NCERT Solutions for Class 8 Maths Exercise 11.2

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NCERT Solutions for Class 8 Maths Exercise 11.2 Class 8 Maths book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 8 Maths chapter wise NCERT solution for Maths Book all the chapters can be downloaded from our website and myCBSEguide mobile app for free.

CERT solutions for Class 8 Maths Mensuration Download as PDF

NCERT Solutions for Class 8 Maths Exercise 11.2

NCERT Solutions for Class 8 Maths Mensuration

Class –VIII Mathematics (Ex. 11.2)
NCERT SOLUTION
1.The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m.

Ans. Here one parallel side of the trapezium  = 1 m

And second side  = 1.2 m and

height  = 0.8 m

 Area of top surface of the table

=  =

=  = 0.88 m2

Hence surface area of the table is 0.88


NCERT Solutions for Class 8 Maths Exercise 11.2

2. The area of a trapezium is 34 and the length of one of the parallel sides is 10 cm and its height is 4 cm.

Find the length of the other parallel side.

Ans. Let the length of the other parallel side be

Length of one parallel side  = 10 am and height  = 4 cm

Area of trapezium =

 

 

 

 

 

 

 

Hence another required parallel

side is 7 cm.


3. Length of the fence of a trapezium shaped field ABCD is 120 m. If BC = 48 m, CD = 17 m and AD = 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.

Ans. Given: BC = 48 m, CD = 17 m,

AD = 40 m and perimeter = 120 m

 Perimeter of trapezium ABCD

= AB + BC + CD + DA

 120 = AB + 48 + 17 + 40

 120 = AB = 105

 AB = 120 – 105 = 15 m

Now Area of the field

=

=

=

= 660

Hence area of the field ABCD is 660.


NCERT Solutions for Class 8 Maths Exercise 11.2

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

Ans. Here  = 13 m,  = 8 m and

AC = 24 m

Area of quadrilateral ABCD

= Area of ABC + Area of ADC

=

=

=

=252

Hence required area of the field

is 252


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

Ans. Given:  =7.5 cm and  = 12 cm

We know that,

Area of rhombus =

=

= 45

Hence area of rhombus is 45 .


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

Ans. Since rhombus is also a kind of parallelogram.

 Area of rhombus

= Base Altitude

= 6  4 = 24

Also Area of rhombus =

 24 =

 

  = 6 cm

Hence the length of the other

diagonal is 6 cm.


NCERT Solutions for Class 8 Maths Exercise 11.2

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 is ` 4.

Ans. Here,  = 45 cm and  = 30 cm

 Area of one tile =

=  = 675

 Area of 3000 tiles

= 675  3000 = 2025000

=

= 202.50

 Cost of polishing the floor per

sq. meter = 4

 Cost of polishing the floor per 202.50 sq. meter = 4  202.50 =  810

Hence the total cost of polishing the floor is 810.


NCERT Solutions for Class 8 Maths Exercise 11.2

8. Mohan wants to buy a trapezium shaped field. Its side along the river is parallel to and twice the sidealong the road. If the area of this field is 10500 m2 and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.

Ans. Given: Perpendicular distance

= 100 m

Area of the trapezium shaped field

= 10500

Let side along the road be  m and side along the river =  m

 Area of the trapezium field

=

10500 =

m

Hence the side along the river =

= 2  70 = 140 m.


NCERT Solutions for Class 8 Maths Exercise 11.2

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. Given: Octagon having eight equal sides, each 5 m.

Construction:  Divided the octagon in 3 figures, two trapeziums whose parallel and perpendicular sides are 11 m and 4 m respectively and third figure is rectangle having length and breadth 11 m and 5 m respectively.

 

 

 

 

Now  Area of two trapeziums = 2 x

= 2  = 4  16 = 64

And Area of rectangle = length  breadth

= 11  5 = 55

 Total area of octagon = 64 + 55

= 119


NCERT Solutions for Class 8 Maths Exercise 11.2

10. There is a pentagonal shaped park as shown in the figure.

For finding its are a Jyoti and Kavita divided it in two different ways.

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

Ans. First way:  By Jyoti’s diagram,

Area of pentagon = Area of trapezium ABCP + Area of trapezium AEDP

=  (AP + BC) x CP +  (ED + AP)  DP      =  (30 + 15 ) x CP +  (15 + 30)  DP

=  (30 + 15) (CP + DP)

=   45 CD

=337.5 m2

Second way:

By Kavita’s diagram

Here, a perpendicular AM drawn to BE.

AM = 30 – 15 = 15 m

Area of pentagon

= Area of ABE + Area of square BCDE

=1515+1515m

= 112.5 + 225.0

= 337.5 m2

Hence total area of pentagon shaped

park = 337.5.


NCERT Solutions for Class 8 Maths Exercise 11.2

11. Diagram of the adjacent picture frame has outer dimensions = 24 cm  28 cm and inner dimensions 16 cm  20 cm. Find the area of each section of theframe, if the width of each section is same.

Ans. Here two of given figures (I) and (II) are similar in dimensions.

And also figures (III) and (IV) are similar in dimensions.

 Area of figure (I) = Area of trapezium

=

=

=

= 96

Also Area of figure (II) = 96 cm2

Now Area of figure (III)

= Area of trapezium =

=  =  = 80

Also Area of figure (IV) = 80

NCERT Solutions for Class 8 Maths Exercise 11.2

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