Area is the measure of the surface covered by a closed shape. Important Questions Class 8 Maths Part 2 Chapter 7 cover the area of triangles, parallelograms, rhombus, trapezium, rectangles, polygons, paths, and unit conversion from Ganita Prakash Part 2.
A triangle drawn inside a rectangle, a path around a park, and a trapezium-shaped field all need the same skill: choosing the correct base and height.
Class 8 Maths Chapter 7 Area teaches students how to measure space covered by different shapes. The chapter becomes easier when students stop memorising formulas blindly and start asking: What is the base? What is the height? Can the shape be divided into simpler parts?
Key Takeaways
| Detail |
Information |
| Chapter |
Class 8 Maths Part 2 Chapter 7 |
| Topic |
Area |
| Book |
Ganita Prakash Class 8 Part 2 |
| Syllabus |
NCERT 2026-27 |
| Question Types |
Formula-based, application, reasoning, unit conversion, MCQs |
| Key Topics |
Triangle, parallelogram, rhombus, trapezium, rectangle, path problems, unit conversion |
Class 8 Maths Part 2 Chapter List
| S.No. |
Chapter Name |
| 1 |
Chapter 1 - Fractions in Disguise |
| 2 |
Chapter 2 - The Baudhayana-Pythagoras Theorem |
| 3 |
Chapter 3 - Proportional Reasoning II |
| 4 |
Chapter 4 - Exploring Some Geometric Themes |
| 5 |
Chapter 5 - Tales by Dots and Lines |
| 6 |
Chapter 6 - Algebra Play |
| 7 |
Chapter 7 - Area |
Class 8 Maths Chapter 7 Area: Topics Covered
Before solving any question from this chapter, identify the shape first. Then mark the base, height, diagonals, or parallel sides needed for the formula.
- Area of rectangle and square
- Area of triangle
- Area of parallelogram
- Area of rhombus
- Area of trapezium
- Area of polygons by splitting into triangles
- Area of paths and crossroads
- Unit conversion in area problems
- Dissection method for area formulas
- Real-life area applications
Important Questions Class 8 Maths Chapter 7 Area with Answers
Important Questions Class 8 Maths Chapter 7 Area should be solved shape by shape. Start with triangle and parallelogram formulas because trapezium, rhombus, and polygon questions often build on them.
These class 8 maths chapter 7 area important questions with answers cover formula-based questions, reasoning questions, MCQs, and real-life area problems.
Class 8 Maths Chapter 7 Area Important Questions with Answers
Each class 8 maths chapter 7 area question below is arranged by topic. Use the formula only after checking whether the given height is perpendicular to the base.
Very Short Answer Important Questions Class 8 Maths Part 2 Chapter 7
Q1. What is area?
Ans. Area is the measure of the surface covered by a closed figure. It is measured in square units like cm², m², or in².
Q2. What is the formula for the area of a rectangle?
Ans. Area of rectangle = Length × Breadth.
Q3. What is the formula for the area of a triangle?
Ans. Area of triangle = 1/2 × base × height.
Q4. What is the formula for the area of a parallelogram?
Ans. Area of parallelogram = base × height.
Q5. What is the formula for the area of a rhombus using diagonals?
Ans. Area of rhombus = 1/2 × d₁ × d₂.
Q6. What is the formula for the area of a trapezium?
Ans. Area of trapezium = 1/2 × height × sum of parallel sides.
Q7. Why is height important in area questions?
Ans. Height gives the perpendicular distance from the base. Slant side should not be used as height unless it is perpendicular.
Q8. How do you find the area of an irregular polygon?
Ans. Divide the polygon into triangles, rectangles, or trapeziums. Find the area of each part and add them.
Area of Triangle Class 8 Questions
Area of triangle class 8 questions depend on base and perpendicular height. The triangle may be acute, right-angled, or obtuse, but the formula remains the same.
Always check whether the given height meets the base at a right angle.
Triangle Area Formula Questions
Q1. Find the area of a triangle with base 4 cm and height 3 cm.
Ans.
Area = 1/2 × base × height
= 1/2 × 4 × 3
= 6 cm²
Q2. Find the area of a triangle with base 5 cm and height 3.2 cm.
Ans.
Area = 1/2 × base × height
= 1/2 × 5 × 3.2
= 8 cm²
Q3. Find the area of a right triangle with base 3 cm and height 4 cm.
Ans.
Area = 1/2 × 3 × 4
= 6 cm²
Q4. In triangle ABC, AX = 4 units, BC = 6 units, and AC = 8 units. Find altitude BY.
Ans.
Using base BC and height AX:
Area of triangle ABC = 1/2 × 6 × 4
= 12 sq. units
Using base AC and height BY:
Area = 1/2 × 8 × BY
= 4BY
So, 4BY = 12
BY = 3 units
Q5. Triangle SUB is isosceles. SE is perpendicular to UB, and area of triangle SEB is 24 sq. units. Find area of triangle SUB.
Ans.
In an isosceles triangle, the perpendicular from the vertex bisects the base.
So, triangle SUE and triangle SEB have equal areas.
Area of triangle SUE = 24 sq. units
Area of triangle SEB = 24 sq. units
Area of triangle SUB = 24 + 24
= 48 sq. units
Q6. Find the area of quadrilateral ABCD if diagonal AC = 22 cm, BM = 3 cm, and DN = 3 cm. Both BM and DN are perpendicular to AC.
Ans.
Diagonal AC divides the quadrilateral into two triangles.
Area of triangle ABC = 1/2 × AC × BM
= 1/2 × 22 × 3
= 33 cm²
Area of triangle ACD = 1/2 × AC × DN
= 1/2 × 22 × 3
= 33 cm²
Total area = 33 + 33
= 66 cm²
Q7. Why cannot perimeter be used as a measure of area?
Ans. Perimeter measures boundary length. Area measures space covered.
A 6 cm × 2 cm rectangle has perimeter 16 cm and area 12 cm². A 5 cm × 3 cm rectangle also has perimeter 16 cm but area 15 cm². Same perimeter can give different areas.
Area of Parallelogram Class 8 Questions
Area of parallelogram class 8 questions test whether students can separate side length from height. The slant side is not the height.
A parallelogram can be cut and rearranged into a rectangle with the same base and height.
Parallelogram Formula Questions
Q1. Find the area of a parallelogram with base 7 cm and height 4 cm.
Ans.
Area = base × height
= 7 × 4
= 28 cm²
Q2. Find the area of a parallelogram with base 5 cm and height 3 cm.
Ans.
Area = 5 × 3
= 15 cm²
Q3. Find the area of a parallelogram with base 4.8 cm and height 5 cm.
Ans.
Area = 4.8 × 5
= 24 cm²
Q4. Find the area of a parallelogram with base 4.4 cm and height 2 cm.
Ans.
Area = 4.4 × 2
= 8.8 cm²
Q5. In parallelogram PQRS, SR = 12 cm and height to SR is 6 cm. If PS = 7.6 cm, find height QN to PS.
Ans.
Area using base SR:
Area = 12 × 6
= 72 cm²
Using base PS:
72 = 7.6 × QN
QN = 72 ÷ 7.6
= 9.47 cm approximately
Q6. A rectangle and a parallelogram have side lengths 5 cm and 4 cm. Which has the greater area?
Ans.
The rectangle has the greater area.
Rectangle area = 5 × 4
= 20 cm²
In the parallelogram, the height is less than the slant side 4 cm. So its area is less than 20 cm².
Q7. What can we say about parallelograms on the same base and between the same parallel lines?
Ans.
All such parallelograms have equal areas.
They have the same base and the same perpendicular height. Since area = base × height, their areas are equal.
Area of Rhombus Class 8 Questions
Area of rhombus class 8 questions use the diagonal formula. A rhombus is also a parallelogram, but diagonal-based questions are more common in this chapter.
The diagonals of a rhombus cut it into four right triangles.
Rhombus Formula Questions
Q1. Find the area of a rhombus whose diagonals are 20 cm and 15 cm.
Ans.
Area = 1/2 × d₁ × d₂
= 1/2 × 20 × 15
= 150 cm²
Q2. The diagonals of a rhombus are 8 cm and 6 cm. Find its area.
Ans.
Area = 1/2 × 8 × 6
= 24 cm²
Q3. Derive the formula for the area of a rhombus in terms of its diagonals.
Ans.
Let the diagonals of the rhombus be AC and BD.
A diagonal divides the rhombus into two triangles. These triangles have the same base BD and heights AO and CO.
Area of triangle ADB = 1/2 × AO × BD
Area of triangle CDB = 1/2 × CO × BD
Total area = 1/2 × BD × (AO + CO)
Since AO + CO = AC,
Area of rhombus = 1/2 × AC × BD
So, area of rhombus = 1/2 × d₁ × d₂.
Area of Trapezium Class 8 Questions
Area of trapezium class 8 questions need three values: both parallel sides and the perpendicular height.
If one of these is missing, use the figure, split the shape, or apply reasoning from parallel lines.
Trapezium Formula Questions
Q1. Find the area of a trapezium with parallel sides 24 m and 36 m, and height 14 m.
Ans.
Area = 1/2 × h × (a + b)
= 1/2 × 14 × (24 + 36)
= 1/2 × 14 × 60
= 420 m²
Q2. Find the area of a trapezium with parallel sides 10 in and 6 in, and height 14 in.
Ans.
Area = 1/2 × 14 × (10 + 6)
= 1/2 × 14 × 16
= 112 in²
Q3. Find the area of a trapezium with parallel sides 12 ft and 18 ft, and height 8 ft.
Ans.
Area = 1/2 × 8 × (12 + 18)
= 1/2 × 8 × 30
= 120 ft²
Q4. Derive the formula for the area of a trapezium using two copies of the same trapezium.
Ans.
Take two identical trapeziums. Rotate one and join it with the other.
The joined shape forms a parallelogram. Its base is the sum of the two parallel sides of the trapezium.
Base of parallelogram = a + b
Height = h
Area of parallelogram = h(a + b)
Since two trapeziums form this parallelogram:
Area of one trapezium = 1/2 × h × (a + b)
Q5. Why does the trapezium formula use the sum of parallel sides?
Ans.
Two identical trapeziums combine to form a parallelogram.
The base of that parallelogram is formed by joining the two parallel sides. That is why the formula uses a + b.
Ganita Prakash Class 8 Part 2 Chapter 7 Area: Rectangle and Path Problems
Ganita Prakash Class 8 Part 2 Chapter 7 Area includes real-life questions on paths, parks, crossroads, and compound shapes.
These questions are easier when students subtract the unused area from the total area.
Area Chapter Class 8 Questions and Answers on Paths
Q1. A path is laid around a rectangular park. What measurements are needed to find the path area?
Ans.
You need the length and breadth of the outer rectangle and the inner rectangle.
Area of path = Area of outer rectangle - Area of inner rectangle
If outer rectangle = 12 m × 10 m and inner rectangle = 8 m × 6 m:
Area of path = 120 - 48
= 72 m²
Q2. Does the path area change if the inner rectangle moves but both rectangles keep the same dimensions?
Ans.
No. The path area does not change.
The area depends only on the area of the outer rectangle and inner rectangle. Position does not change their areas.
Q3. A rectangular plot is 14 m by 12 m. Two crosspaths of width w m run through it. Find the path area.
Ans.
Area of crosspath = 14w + 12w - w²
The central square w² is subtracted because it is counted twice.
So, area = w(14 + 12 - w)
If w = 2 m:
Area = 2(14 + 12 - 2)
= 2 × 24
= 48 m²
Q4. A spiral tube has width 1 unit throughout. Its strip areas are 20, 19, 19, 14, 14, 9, 9, 4, and 4 sq. units. Find total area.
Ans.
Total area = 20 + 19 + 19 + 14 + 14 + 9 + 9 + 4 + 4
= 112 sq. units
If the tube is made straight with width 1 unit, its length will be 112 units.
Class 8 Maths Chapter 7 Area Unit Conversion Questions
Unit conversion questions in area class 8 need care because length units and square units convert differently.
If 1 inch = 2.54 cm, then 1 square inch = 2.54 × 2.54 square cm.
Unit Conversion Questions from Ganita Prakash Part 2 Chapter 7
Q1. Express 5 inches and 7.4 inches in centimetres.
Ans.
1 inch = 2.54 cm
5 inches = 5 × 2.54
= 12.7 cm
7.4 inches = 7.4 × 2.54
= 18.796 cm
Q2. Express 5.08 cm and 11.43 cm in inches.
Ans.
1 inch = 2.54 cm
5.08 cm = 5.08 ÷ 2.54
= 2 inches
11.43 cm = 11.43 ÷ 2.54
= 4.5 inches
Q3. How many cm² are there in 1 in²?
Ans.
1 inch = 2.54 cm
1 in² = 2.54² cm²
= 6.4516 cm²
Q4. Convert 161.29 cm² to in².
Ans.
1 in² = 6.4516 cm²
161.29 cm² = 161.29 ÷ 6.4516
= 25 in²
Class 8 Maths Extra Questions Chapter 7
Class 8 maths extra questions chapter 7 usually test reasoning more than direct formula use.
For such questions, draw the figure again, split it into known shapes, and compare base-height pairs.
HOTS and Application-Based Questions
Q1. If M and N are the midpoints of XY and XZ in triangle XYZ, what fraction of the area of triangle XYZ is triangle XMN?
Ans.
MN is parallel to YZ and is half of YZ.
The height of triangle XMN is also half the height of triangle XYZ.
Area of triangle XMN = 1/2 × half base × half height
So, its area is 1/4 of the area of triangle XYZ.
Q2. If the side length of a square is doubled, how does its area change?
Ans.
Let the original side be s.
Original area = s²
New side = 2s
New area = (2s)²
= 4s²
The area becomes four times the original area.
Q3. Three identical squares form a figure. If the red triangular region has area 49 sq. units and the blue triangular region has the same base and height, find the blue area.
Ans.
Triangles with the same base and height have equal areas.
So, area of blue region = 49 sq. units.
Q4. A rectangular park is 80 m long and 50 m wide. Two crossroads of width 4 m run through its centre. Find the area of the roads.
Ans.
Area of roads = length road + breadth road - overlap
= 80 × 4 + 50 × 4 - 4 × 4
= 320 + 200 - 16
= 504 m²
Important MCQs for Class 8 Maths Chapter 7 Area
MCQs in this chapter often test direct formula use. Check whether the question gives height, diagonal, or parallel sides before choosing the formula.
Q1. The area of a triangle with base 6 cm and height 4 cm is:
(a) 24 cm²
(b) 12 cm²
(c) 10 cm²
(d) 20 cm²
Ans. (b) 12 cm²
Area = 1/2 × 6 × 4 = 12 cm²
Q2. The area of a parallelogram with base 8 cm and height 5 cm is:
(a) 20 cm²
(b) 40 cm²
(c) 13 cm²
(d) 80 cm²
Ans. (b) 40 cm²
Area = 8 × 5 = 40 cm²
Q3. The area of a rhombus with diagonals 10 cm and 6 cm is:
(a) 60 cm²
(b) 15 cm²
(c) 30 cm²
(d) 16 cm²
Ans. (c) 30 cm²
Area = 1/2 × 10 × 6 = 30 cm²
Q4. The area of a trapezium with parallel sides 8 cm and 12 cm, and height 5 cm is:
(a) 50 cm²
(b) 100 cm²
(c) 25 cm²
(d) 60 cm²
Ans. (a) 50 cm²
Area = 1/2 × 5 × (8 + 12)
= 50 cm²
Q5. If 1 inch = 2.54 cm, how many cm² are there in 10 in²?
(a) 25.4 cm²
(b) 6.4516 cm²
(c) 64.516 cm²
(d) 254 cm²
Ans. (c) 64.516 cm²
10 in² = 10 × 6.4516
= 64.516 cm²
Maths Questions for Class 8 with Answers on Area
These maths questions for class 8 with answers help students practise mixed formulas from the chapter.
Do not apply the same formula to every shape. Match the formula to the shape first.
Mixed Practice Questions
Q1. Find the area of a rectangle with length 12 cm and breadth 7 cm.
Ans.
Area = length × breadth
= 12 × 7
= 84 cm²
Q2. Find the area of a square with side 9 cm.
Ans.
Area = side × side
= 9 × 9
= 81 cm²
Q3. Find the area of a triangle with base 15 cm and height 4 cm.
Ans.
Area = 1/2 × 15 × 4
= 30 cm²
Q4. Find the area of a parallelogram with base 11 cm and height 2 cm.
Ans.
Area = base × height
= 11 × 2
= 22 cm²
Q5. Find the area of a trapezium with parallel sides 6 cm and 10 cm, and height 5 cm.
Ans.
Area = 1/2 × 5 × (6 + 10)
= 1/2 × 5 × 16
= 40 cm²
Class 8 Maths Ganita Prakash Solutions: Area Formula Practice
Class 8 Maths Ganita Prakash solutions for this chapter focus on formula meaning, not only formula memorisation.
Students should know why each formula works through dissection, base-height comparison, or splitting into known shapes.
Ganita Prakash Part 2 Chapter 7 Formula-Based Questions
Q1. Why does the triangle formula have 1/2?
Ans.
Two identical triangles can form a parallelogram.
Since area of parallelogram = base × height, area of one triangle is half of that.
So, area of triangle = 1/2 × base × height.
Q2. Why is the area of a parallelogram not side × side?
Ans.
The slant side is not the height.
Area depends on perpendicular height, not tilted side length. So, area = base × height.
Q3. Why does the rhombus formula use diagonals?
Ans.
The diagonals split the rhombus into triangles.
Adding those triangle areas gives 1/2 × product of diagonals.
Q4. How can any polygon’s area be found?
Ans.
Divide the polygon into triangles, rectangles, or trapeziums.
Find each smaller area and add them.
All Formulas from Class 8 Maths Chapter 7 Area
| Shape |
Formula |
| Rectangle |
Length × Breadth |
| Square |
Side × Side |
| Triangle |
1/2 × base × height |
| Parallelogram |
Base × height |
| Rhombus |
1/2 × d₁ × d₂ |
| Trapezium |
1/2 × height × sum of parallel sides |
| Any polygon |
Divide into simpler shapes and add areas |