Open App

Class 8 Exam > Class 8 Notes > Mathematics (Maths) Class 8 > Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.3

Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.3 | Mathematics (Maths) Class 8 PDF Download

Download, print and study this document offline

Please wait while the PDF view is loading

 Page 1


 
 
 
 
RD Sharma Solutions for Class 8 Maths Chapter 20 – 
Mensuration – I (Area of a Trapezium and a Polygon) 
 
EXERCISE 20.3                                                  PAGE NO: 20.28 
 
1. Find the area of the pentagon shown in fig. 20.48, if AD = 10 cm, AG = 8 
cm, AH = 6 cm, AF = 5 cm, BF = 5 cm, CG = 7 cm and EH = 3 cm. 
 
 
Solution: 
GH = AG – AH = 8 – 6 = 2 cm 
HF = AH – AF = 6 – 5 = 1 cm 
GD = AD – AG = 10 – 8 = 2 cm 
From the figure we can write, 
Area of given figure = Area of triangle AFB + Area of trapezium BCGF + Area of 
triangle CGD + Area of triangle AHE + Area of triangle EGD 
We know that, 
Area of right angled triangle = 1/2 × base × altitude 
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude 
Area of given pentagon = 1/2 × AF × BF + 1/2 (CG + BF) × FG + 1/2 × GD × CG + 1/2 
× AH × EH + 1/2 × HD × EH 
Area of given pentagon = 1/2 × 5 × 5 + 1/2 (7 + 5) × 3 + 1/2 × 2 × 7 + 1/2 × 6 × 3 + 1/2 × 
4 × 3 
Area of given pentagon = 12.5 + 18 + 7 + 9 + 6 = 52.5 
? Area of given pentagon = 52.5 cm
2
 
 
2. Find the area enclosed by each of the following figures [fig. 20.49 (i)-(ii)] as the 
sum of the areas of a rectangle and a trapezium. 
Page 2


 
 
 
 
RD Sharma Solutions for Class 8 Maths Chapter 20 – 
Mensuration – I (Area of a Trapezium and a Polygon) 
 
EXERCISE 20.3                                                  PAGE NO: 20.28 
 
1. Find the area of the pentagon shown in fig. 20.48, if AD = 10 cm, AG = 8 
cm, AH = 6 cm, AF = 5 cm, BF = 5 cm, CG = 7 cm and EH = 3 cm. 
 
 
Solution: 
GH = AG – AH = 8 – 6 = 2 cm 
HF = AH – AF = 6 – 5 = 1 cm 
GD = AD – AG = 10 – 8 = 2 cm 
From the figure we can write, 
Area of given figure = Area of triangle AFB + Area of trapezium BCGF + Area of 
triangle CGD + Area of triangle AHE + Area of triangle EGD 
We know that, 
Area of right angled triangle = 1/2 × base × altitude 
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude 
Area of given pentagon = 1/2 × AF × BF + 1/2 (CG + BF) × FG + 1/2 × GD × CG + 1/2 
× AH × EH + 1/2 × HD × EH 
Area of given pentagon = 1/2 × 5 × 5 + 1/2 (7 + 5) × 3 + 1/2 × 2 × 7 + 1/2 × 6 × 3 + 1/2 × 
4 × 3 
Area of given pentagon = 12.5 + 18 + 7 + 9 + 6 = 52.5 
? Area of given pentagon = 52.5 cm
2
 
 
2. Find the area enclosed by each of the following figures [fig. 20.49 (i)-(ii)] as the 
sum of the areas of a rectangle and a trapezium. 
 
 
 
 
RD Sharma Solutions for Class 8 Maths Chapter 20 – 
Mensuration – I (Area of a Trapezium and a Polygon) 
 
 
 
Solution: 
Figure (i) 
From the figure we can write, 
Area of figure = Area of trapezium + Area of rectangle 
Area of figure = 1/2 (Sum of lengths of parallel sides) × altitude + Length × Breadth 
Area of figure = 1/2 (18 + 7) × 8 + 18 × 18 
Area of figure = 1/2 (25) × 8 + 18 × 18 
Area of figure =  
? Area of figure is 424 cm
2
 
 
Figure (ii) 
From the figure we can write, 
Area of figure = Area of trapezium + Area of rectangle 
Area of figure = 1/2 (Sum of lengths of parallel sides) × altitude + Length × Breadth 
Area of given figure = 1/2 (15 + 6) × 8 + 15 × 20 
Area of given figure = 84 + 300 = 384 
? Area of figure is 384 cm
2
 
 
Figure (iii) 
Using Pythagoras theorem in the right angled triangle, 
5
2
 = 4
2
 + x
2
 
x
2
 = 25 – 16 
x
2
 = 9 
x = 3 cm 
From the figure we can write, 
Area of figure = Area of trapezium + Area of rectangle 
Area of figure = 1/2 (Sum of lengths of parallel sides) × altitude + Length × Breadth 
Page 3


 
 
 
 
RD Sharma Solutions for Class 8 Maths Chapter 20 – 
Mensuration – I (Area of a Trapezium and a Polygon) 
 
EXERCISE 20.3                                                  PAGE NO: 20.28 
 
1. Find the area of the pentagon shown in fig. 20.48, if AD = 10 cm, AG = 8 
cm, AH = 6 cm, AF = 5 cm, BF = 5 cm, CG = 7 cm and EH = 3 cm. 
 
 
Solution: 
GH = AG – AH = 8 – 6 = 2 cm 
HF = AH – AF = 6 – 5 = 1 cm 
GD = AD – AG = 10 – 8 = 2 cm 
From the figure we can write, 
Area of given figure = Area of triangle AFB + Area of trapezium BCGF + Area of 
triangle CGD + Area of triangle AHE + Area of triangle EGD 
We know that, 
Area of right angled triangle = 1/2 × base × altitude 
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude 
Area of given pentagon = 1/2 × AF × BF + 1/2 (CG + BF) × FG + 1/2 × GD × CG + 1/2 
× AH × EH + 1/2 × HD × EH 
Area of given pentagon = 1/2 × 5 × 5 + 1/2 (7 + 5) × 3 + 1/2 × 2 × 7 + 1/2 × 6 × 3 + 1/2 × 
4 × 3 
Area of given pentagon = 12.5 + 18 + 7 + 9 + 6 = 52.5 
? Area of given pentagon = 52.5 cm
2
 
 
2. Find the area enclosed by each of the following figures [fig. 20.49 (i)-(ii)] as the 
sum of the areas of a rectangle and a trapezium. 
 
 
 
 
RD Sharma Solutions for Class 8 Maths Chapter 20 – 
Mensuration – I (Area of a Trapezium and a Polygon) 
 
 
 
Solution: 
Figure (i) 
From the figure we can write, 
Area of figure = Area of trapezium + Area of rectangle 
Area of figure = 1/2 (Sum of lengths of parallel sides) × altitude + Length × Breadth 
Area of figure = 1/2 (18 + 7) × 8 + 18 × 18 
Area of figure = 1/2 (25) × 8 + 18 × 18 
Area of figure =  
? Area of figure is 424 cm
2
 
 
Figure (ii) 
From the figure we can write, 
Area of figure = Area of trapezium + Area of rectangle 
Area of figure = 1/2 (Sum of lengths of parallel sides) × altitude + Length × Breadth 
Area of given figure = 1/2 (15 + 6) × 8 + 15 × 20 
Area of given figure = 84 + 300 = 384 
? Area of figure is 384 cm
2
 
 
Figure (iii) 
Using Pythagoras theorem in the right angled triangle, 
5
2
 = 4
2
 + x
2
 
x
2
 = 25 – 16 
x
2
 = 9 
x = 3 cm 
From the figure we can write, 
Area of figure = Area of trapezium + Area of rectangle 
Area of figure = 1/2 (Sum of lengths of parallel sides) × altitude + Length × Breadth 
 
 
 
 
RD Sharma Solutions for Class 8 Maths Chapter 20 – 
Mensuration – I (Area of a Trapezium and a Polygon) 
 
Area of given figure = 1/2 (14 + 6) × 3 + 4 × 6 
Area of given figure = 30 + 24 = 54 
? Area of figure is 54 cm
2
 
 
3. There is a pentagonal shaped park as shown in Fig. 20.50. Jyoti and Kavita 
divided it in two different ways. 
Find the area of this park using both ways. Can you suggest some another way of 
finding its area? 
 
Solution: 
From the figure we can write, 
Area of figure = Area of trapezium + Area of rectangle 
Area of Jyoti’s diagram = 2 × 1/2 (Sum of lengths of parallel sides) × altitude 
Area of figure = 2 × 1/2 × (15 + 30) × 7.5 
Area of figure = 45 × 7.5 = 337.5 
Therefore, Area of figure = 337.5 cm
2
 
We also know that, 
Area of Pentagon = Area of triangle + area of rectangle 
Area of Pentagon = 1/2 × Base × Altitude + Length × Breadth  
Area of Pentagon = 1/2 × 15 × 15 + 15 × 15 
Area of Pentagon = 112.5 + 225 = 337.5 
? Area of pentagon is 337.5 m
2
 
 
4. Find the area of the following polygon, if AL = 10 cm, AM = 20 cm, AN = 50 
cm. AO = 60 cm and AD = 90 cm. 
Page 4


 
 
 
 
RD Sharma Solutions for Class 8 Maths Chapter 20 – 
Mensuration – I (Area of a Trapezium and a Polygon) 
 
EXERCISE 20.3                                                  PAGE NO: 20.28 
 
1. Find the area of the pentagon shown in fig. 20.48, if AD = 10 cm, AG = 8 
cm, AH = 6 cm, AF = 5 cm, BF = 5 cm, CG = 7 cm and EH = 3 cm. 
 
 
Solution: 
GH = AG – AH = 8 – 6 = 2 cm 
HF = AH – AF = 6 – 5 = 1 cm 
GD = AD – AG = 10 – 8 = 2 cm 
From the figure we can write, 
Area of given figure = Area of triangle AFB + Area of trapezium BCGF + Area of 
triangle CGD + Area of triangle AHE + Area of triangle EGD 
We know that, 
Area of right angled triangle = 1/2 × base × altitude 
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude 
Area of given pentagon = 1/2 × AF × BF + 1/2 (CG + BF) × FG + 1/2 × GD × CG + 1/2 
× AH × EH + 1/2 × HD × EH 
Area of given pentagon = 1/2 × 5 × 5 + 1/2 (7 + 5) × 3 + 1/2 × 2 × 7 + 1/2 × 6 × 3 + 1/2 × 
4 × 3 
Area of given pentagon = 12.5 + 18 + 7 + 9 + 6 = 52.5 
? Area of given pentagon = 52.5 cm
2
 
 
2. Find the area enclosed by each of the following figures [fig. 20.49 (i)-(ii)] as the 
sum of the areas of a rectangle and a trapezium. 
 
 
 
 
RD Sharma Solutions for Class 8 Maths Chapter 20 – 
Mensuration – I (Area of a Trapezium and a Polygon) 
 
 
 
Solution: 
Figure (i) 
From the figure we can write, 
Area of figure = Area of trapezium + Area of rectangle 
Area of figure = 1/2 (Sum of lengths of parallel sides) × altitude + Length × Breadth 
Area of figure = 1/2 (18 + 7) × 8 + 18 × 18 
Area of figure = 1/2 (25) × 8 + 18 × 18 
Area of figure =  
? Area of figure is 424 cm
2
 
 
Figure (ii) 
From the figure we can write, 
Area of figure = Area of trapezium + Area of rectangle 
Area of figure = 1/2 (Sum of lengths of parallel sides) × altitude + Length × Breadth 
Area of given figure = 1/2 (15 + 6) × 8 + 15 × 20 
Area of given figure = 84 + 300 = 384 
? Area of figure is 384 cm
2
 
 
Figure (iii) 
Using Pythagoras theorem in the right angled triangle, 
5
2
 = 4
2
 + x
2
 
x
2
 = 25 – 16 
x
2
 = 9 
x = 3 cm 
From the figure we can write, 
Area of figure = Area of trapezium + Area of rectangle 
Area of figure = 1/2 (Sum of lengths of parallel sides) × altitude + Length × Breadth 
 
 
 
 
RD Sharma Solutions for Class 8 Maths Chapter 20 – 
Mensuration – I (Area of a Trapezium and a Polygon) 
 
Area of given figure = 1/2 (14 + 6) × 3 + 4 × 6 
Area of given figure = 30 + 24 = 54 
? Area of figure is 54 cm
2
 
 
3. There is a pentagonal shaped park as shown in Fig. 20.50. Jyoti and Kavita 
divided it in two different ways. 
Find the area of this park using both ways. Can you suggest some another way of 
finding its area? 
 
Solution: 
From the figure we can write, 
Area of figure = Area of trapezium + Area of rectangle 
Area of Jyoti’s diagram = 2 × 1/2 (Sum of lengths of parallel sides) × altitude 
Area of figure = 2 × 1/2 × (15 + 30) × 7.5 
Area of figure = 45 × 7.5 = 337.5 
Therefore, Area of figure = 337.5 cm
2
 
We also know that, 
Area of Pentagon = Area of triangle + area of rectangle 
Area of Pentagon = 1/2 × Base × Altitude + Length × Breadth  
Area of Pentagon = 1/2 × 15 × 15 + 15 × 15 
Area of Pentagon = 112.5 + 225 = 337.5 
? Area of pentagon is 337.5 m
2
 
 
4. Find the area of the following polygon, if AL = 10 cm, AM = 20 cm, AN = 50 
cm. AO = 60 cm and AD = 90 cm. 
 
 
 
 
RD Sharma Solutions for Class 8 Maths Chapter 20 – 
Mensuration – I (Area of a Trapezium and a Polygon) 
 
 
 
Solution: 
Given that, 
AL = 10 cm; AM = 20 cm; AN = 50 cm; AO = 60 cm; AD = 90 cm 
LM = AM – AL = 20 – 10 = 10 cm 
MN = AN – AM = 50 – 20 = 30 cm 
OD = AD – AO = 90 – 60 = 30 cm 
ON = AO – AN = 60 – 50 = 10 cm 
DN = OD + ON = 30 + 10 = 40 cm 
OM = MN + ON = 30 + 10 = 40 cm 
LN = LM + MN = 10 + 30 = 40 cm 
From the figure we can write, 
Area of figure = Area of triangle AMF + Area of trapezium FMNE + Area of triangle 
END + Area of triangle ALB + Area of trapezium LBCN + Area of triangle DNC 
We know that, 
Area of right angled triangle = 1/2 × base × altitude 
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude 
Area of given hexagon = 1/2 × AM × FM + 1/2 (MF + OE) × OM + 1/2 × OD × OE + 
1/2 × AL × BL + 1/2 × (BL + CN) × LN + 1/2 × DN × CN 
Area of given hexagon = 1/2 × 20 × 20 + 1/2 (20 + 60) × 40 + 1/2 × 30 × 60 + 1/2 × 10 × 
30 + 1/2 × (30 + 40) × 40 + 1/2 × 40 × 40 
Area of given hexagon = 200 + 1600 + 900 + 150 + 1400 + 800 = 5050 
? Area of given hexagon is 5050 cm
2
 
 
5. Find the area of the following regular hexagon. 
Page 5


 
 
 
 
RD Sharma Solutions for Class 8 Maths Chapter 20 – 
Mensuration – I (Area of a Trapezium and a Polygon) 
 
EXERCISE 20.3                                                  PAGE NO: 20.28 
 
1. Find the area of the pentagon shown in fig. 20.48, if AD = 10 cm, AG = 8 
cm, AH = 6 cm, AF = 5 cm, BF = 5 cm, CG = 7 cm and EH = 3 cm. 
 
 
Solution: 
GH = AG – AH = 8 – 6 = 2 cm 
HF = AH – AF = 6 – 5 = 1 cm 
GD = AD – AG = 10 – 8 = 2 cm 
From the figure we can write, 
Area of given figure = Area of triangle AFB + Area of trapezium BCGF + Area of 
triangle CGD + Area of triangle AHE + Area of triangle EGD 
We know that, 
Area of right angled triangle = 1/2 × base × altitude 
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude 
Area of given pentagon = 1/2 × AF × BF + 1/2 (CG + BF) × FG + 1/2 × GD × CG + 1/2 
× AH × EH + 1/2 × HD × EH 
Area of given pentagon = 1/2 × 5 × 5 + 1/2 (7 + 5) × 3 + 1/2 × 2 × 7 + 1/2 × 6 × 3 + 1/2 × 
4 × 3 
Area of given pentagon = 12.5 + 18 + 7 + 9 + 6 = 52.5 
? Area of given pentagon = 52.5 cm
2
 
 
2. Find the area enclosed by each of the following figures [fig. 20.49 (i)-(ii)] as the 
sum of the areas of a rectangle and a trapezium. 
 
 
 
 
RD Sharma Solutions for Class 8 Maths Chapter 20 – 
Mensuration – I (Area of a Trapezium and a Polygon) 
 
 
 
Solution: 
Figure (i) 
From the figure we can write, 
Area of figure = Area of trapezium + Area of rectangle 
Area of figure = 1/2 (Sum of lengths of parallel sides) × altitude + Length × Breadth 
Area of figure = 1/2 (18 + 7) × 8 + 18 × 18 
Area of figure = 1/2 (25) × 8 + 18 × 18 
Area of figure =  
? Area of figure is 424 cm
2
 
 
Figure (ii) 
From the figure we can write, 
Area of figure = Area of trapezium + Area of rectangle 
Area of figure = 1/2 (Sum of lengths of parallel sides) × altitude + Length × Breadth 
Area of given figure = 1/2 (15 + 6) × 8 + 15 × 20 
Area of given figure = 84 + 300 = 384 
? Area of figure is 384 cm
2
 
 
Figure (iii) 
Using Pythagoras theorem in the right angled triangle, 
5
2
 = 4
2
 + x
2
 
x
2
 = 25 – 16 
x
2
 = 9 
x = 3 cm 
From the figure we can write, 
Area of figure = Area of trapezium + Area of rectangle 
Area of figure = 1/2 (Sum of lengths of parallel sides) × altitude + Length × Breadth 
 
 
 
 
RD Sharma Solutions for Class 8 Maths Chapter 20 – 
Mensuration – I (Area of a Trapezium and a Polygon) 
 
Area of given figure = 1/2 (14 + 6) × 3 + 4 × 6 
Area of given figure = 30 + 24 = 54 
? Area of figure is 54 cm
2
 
 
3. There is a pentagonal shaped park as shown in Fig. 20.50. Jyoti and Kavita 
divided it in two different ways. 
Find the area of this park using both ways. Can you suggest some another way of 
finding its area? 
 
Solution: 
From the figure we can write, 
Area of figure = Area of trapezium + Area of rectangle 
Area of Jyoti’s diagram = 2 × 1/2 (Sum of lengths of parallel sides) × altitude 
Area of figure = 2 × 1/2 × (15 + 30) × 7.5 
Area of figure = 45 × 7.5 = 337.5 
Therefore, Area of figure = 337.5 cm
2
 
We also know that, 
Area of Pentagon = Area of triangle + area of rectangle 
Area of Pentagon = 1/2 × Base × Altitude + Length × Breadth  
Area of Pentagon = 1/2 × 15 × 15 + 15 × 15 
Area of Pentagon = 112.5 + 225 = 337.5 
? Area of pentagon is 337.5 m
2
 
 
4. Find the area of the following polygon, if AL = 10 cm, AM = 20 cm, AN = 50 
cm. AO = 60 cm and AD = 90 cm. 
 
 
 
 
RD Sharma Solutions for Class 8 Maths Chapter 20 – 
Mensuration – I (Area of a Trapezium and a Polygon) 
 
 
 
Solution: 
Given that, 
AL = 10 cm; AM = 20 cm; AN = 50 cm; AO = 60 cm; AD = 90 cm 
LM = AM – AL = 20 – 10 = 10 cm 
MN = AN – AM = 50 – 20 = 30 cm 
OD = AD – AO = 90 – 60 = 30 cm 
ON = AO – AN = 60 – 50 = 10 cm 
DN = OD + ON = 30 + 10 = 40 cm 
OM = MN + ON = 30 + 10 = 40 cm 
LN = LM + MN = 10 + 30 = 40 cm 
From the figure we can write, 
Area of figure = Area of triangle AMF + Area of trapezium FMNE + Area of triangle 
END + Area of triangle ALB + Area of trapezium LBCN + Area of triangle DNC 
We know that, 
Area of right angled triangle = 1/2 × base × altitude 
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude 
Area of given hexagon = 1/2 × AM × FM + 1/2 (MF + OE) × OM + 1/2 × OD × OE + 
1/2 × AL × BL + 1/2 × (BL + CN) × LN + 1/2 × DN × CN 
Area of given hexagon = 1/2 × 20 × 20 + 1/2 (20 + 60) × 40 + 1/2 × 30 × 60 + 1/2 × 10 × 
30 + 1/2 × (30 + 40) × 40 + 1/2 × 40 × 40 
Area of given hexagon = 200 + 1600 + 900 + 150 + 1400 + 800 = 5050 
? Area of given hexagon is 5050 cm
2
 
 
5. Find the area of the following regular hexagon. 
RD Sharma Solutions for Class 8 Maths Chapter 20 – 
Mensuration – I (Area of a Trapezium and a Polygon)
Solution: 
Given that, 
NQ = 23 cm 
NA = BQ = 10/2 = 5 cm 
MR = OP = 13 cm
In the right triangle BPQ 
PQ
2
 = BQ
2
 + BP
2
 
Substituting the values 
(13)
2
 = (5)
2
 + BP
2
169 = 25 + BP
2
So we get
BP
2
 = 169 - 25 = 144
BP = 12 cm
Here
PR = MO = 2 × 12 = 24 cm
Area of rectangle RPOM = RP × PO = 24 × 13 = 321 cm
2
Area of triangle PRQ = 1/2 × PR × BQ
= 1/2 × 24 × 5
= 60 cm
2
Area of triangle MON = 60 cm
2
Area of hexagon = 312 + 60 + 60 = 432 cm
2
Therefore, area of given hexagon is 432 cm
2

	Mathematics (Maths) Class 8 79 videos\|408 docs\|31 tests

Mathematics (Maths) Class 8

79 videos|408 docs|31 tests

Join Course for Free

Top Courses for Class 8

View all

Related Exams

Class 8

About this Document

	4.93/5 Rating
	Nov 24, 2024 Last updated

Document Description: Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.3 for Class 8 2024 is part of Mathematics (Maths) Class 8 preparation. The notes and questions for Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.3 have been prepared according to the Class 8 exam syllabus. Information about Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.3 covers topics like and Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.3 Example, for Class 8 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.3.

Introduction of Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.3 in English is available as part of our Mathematics (Maths) Class 8 for Class 8 & Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.3 in Hindi for Mathematics (Maths) Class 8 course. Download more important topics related with notes, lectures and mock test series for Class 8 Exam by signing up for free. Class 8: Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.3 | Mathematics (Maths) Class 8

Description

Full syllabus notes, lecture & questions for Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.3 | Mathematics (Maths) Class 8 - Class 8 | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) Class 8 | Best notes, free PDF download

Information about Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.3

In this doc you can find the meaning of Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.3 defined & explained in the simplest way possible. Besides explaining types of Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.3 theory, EduRev gives you an ample number of questions to practice Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.3 tests, examples and also practice Class 8 tests

	Mathematics (Maths) Class 8 79 videos\|408 docs\|31 tests

Mathematics (Maths) Class 8

79 videos|408 docs|31 tests

Join Course for Free

Download as PDF

Explore Courses for Class 8 exam

Top Courses for Class 8

Explore Courses

Signup for Free!

Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.

Start learning for Free

10M+ students study on EduRev

Summary

Exam

Important questions

Semester Notes

past year papers

Previous Year Questions with Solutions

Objective type Questions

Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.3 | Mathematics (Maths) Class 8

pdf

Free

mock tests for examination

video lectures

Extra Questions

study material

MCQs

practice quizzes

Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.3 | Mathematics (Maths) Class 8

Viva Questions

shortcuts and tricks

Sample Paper

Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.3 | Mathematics (Maths) Class 8

ppt

;

Additional Information about Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.3 for Class 8 Preparation

Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.3 Free PDF Download

The Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.3 is an invaluable resource that delves deep into the core of the Class 8 exam. These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective. With the help of these notes, you can grasp complex subjects quickly, revise important points easily, and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner, allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material, or toppers' notes, this PDF has got you covered. Download the Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.3 now and kickstart your journey towards success in the Class 8 exam.

Importance of Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.3

The importance of Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.3 cannot be overstated, especially for Class 8 aspirants. This document holds the key to success in the Class 8 exam. It offers a detailed understanding of the concept, providing invaluable insights into the topic. By knowing the concepts well in advance, students can plan their preparation effectively. Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.

Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.3 Notes

Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.3 Notes offer in-depth insights into the specific topic to help you master it with ease. This comprehensive document covers all aspects related to Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.3. It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation. Practice papers and question papers enable you to assess your progress effectively. Additionally, the paper analysis provides valuable tips for tackling the exam strategically. Access to Toppers' notes gives you an edge in understanding complex concepts. Whether you're a beginner or aiming for advanced proficiency, Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.3 Notes on EduRev are your ultimate resource for success.

Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.3 Class 8 Questions

The "Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.3 Class 8 Questions" guide is a valuable resource for all aspiring students preparing for the Class 8 exam. It focuses on providing a wide range of practice questions to help students gauge their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation. The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level. Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.

Study Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.3 on the App

Students of Class 8 can study Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.3 alongwith tests & analysis from the EduRev app, which will help them while preparing for their exam. Apart from the Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.3, students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc. The EduRev App will make your learning easier as you can access it from anywhere you want. The content of Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.3 is prepared as per the latest Class 8 syllabus.

Education Revolution