. Be careful while doing the calculation part. Check your answer twice after solving it.