Mathematics – Class 8 – Chapter 11 – Mensuration – Exercise 11.2 – NCERT Exercise Solution

1.The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m.

Mathematics - Class 8 - Chapter 11 - Exercise 11.2 - Mensuration - NCERT Exercise Solution

Solution:

Given,

One parallel side of the trapezium (a) = 1 m

Second side (b) = 1.2 m and

Height (h) = 0.8 m

According to question:

Area of top surface of the table= (½) × (a+b) h

                                                  = (½) × (1+1.2) 0.8

                                                  = (½) × 2.2 × 0.8 = 0.88

Hence, area of top surface of the table is 0.88 m2.

2. 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.

Mathematics - Class 8 - Chapter 11 - Exercise 11.2 - Mensuration - NCERT Exercise Solution

Solution:

Given, Length of one parallel side, a = 10 cm

height, (h) = 4 cm and

Area of a trapezium is 34 cm2

Now,

Let the length of the other parallel side be b

Area of trapezium = (1/2) × (a+b) h

34 = ½(10+b)×4

34 = 2 (10 + b)

34 = 20 + 2b

2b = 34 – 20 = 14

b = 7

Hence, required parallel side is 7 cm.

3. 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.

Solution:

Given:

BC = 48 m, CD = 17 m,

AD = 40 m

Perimeter = 120 m

So,

Perimeter of trapezium ABCD = AB+BC+CD+DA

120 = AB+48+17+40

120 = AB = 105

AB = 120–105 = 15 m

Now,

Area of the field = (½)×(BC+AD)×AB

                          = (½)×(48 +40)×15

                        = (½)×88×15

                         = 660

Hence, area of the field ABCD is 660m2.

4. The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.

Mathematics - Class 8 - Chapter 11 - Exercise 11.2 - Mensuration - NCERT Exercise Solution

Solution:

Given,

AE (h1) = 13 m

CF (h2) = 8 m

Diagonal BD (b) = 24 m

So,

Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ADC

                                            = ½(bh1) + ½(bh2)

                                            = ½ ×b (h1+h2)

                                            = (½)×24 × (13+8)

                                            = (½)×24×21 = 252

Hence, required area of the field is 252 m2.

5. The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.

Solution:

Given: 

Diagonal of rhombus

d1 = 7.5 cm and d2 = 12 cm

Now,

Area of rhombus = (½) × d1 × d2

                               = (½)×7.5×12 = 45

Hence, area of rhombus is 45 cm2.

6. Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm. If one of the diagonals is 8 cm long, find the length of the other diagonal.

Solution:

Given,

Side = 5 cm

Altitude = 4.8 cm

Length of one diagonal = 8 cm

So,

Area of rhombus = base x altitude

                               = 5 x 4.8 = 24 cm2

Now, we also know,

Area of rhombus = (½) × d1d2

After substituting the values, we get

24 = (½)×8×d2

d2 = 6

Hence, the length of the other diagonal is 6 cm.

7. The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m2is Rs. 4.

Solution:

Given,

Length of diagonal of tiles which are rhombus shaped,

d1 = 45 cm and d2= 30 cm

So,

Area of one tile = (½)d1d2

                         = (½)×45×30 = 675

Area of one tile is 675 cm2

Now, Area of 3000 tiles is

= 675×3000

= 2025000 cm2

We know that: 1m2 = 10000 cm2

 So,

2025000 cm2 = 2025000/10000 m2

                      = 202.50 m2

Cost of polishing the floor per sq. meter = 4

Cost of polishing the floor per 202.50 sq. meter = 4×202.50 = 810

Hence, the total cost of polishing the floor is Rs. 810.

8. Mohan wants to buy a trapezium shaped field. Its side along the river is parallel to and twice the side along the road. If the area of this field is 10500 m2 and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.

Mathematics - Class 8 - Chapter 11 - Exercise 11.2 - Mensuration - NCERT Exercise Solution

Solution:

Given,

Perpendicular distance (h) = 100 m

Area of the trapezium shaped field = 10500 m2

Let side along the road be ‘x’ m

Then, side along the river = ‘2x’ m

Now,

Area of the trapezium field = (½)×(a+b)×h

10500 = (½)×(x+2x)×100

10500 = 3x × 50

x = 70,

so, 2x = 140

Hence, the side along the river = 140 m.

9. 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.

Mathematics - Class 8 - Chapter 11 - Exercise 11.2 - Mensuration - NCERT Exercise Solution

Solution:

Octagon has 8 side, each given 5 m

According to question:

Area of the octagonal surface = area of trapezium ABCH + area of rectangle HCDG + area of trapezium GDEF

So,

Area of trapezium ABCH = Area of trapezium GDEF

= 12 (a + b) × h

= 12 (11 + 5) × 4

= 12 × 16 × 4

= 32 m2

Now,

Area of rectangle HCDG = l × b = 11 m × 5 m = 55 m2

Area of the octagonal surface = 32 m2 + 55 m2 + 32 m2 = 119 m2

Hence, the required area = 119 m2 = 119 m2

10. There is a pentagonal shaped park as shown in the figure. For finding its area Jyoti and Kavita divided it in two different ways.

Mathematics - Class 8 - Chapter 11 - Exercise 11.2 - Mensuration - NCERT Exercise Solution

Find the area of this park using both ways. Can you suggest some other way of finding its area?

Solution:

First way: 

Mathematics - Class 8 - Chapter 11 - Exercise 11.2 - Mensuration - NCERT Exercise Solution

According to Jyoti’s diagram,

Area of pentagon = Area of trapezium ABCP + Area of trapezium AEDP

= (½)(AP+BC) × CP+(1/2) × (ED+AP) × DP

= (½) (30+15) × CP+(1/2) × (15+30) × DP

= (½) × (30+15) × (CP+DP)

= (½) × 45 × CD

= (1/2) × 45 × 15

=337.5 m2

Hence, Area of pentagon is 337.5 m2

Second way:

According to Kavita’s diagram

Here, a perpendicular AM drawn to BE.

AM = 30–15 = 15 m

Mathematics - Class 8 - Chapter 11 - Exercise 11.2 - Mensuration - NCERT Exercise Solution

from figure:

Area of pentagon = Area of triangle ABE + Area of square BCDE

= (½)×15×15+(15×15)

= 112.5+225.0

= 337.5

Hence, total area of pentagon shaped park = 337.5 m2

Yes, we can also find the other way to calculate the area of the given pentagonal shape.

11. Diagram of the adjacent picture frame has outer dimensions = 24 cm×28 cm and inner dimensions 16 cm×20 cm. Find the area of each section of the frame, if the width of each section is same.

Mathematics - Class 8 - Chapter 11 - Exercise 11.2 - Mensuration - NCERT Exercise Solution

Solution:

According to question:

h1 = ½ (28 -20)

      = ½ x 8 = 4 cm

h2 = ½ (24 -16) = 4 cm

Now,

 Area of trapezium P = (½)×(a+b)×h1

                                  = (½)×(24+16)4

                                   = (½)×40×4 = 80 cm2

So, Area of trapezium R = 80 cm2

Now,

Area of trapezium Q = area of trapezium S

= ½ x (28 + 20) x 4

=96 cm2

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