NCERT Solutions for Class 8 Maths Chapter 11 - Exercise

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.

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