Important Questions Class 8 Maths Chapter 11 Mensuration

Some of the most important mensuration class 8 questions i.e chapter 11 are given here. These chapter 11 class 8 maths questions cover several short answer type questions, long answer type questions and HOTS questions that are crucial for CBSE class 8 exams. Here, some of the important mensuration questions from NCERT class 8 are also included.

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Mensuration Important Questions For Class 8 (Chapter 11)

These class 8 mensuration questions are categorized into short answer type questions and long answer type questions. These questions cover various concepts which will help class 8 students to develop problem-solving skills for the exam.

Short Answer Type Questions:

1. The parallel sides of a trapezium measure 12 cm and 20 cm. Calculate its area if the distance between the parallel lines is 15 cm.

Solution:

Area of trapezium = ½ × distance between parallel sides × sum of parallel sides

= ½ × 15 × (24+20)

= 330 cm2

2. Calculate the height of a cuboid which has a base area of 180 cm2 and volume is 900 cm3.

Solution:

Volume of cuboid = base area × height

900 = 180 × height

So, height = 900/180 = 5 cm

3. A square and a rectangle have the same perimeter. Calculate the area of the rectangle if the side of the square is 60 cm and the length of the rectangle is 80 cm.

Solution:

Perimeter of square = 4 (side of the square) = 4 (60) = 240 m

Perimeter of rectangle = 2 (Length + Breadth) = 2 (80 + Breadth)

= 160 + 2 × Breadth

Now, as the perimeter of square and rectangle are the same,

160 + 2 × Breadth = 240

Or, Breadth of rectangle = 40m

Now, area of rectangle = Length × Breadth = (80 × 40) = 3200m2

4. A lawnmower takes 750 complete revolutions to cut grass on a field. Calculate the area of the field if the diameter of the lawnmower is 84 cm and length is 1 m.

Solution:

Given, length of lawnmower = 1m = 100cm

Its circumference = π × D = 22/7 × 84 = 264 cm

Now, length of field = 264 × 750 = 198000 cm

Here, the width of field = length of the lawnmower i.e. 100 cm

So, area of field = 198000 × 100 = 19,800,000 cm²

Or, 1980 m²

5. The area of a rhombus is 16 cm2 and the length of one of its diagonal is 4 cm. Calculate the length of other the diagonal.

Solution:

Area of rhombus = ½ × d1 × d2

⇒ 16 = ½ × 4 × d2

So, d2 = 32/4 = 8 cm

Long Answer Type Questions:

6. From a circular sheet of radius 4 cm, a circle of radius 3 cm is cut out. Calculate the area of the remaining sheet after the smaller circle is removed.

Solution:

The area of the remaining sheet after the smaller circle is removed will be = Area of the entire circle with radius 4 cm – Area of the circle with radius 3 cm

We know,

Area of circle = πr²

So,

Area of the entire circle = π(4)² = 16π

And,

Area of the circle with radius 3 cm which is cut out = π(3)² = 9π

Thus, the remaining area = 16π – 9π

7. A cuboidal box of dimensions 1 m × 2 m × 1.5 m is to be painted except its bottom. Calculate how much area of the box has to be painted.

Solution:

Given,

Length of box = 2 m,

Breadth of box = 1 m

Height of box = 1.5 m

We know the surface area of cuboid = 2(lb + lh + bh)

But here the bottom part is not to be painted.

So,

Surface area of box = lb + 2(bh + hl)

= 2 × 1 + 2 (1 × 1.5 + 1.5 × 2)

= 2 + 2 (1.5 + 3.0) = 2 + 9.0 = 11

Thus, the required surface area of the box = 11m2.

8. In a trapezium, the parallel sides measure 40 cm and 20 cm. Calculate the area of the trapezium if its non-parallel sides are equal having the lengths of 26 cm.

Solution:

From the question statement draw the diagram.

Consider a trapezium of ABCD. Let AB and DC be the parallel sides as shown in the figure.

Important Mensuration Questions for Class 8 Maths Chapter 11

Now, CM will be the distance between the two parallel sides or the height of the trapezium.

We know,

Area of trapezium = ½ × sum of parallel sides × height.

So, height has to be found.

In the diagram, draw CL || AD

Important Questions Class 8 Maths Chapter 11 Mensuration

Now, ALCD is a parallelogram ⇒ AL = CD = 20 cm and CL = AD = 26 cm

As AD = CB,

CL = CB ⇒ ΔCLB is an isosceles triangle with CB as its height.

Here, BL = AB – AL = (40 – 20) = 20 cm. So,

LM = MB = ½ BL = ½ × 20 = 10 cm

Now, in ΔCLM,

CL2 = CM2 + LM2 (Pythagoras Theorem)

262 = CM2 + 102

CM2 = 262 – 102 = (26 – 10) (26 + 10) = 16 × 36 = 576

CM = √576 = 24 cm

Now, the area of trapezium can be calculated.

Area of trapezium ABCD = ½ × (AB + CD) × CM

= ½ × (20 + 40) × 24

Or, Area of trapezium ABCD = 720 cm2

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1 Comment

  1. Thanks for sending these important questions. These will help me to attend this government exam. Thanks a lot. I already had joined in your app. Byjus app is more helpful for me. I am improving more. Thanks for this.

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