Area Class 8 Extra Questions Maths Part 2 Chapter 7 - #NCSOLVE 📚

During revision, students quickly go through Class 8 Maths Extra Questions Part 2 Chapter 7 Area Class 8 Extra Questions with Answers for clarity.

Class 8 Area Extra Questions

Class 8 Maths Chapter 7 Area Extra Questions

Area Extra Questions Class 8

Question 1.
Each of the following rectangles of length 6 cm and breadth 4 cm is composed of congruent polygons. Find the area of each polygon.

Solution:
Given, Length = 6 cm and Breadth = 4 cm
So, Area of each rectangle = 6 × 4 = 24 cm²
(a) It is divided into 6 congruent polygons.
Area of each polygon = \(\frac {24}{6}\) = 4 cm²
(b) It is divided into 2 congruent polygons.
Area of each polygon = \(\frac {24}{2}\) = 12 cm²

Question 2.
Two cross roads, each of width 4 m, run at right angles through the centre of a rectangular park of length 80 m and breadth 50 m, parallel to its sides. Find the area of the roads and the cost of constructing them at the rate of ₹ 120 per m².

Solution:
Area of the crossroads is the area of the shaded portion, i.e., the area of the rectangle PQRS and the area of the rectangle EFGH.
But while doing this, the area of the square KLMN is taken twice, which is to be subtracted.
Now, PQ = 4 m and PS = 50 m
EH = 4 m and EF = 80 m
KL = 4m and KN = 4 m
Area of the roads = Area of the rectangle PQRS + area of the rectangle EFGH – Area of the square KLMN
= PS × PQ + EF × EH – KL × KN
= (50 × 4 + 80 × 4 – 4 × 4) m²
= (200 + 320 – 16) m²
= 504 m²
Cost of constructing the path = ₹ 120 × 504 = ₹ 60,480

Question 3.
The two sides of the given parallelogram are 8 cm and 6 cm. The height corresponding to the base CD is 3 cm. Find the
(a) Area of the parallelogram.
(b) The height corresponding to the base AD.
Solution:
In the given figure,

(a) Base of the parallelogram, b = 8 cm
Height of the parallelogram, h = 3 cm
We have, area of parallelogram = b × h
= 8 cm × 3 cm
= 24 cm²

(b) For base AD = 6 cm, let height = h
∴ Area of the parallelogram = Base (AD) × height (h)
⇒ 24 cm² = 6 cm × h
⇒ h = 4 cm
Thus, the height corresponding to base AD is 4 cm.

Question 4.
In the given diagram, AC = 8 cm, BC = 4 cm, and AD = 5 cm.

Find:
(a) the area of the ΔABC
(b) the length of side BE
Solution:
In the given ΔABC,
base of the triangle, BC = 4 cm,
height of the triangle, AD = 5 cm
Therefore,
(a) Area of the ΔABC = \(\frac {1}{2}\) × base × height
= \(\frac {1}{2}\) × BC × AD
= \(\frac {1}{2}\) × 4 cm × 5 cm
= 10 cm²

(b) Also, area of the ΔABC = \(\frac {1}{2}\) × AC × BE
Therefore, 10 cm² = \(\frac {1}{2}\) × 8 cm × BE
BE = \(\frac {10}{4}\) cm = \(\frac {5}{2}\) cm = 2.5 cm
Thus, BE = 2.5 cm

Question 5.
In the given figure, calculate the area of the shaded part, if it is given that O is a point on CD and OC = 10 cm and OD = 5 cm. The area of ABDC is 120 cm².

Solution:
Given OC = 10 cm, OD = 5 cm
and area of rectangle ABDC = 120 cm²
Now, CD = OC + OD
= 10 + 5
= 15 cm
Area of rectangle ABDC = CD × AC
⇒ 120 = 15 × AC
⇒ AC = 8 cm
Area of triangle AOB = \(\frac {1}{2}\) × AB × AC
= \(\frac {1}{2}\) × 8 × 15
= 60 cm²
Area of shaded part = Area of rectangle ABDC – Area of triangle AOB
= 120 – 60
= 60 cm²

Question 6.
The cross-section of a canal is a trapezium in shape. If the canal is 10 m wide at the top, 6 m wide at the bottom, and the area of cross-section is 88 sq. m, determine its depth.
Solution:
Let h be the depth of the cross-section.

Area of trapezium = 88 sq. m
⇒ \(\frac {1}{2}\) × h × (a + b) = 88
⇒ \(\frac {1}{2}\) × h × (10 + 6) = 88
⇒ h = \(\frac{88 \times 2}{10+6}=\frac{176}{16}\) = 11 m
Hence, depth of canal = 11 m

Question 7.
In the figure given below, the area of ∆AFB is equal to the area of the parallelogram ABCD. If the altitude EF is 16 cm long, find the altitude of the parallelogram to the base AB of length 10 cm. What is the area of ∆DAO, where O is the midpoint of DC?

Solution:

Given, Area of ∆AFB = Area of parallelogram ABCD
⇒ \(\frac {1}{2}\) × AB × EF = CD × corresponding height
⇒ \(\frac {1}{2}\) × AB × EF = CD × EG
Let the corresponding height be h.
Then, \(\frac {1}{2}\) × 10 × 16 = 10 × h [∵ AB = CD, EF = 6 cm]
⇒ h = 8 cm
In ∆DAO, DO = 5 cm [∵ O is the midpoint of CD]
Area of ∆DAO = \(\frac {1}{2}\) × OD × h
= \(\frac {1}{2}\) × 5 × 8
= 20 cm²

Question 8.
A floor is 5 m long and 4 m wide. A square carpet of sides 3 m is laid on the floor. Find the area of the floor that is not carpeted.
Solution:
Length of the floor = 5 m
Width of the floor = 4 m
Area of the floor = length × width
= 5 m × 4 m
= 20 sq m.
Length of the square carpet = 3 m
Area of the carpet = length × length
= 3 m × 3 m
= 9 sq m.
Hence, the area of the floor laid with carpet is 9 sq m.
Area of the floor that is not carpeted = area of the floor – the area of the carpet
= 20 sq m – 9 sq m
= 11 sq m.

Question 9.
Four square flowerbeds, each of side 4 m are in four corners on a piece of land 12 m long and 10 m wide. Find the area of the remaining part of the land.
Solution:
Length of the land (l) = 12 m
Width of the land (w) = 10 m
Area of the whole land = l × w
= 12 m × 10 m
= 120 sq m.
The sidelength of each of the four square flowerbeds (s) = 4 m
Area of one flowerbed = s × s
= 4 m × 4 m
= 16 sq m.
Hence, the area of the four flowerbeds = 4 × 16 sq m = 64 sq m.
Area of the remaining part of the land = area of the complete land – the area of all four flowerbeds
= 120 sq m – 64 sq m
= 56 sq m.

Question 10.
Find the area of a triangle with a base equal to 10 cm and height equal to 8 cm.
Solution:
For the given triangle, base = 10 cm and height = 8 cm
We know that,
Area of a triangle = \(\frac {1}{2}\) × (Product of base and height of the triangle)
So, Area of the given triangle = \(\frac {1}{2}\) × (10 × 8)
= \(\frac {1}{2}\) × (80)
= 40 cm²

Question 11.
Identify if the given shape is a triangle or not, and also give reasons.


Solution:
(a) The given figure is a triangle as it has three straight sides and is a closed figure.
(b) The shape is not a triangle as it has four sides.
(c) The shape is not a triangle as it is an open figure with three open sides.

Question 12.
Is it possible to form a triangle with sides measuring 2 cm, 3 cm, and 6 cm?
Solution:
To form a triangle, the sum of any two sides should always be greater than the third side.
For the given measurements,
3 cm + 6 cm = 9 cm is greater than 2 cm,
6 cm + 2 cm = 8 cm is greater than 3 cm,
but 2 cm + 3 cm = 5 cm is lesser than 6 cm.
So, the given measures cannot form a triangle.

Question 13.
If the area of a triangle is 20 cm2 and the length of its base is 5 cm, find the triangle’s height.
Solution:
We know that,
Area of a triangle = \(\frac {1}{2}\) × base × height
So, 20 = \(\frac {1}{2}\) × 5 × height
Height of the triangle = \(\frac{(20 \times 2)}{5}\) = 8 cm

Question 14.
Find the height of the trapezium if its area is 128 sq in and the lengths of the parallel sides are 18 in and 14 in.
Solution:
Given:
Lengths of parallel sides of the trapezium = 18 in and 14 in
Area = 128 sq in
Let the height of the trapezium be ‘h’ units.
Using the trapezium formula, we know,
Area = \(\frac {1}{2}\) × (Sum of parallel sides) × height
Substitute all these values in the area of the trapezium formula,
128 = \(\frac {1}{2}\) × (18 + 14) × h
⇒ h = 128 × \(\frac {2}{32}\)
⇒ h = 8 in
The height of the given trapezium = 8 in.

Question 15.
Find the area of an isosceles trapezium with the length of each leg to be 18 units and the bases of lengths 22 units and 12 units.
Solution:
Given:
Bases, b = 22 units; a = 12 units.
Let us assume that its height is h.
We can divide the given trapezium into two congruent right triangles and rectangles as follows:
From the above figure,
x + x + 12 = 22
⇒ 2x + 12 = 22
⇒ 2x = 10
⇒ x = 5

Using Pythagoras Theorem
x² + h² = (13)²
⇒ (5)² + h² = 169
⇒ 25 + h² = 169
⇒ h² = 144
⇒ h = √144 = 12
The area of the given trapezium is,
A = \(\frac {1}{2}\) × (a + b) × h
⇒ A = \(\frac {1}{2}\) × (12 + 22) × (12)
⇒ A = 204 square units
The area of the given trapezium = 204 square units.

Question 16.
Find the area of a trapezium whose bases are 25 in and 10 in and whose height is 6 in.
Solution:
Given:
The bases are a = 25 in; b = 10 in.
The height is h = 6 in.
The area of the trapezium
A = \(\frac {1}{2}\) × (a + b) × h
⇒ A = \(\frac {1}{2}\) × (25 + 10) × 6
⇒ A = \(\frac {1}{2}\) × 35 × 6
⇒ A = 105 in².
Area of the given trapezium is 105 in².

The post Area Class 8 Extra Questions Maths Part 2 Chapter 7 appeared first on Learn CBSE.

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