# NCERT Solutions for Class 8 Maths Chapter 11

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Exercise 11.1, Exercise 11.2, Exercise 11.3 and Exercise 11.4 in English Medium and Hindi Medium updated for new academic session 2020-2021 following new syllabus.

Download Prashnavali 11.1, Prashnavali 11.2, Prashnavali 11.3 and Prashnavali 11.4 in Hindi Medium to study online or free PDF download. All NCERT Solutions 2020-21 are updated as per the latest CBSE Syllabus for the academic session 2020-2021. Students can download NCERT Books 2020-21 in the PDF file format to use it offline.

## NCERT Solutions for Class 8 Maths Chapter 11

 Class: 8 Subject: Maths – गणित Chapter 11: Mensuration

### Class 8 Maths Chapter 11 Solutions

Class 8 Maths Chapter 11 Mensuration all exercises in English Medium as well as Hindi Medium are given below to study online or download in PDF form. Download NCERT Solutions offline apps, which work without internet, once downloaded. All the solutions given in apps are updated for academic session 2020-21.

• ### 8 Maths Chapter 11 Solutions in Hindi Medium

#### Class 8 Maths Exercise 11.1 & 11.2 Solutions in Video

Class 8 Maths Exercise 11.1 Solutions in Video
Class 8 Maths Exercise 11.2 Solutions in Video

#### Class 8 Maths Exercise 11.3 & 11.4 Solutions in Video

Class 8 Maths Exercise 11.3 Solutions in Video
Class 8 Maths Exercise 11.4 Solutions in Video

#### Important Terms about Class 8 Maths Chapter 11

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

Base of flooring tile = 24 cm = 0.24 m
Corresponding height of a flooring tile = 10 cm = 0.10 m
Now
Area of flooring tile = Base x Altitude
= 0.24 x 0.10
= 0.024 m^2

Number of tiles required to cover the floor
= Area of Floor/Area of one tile
= 1080/0.024
= 45000 tiles
Hence 45000 tiles are required to cover the floor.

##### 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.

Since rhombus is also a kind of parallelogram.
Area of rhombus = Base x Altitude = 6 x 4 = 24 cm^2
Also, Area of rhombus = 1/2 x d1 x d2
According to question,
24 = 1/2 x 8 x d2
So, 4d2 = 24 and d2 = 6 cm
Hence, the length of the other diagonal is 6 cm.

##### किसी समचतुर्भुज के विकर्ण 7.5 cm एवं 12 cm है। इसका क्षेत्रफल ज्ञात कीजिए।

दिया है: d1 =7.5 cm और d2 = 12 cm
हम जानते हैं कि, समचतुर्भुज का क्षेत्रफल
= 1/2 x d1 x d2
= 1/2 x 7.5 x 12
= 45 cm^2
अतः, समचतुर्भुज का क्षेत्रफल 45 cm^2 है।

##### एक ऐसे घन की भुजा ज्ञात कीजिए जिसका पृष्ठीय क्षेत्रफल 600 cm^2 है।

दिया है: घन का पृष्ठीय क्षेत्रफल = 600 cm^2
6x^2 = 600
x^2 = 100
x = 10 cm
अतः, घन की भुजा 10 cm है।

In Chapter 11 Mensuration, we have to find the Areas, Perimeter, missing dimensions of the geometrical figures. Area of triangle, square, parallelogram, trapezium on the basis of given information and applying simple formula. Area of various specific quadrilateral like parallelogram, rhombus, rectangle, square and general quadrilateral in which we divide the quadrilateral into two triangles and then find the area of triangles and add them. Similarly we can find the area of polygon by dividing it into triangles and adding the areas of all triangles. Surface area and volumes of solid figures like Cube, Cuboids, Cylinders, Cones, etc. are given to find one the basis of application of formulae.

##### Do you know?

इस पाठ में प्रयुक्त विधियां हमें कक्षा ९ तथा १० में भी प्रयोग करनी पड़ती हैं। अतः यह पाठ आने वाली कक्षाओं की दृष्टि से भी महत्वपूर्ण है। विद्यार्थी इस पाठ में प्रयोग होने वाले सभी तथ्यों की ध्यान से समझें ताकि उत्तरोत्तर कक्षाओं में कोई भी परेशानी न हो।                            