# NCERT Solutions for Class 8 Maths Chapter 11 Mensuration (EX 11.2) Exercise 11.2

## NCERT Solutions for Class 8 Maths Chapter 11 Mensuration (EX 11.2) Exercise 11.2

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## Access NCERT solutions for Class 8 Chapter 11 - Mensuration

Exercise 11.2

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.

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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 ${\text{BC}}$.

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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.

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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 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.

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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.

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

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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.

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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.

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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.

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

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

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

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

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.

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Ans:Divide and label the frame as follows.

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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 $96\,{\text{c}}{{\text{m}}^2}$.

## NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Exercise 11.2

Opting for the NCERT solutions for Ex 11.2 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 11.2 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 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 11.2 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.

Besides these NCERT solutions for Class 8 Maths Chapter 11 Exercise 11.2, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

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## FAQs on NCERT Solutions for Class 8 Maths Chapter 11 Mensuration (EX 11.2) Exercise 11.2

1. Can I avail NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Exercise 11.2 online?

Yes, you can definitely avail NCERT Solutions for Class 8 CBSE Mathematics Chapter 11 Mensuration Exercise 11.2 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 11 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 Maths Ex 11.2 Mensuration.

2. What will students learn from Exercise 11.2 of Class 8 Maths Chapter 11 Mensuration?

Students will learn how to calculate areas of different shapes including 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 Maths Chapter 11 Mensuration Ex 11.2. For having a clear understanding of the various topics associated with the exercise, students must download the free PDF of NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Exercise 11.2 available on Vedantu.

3. How can NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Exercise 11.2 ensure optimum revision?

NCERT Solutions for Exercise 11.2 Chapter 11 Mensuration of CBSE Class 8 is a great material for revision. The free PDF of NCERT Solutions for Class 8 Maths Chapter 11 Ex 11.2 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 11 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 11.2 of Chapter 11 of Class 8 Maths?

Exercise 11.2 of Chapter 11 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 11.2 of Chapter 11 of Class 8 Maths?

The topic “Trapezium” is discussed in Exercise 11.2 of Chapter 11 “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 11.2 of Chapter 11 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 11.2 of Chapter 11 Class 8 Maths?

The NCERT Solutions of Exercise 11.2 of Chapter 11 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 11.2 of Chapter 11 of Class 8 Maths.

• On the screen of your device, you will find that there are the NCERT Solutions of Exercise 11.2 of Chapter 11 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 11.2 of Chapter 11 Class 8 Maths?

Chapter 11 “Mensuration” of Class 8 Maths requires lots of calculation. Students need to practice more in order to score well in Exercise 11.2 of Chapter 11 of Class 8 Maths. First of all, read the theory and learn the formulas used in Exercise11.2. 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. SHARE TWEET SHARE SUBSCRIBE