Important Questions Class 8 Maths Part 2 Chapter 7 Area
Important Questions Class 8 Maths Part 2 Chapter 7 Area
Area is Chapter 7 of the Class 8 Maths book Ganita Prakash Part 2. It is about finding the area of plane figures. Students find the area of triangles, parallelograms, trapeziums, rhombuses and other polygons by breaking them into simpler shapes and using formulas.
This page gives 12 important questions from Class 8 Maths Part 2 Chapter 7 Area with step-by-step answers prepared by Extramarks experts. The questions cover area of a trapezium, rhombus and other polygons, and real-life area problems. Solve each question on paper first and then check your working with the answer.
Q1. The parallel sides of a trapezium are 25 cm and 13 cm. Its non-parallel sides are equal, each being 10 cm. Find the area of the trapezium.
Answer:
Q2. The area of a trapezium shaped field is 640 m2, the height of the field is 16 m and one of its parallel sides is 20 m. Find
Answer:
Q3. Find the area of the shown polygon.
Answer: Mark the vertices of polygon as A, B, C, D and E.
Q4. 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.
Answer:
Side of regular octagon = 5 m
Area of trapezium ABCH = Area of trapezium DEFG
Area of rectangle HGDC = 11 × 5 = 55 m2
Area of octagon = Area of trapezium ABCH + Area of trapezium DEFG + Area of rectangle HGDC
= 32 m2 + 32 m2 + 55 m2 = 119 m2
Therefore, the area of octagonal surface is 119 m2
Q5. 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.
Answer:
Q6. The area of a trapezium is 34 cm2 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.
Answer:
Q7. The area of a trapezium is 384 cm2. Its parallel sides are in the ratio 3:5 and the distance between them is 12 cm. Find the length of each parallel side.
Answer:
Q8. The perimeter of a trapezium is 52 cm. Its non-parallel sides are 10 cm each and the distance between two parallel sides is 8 cm. Find the area of the trapezium.
Answer: Perimeter of trapezium = 52 cm
Sum of the parallel sides + Sum of non- parallel sides = 52 cm
Sum of the parallel sides + 2(10) = 52 cm
Sum of the parallel sides = 52 - 20
= 32 cm
Area of trapezium= 128 cm2
Q9. Find the area of the following trapezium
Answer:
Q10. Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal.
Answer:
Q11. Find the area of a rhombus whose diagonals are of lengths 20 cm and 16 cm.
Answer: Area of a rhombus = ½ × d1 × d2
= ½ × 20 × 16
= 160 cm2.
Q12. The diagonals of a rhombus are 16 cm and 12 cm, find
a) Its area
b) Its length of the side
c) Its perimeter
Answer: The diagonals of the rhombus are d1 = 16 cm and d2 = 12 cm
CD² = OD² + OC²
Thus each side = 10 cm as all the sides are equal
c) Perimeter of the rhombus = 4 x side
= 4 x 10 = 40 cm
Related Study Material on Extramarks
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Q.1 Divide 39x3 (50x2 98) by 26x2(5x + 7).
Marks:4
Ans
Q.2 Factorise the polynomial given below.
7 + 10(ab) 8(ab)2
Marks:4
Ans
We have,
7 + 10(ab) 8(ab)2
Let x = (ab), then we get
7 + 10(ab) 8(ab)2 = 7 + 10x 8x2
We now find two numbers such that their sum is 10 and product is -56.
The numbers are 14 and -4.
Therefore, 7 + 10x 8x2 = 7 + 14x 4x 8x2
= 7(1 + 2x) 4x(1 + 2x)
= (1 + 2x)(7 4x)
= (1 + 2(a b))(7 4(a b))
[Put x = (a b)]
= (1 + 2a 2b)(7 4a + 4b)
Q.3 Factorise ax2 + by2 + bx2 + ay2.
Marks:2
Ans
ax2 + by2 + bx2 + ay2
= ax2 + bx2 + by2 + ay2
= x2(a + b) +y2(a + b) [Taking x2 and y2 as common]
= (x2 + y2)(a + b)
Q.4 Factorise the polynomial given below
(2r 3s)2 7(2r 3s) 30
Marks:3
Ans
We have,
(2r 3s)2 7(2r 3s) 30
Let x = (2r 3s), then we get
(2r 3s)2 7(2r 3s) 30= x2 7x 30
We now find two numbers such that their sum is 7 and product is 30.
The numbers are 10 and 3.
Therefore, x2 7x 30 = x2 10x + 3x 30
= x(x 10) + 3(x 10)
= (x + 3)(x 10)
= (2r 3s + 3)(2r 3s 10) [Putting x = (2r 3s)]
Q.5 Find the factors of 25x2 4y2 + 28yz 49z2.
Marks:5
Ans
25x2 4y2 + 28yz 49z2
= 25x2 [4y2 28yz + 49z2]
= 25x2 [(2y)2 2 x 2y x 7z + (7z)2]
= (5x)2 (2y 7z)2
= (5x + 2y 7z)(5x 2y + 7z) [Using a2 b2 = (a+b)(a b)]
CBSE Class 8 Maths Important Questions
FAQs (Frequently Asked Questions)
Triangles, parallelograms, trapeziums, path problems, and formula derivations carry the most marks. Area of trapezium class 8 questions can appear as 3-mark or 5-mark answers. Area of triangle class 8 and area of parallelogram class 8 questions are common in formula-based tests.
The area of rhombus class 8 formula is 1/2 × d₁ × d₂, where d₁ and d₂ are the diagonals. This formula comes from splitting the rhombus into triangles and using the diagonals as base and height parts.
For a path around a rectangle, subtract the inner rectangle area from the outer rectangle area. For crossroads, add both road strips and subtract the overlapping middle square once because it gets counted twice.
Yes. Unit conversion questions are important in class 8 maths chapter 7 area. Students should remember 1 inch = 2.54 cm and 1 in² = 6.4516 cm². These questions usually appear as 1-mark or 2-mark problems.
Dissection means cutting a shape into parts and rearranging them to form another shape with the same area. In ganita prakash class 8 part 2 chapter 7 area, dissection helps derive formulas for triangles, parallelograms, rhombus, and trapezium.