# Samacheer Kalvi Class 9 Maths Chapter 7 Mensuration Solutions

## Samacheer Kalvi Class 9 Maths Chapter 7 Mensuration Solutions

Welcome to NCTB Solutions. Here with this post we are going to help 9th class students by providing Solutions for Samacheer Kalvi Class 9 Maths chapter 7 Mensuration. Here students can easily find all the solutions for Mensuration Exercise 7.1, 7.2, 7.3 and 7.4. Also here our Expert Maths Teacher’s solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 7 solutions. Here all the solutions are based on Tamil Nadu State Board latest syllabus. #### Mensuration Exercise 7.1 Solutions :

(1) Using Heron’s formula, find the area of a triangle whose sides are

(i) 10 cm, 24 cm, 26 cm

(ii) 1.8m, 8 m, 8.2 m

Solution :

(i) Side = 10 + 24 + 26/2

= 60/2

= 30

So, Area = √30 (30 – 10) (30 – 24) (30 – 26)

= √30 × 20 × 6× 4

= √14400

= 120

Therefore, the area of a triangle is 120.

(ii) Side = 1.8 + 8 + 8.2/2

= 9

So, Area = √9 (9-1.8) (9-8) (9-8.2)

= √51.84

= 7.2

Thus, the area of a triangle is 7.2

(2) The sides of the triangular ground are 22 m, 120 m and 122 m. Find the area and cost of levelling the ground at the rate of ₹ 20 per m2.

Solution :

Side = 22 + 120 + 122/2 = 132

Area = √132 (132 – 22) + (132 – 120)(132 – 122)

= √132 x 110 x 12 x 10

= 1320 m

Therefore, total Cost,

= 1320 x 20

= Rs 26400

(3) The perimeter of a triangular plot is 600 m. If the sides are in the ratio 5 : 12 : 13, then find the area of the plot.

Solution :

Let, the sides are 5x, 12x and 13x

5x + 12x + 13x = 30x

so, 30x/2 = 600

x = 40

Sides are 200, 480, 520

Area = √600(600 – 200)(600 – 480)(600 – 520)

= √600 x 400 x 120 x 80

= √2304000000

= 48000

(4) Find the area of an equilateral triangle whose perimeter is 180 cm.

Solution :

Perimeter = 180 cm

So, one side = 180/3 = 60

Area = √3/4 x 60^2

= √3/4 x 3600

= 900√3

= 1558.8 m^2

(5) An advertisement board is in the form of an isosceles triangle with perimeter 36 m and each of the equal sides are 13 m. Find the cost of painting it at ₹ 17.50 per square metre.

Solution :

Weight of triangle = n

= √13^2 – 5^2

= √144

= 12

So Area of the triangle board

= 1/2 x 10 x 12

= 60 m^2

Now the total cost,

= 60 x 17.50

= 1050 Rs

(6) Find the area of the unshaded region.

Solution :

AB² = 12² + 16² = 400

= AB = 20

S = 34 + 42 + 20/2 = 48

Area of the △ABC = √48 (48-34) (48-20) (48-42)

= √48 × 14 × 28 × 6

= √112896

= 336

Area of the ABD,

= 1/2 × 12 × 16

= 96 cm²

= 336 – 96

= 240 cm²

Therefore, the area of the unshaded region is 240 cm²

(7) Find the area of a quadrilateral ABCD whose sides are AB = 13 cm, BC = 12 cm, CD = 9, cm AD = 14 cm and diagonal BD = 15 cm.

Solution :

Side = 13 + 14 + 15/2 = 21

Area of ABD = √21 (21 – 13) (21 – 14) (21 – 15)

= √21 × 8 × 7 × 6

= 84

Side = 12 + 9 + 15/2 = 18

Area of BC = √18 (18 – 12) (18 – 9) (18 – 15)

= √18 × 6 × 3 × 3

= √2916 = 54

∴ Total area = 84 + 54 = 138 cm²

(8) A park is in the shape of a quadrilateral. The sides of the park are 15 m, 20 m, 26 m and 17 m and the angle between the first two sides is a right angle. Find the area of the park.

Solution :

Area of △ABD = 1/2 × 20 × 15 = 150 m²

In △ABD BD² = AD² + AB²

= BD² = 15² + 20² = 625

= BD = 25

In △BCD = S = 25 + 26 + 17/2 = 34

Area = √34 (34 – 25) (34 – 26) (34 – 17)

= √34 × 9 × 8 × 17

= √41616

= 204

∴ Total area,

= (204 + 150)

= 354 m²

(9) A land is in the shape of rhombus. The perimeter of the land is 160 m and one of the diagonal is 48 m. Find the area of the land.

Solution :

4a = 160

a = 40

Sides = 40 m

Diagonal = 48 m

In △ABC

S = 40 + 40 + 48/2 = 64

Area = √64 (64 – 40) (64 – 40) (64 – 48)

= √64 × 24 × 2416

= 24 × 32

= 768

∴ Total area,

= 2 × 768

= 1536 m²

Hence, the area of the land will be 1536 m²

(10) The adjacent sides of a parallelogram measures 34 m, 20 m and the measure of one of the diagonal is 42 m. Find the area of parallelogram.

Solution :

In △ABC

S = 34 + 20 + 42/2 = 48

Area = √48 (48 – 34) (48 – 20) (48 – 42)

= √48 × 14 × 28 × 6

= √112896

= 336

Therefore, total area of the parallelogram,

= 2 × 336

= 672 m²

#### Mensuration Exercise 7.2 Solutions :

(1) Find the Total Surface Area and the Lateral Surface Area of a cuboid whose dimensions are : length = 20 cm, breadth = 15 cm and height = 8 cm

Solution :

Given length = 20 cm

Height = 8 cm

Total surface area of cuboid,

= 2(20 x 15 + 15 x 8 + 8 x 20)

= 2 (300 + 120 + 160)

= 1160 cm

lateral surface area of a cuboid,

= 2 x 8 (20 + 15)

= 16 x 35

= 560 cm

Therefore the surface are will be 560 cm.

(2) The dimensions of a cuboidal box are 6 m × 400 cm × 1.5 m. Find the cost of painting its entire outer surface at the rate of ₹22 per m2.

Solution :

Total Surface area of the cuboid,

= 2 (6 × 4 + 4 × 1.5 + 1.5 × 6)

= 2 (24 + 6 + 9)

= 78

Now, the total cost,

= 78 x 22

= 1716 Rs.

(3) The dimensions of a hall is 10 m × 9 m × 8 m. Find the cost of white washing the walls and ceiling at the rate of ₹8.50 per m2.

Solution :

Area to be washed,

= 2 x (10 × 5 + 9 × 8) + 10 x 9

= 2 (80 + 72) + 90

= 304 + 90

= 394

So, total cost for washing,

= 394 x 8.50

= 3349 Rs

(4) Find the TSA and LSA of the cube whose side is (i) 8 m (ii) 21 cm (iii) 7.5 cm

Solution :

TSA, of the cube,

= 6a² = 6×64

= 384 m²

LSA, of the cube,

= 4a² = 4×64

= 256 m²

(5) If the total surface area of a cube is 2400 cm² then, find its lateral surface area.

Solution :

We know, total surface area 6a²

So, 6a² = 2400

= a² = 400

∴ Lateral surface area = 4a²

= 4 × 400

= 1600

(6) A cubical container of side 6.5 m is to be painted on the entire outer surface. Find the area to be painted and the total cost of painting it at the rate of ₹24 per m2.

Solution :

Side = 6.5 m

Area to be painted

= 6 × 6.5²

= 253.5 m²

And total cost for painting,

= 24 × 253.5

= 6084 RS

Therefore, the total cost of painting will be Rs 6084.

(7) Three identical cubes of side 4 cm are joined end to end. Find the total surface area and lateral surface area of the new resulting cuboid.

Solution :

Give a = 4 cm

If three cube are joined then

It becomes l = 4 + 4 + 4 = 12

b = 4

h = 4

∴ Total surface area,

= 2 (12 × 4 + 4 × 4 + 4 × 12)

= 2 (48 + 16 + 48)

= 224

∴ Lateral surface area,

= 2 × 4 (12 + 4)

= 8 × 16

= 128

hence, the total surface area = 224 cm

and lateral surface area = 128 cm.

#### Mensuration Exercise 7.3 Solutions :

(1) Find the volume of a cuboid whose dimensions are

(i) length = 12 cm, breadth = 8 cm, height = 6 cm

(ii) length = 60 m, breadth = 25 m, height = 1.5 m

Solution :

(i) Given, Length = 12 cm

Height = 6 cm

Therefore, Volume of the cuboid,

= 12 × 8 × 6

= 576 cm²

So, the volume of a cuboid is 576 cm²

(ii) As per the question,

Length = 60 m

Height = 1.5 m

Therefore, Volume of the cuboid,

= 60 × 25 × 1.5

= 2250 m³

Thus, the volume of a cuboid is 2250 m³

(2) The dimensions of a match box are 6 cm × 3.5 cm × 2.5 cm. Find the volume of a packet containing 12 such match boxes.

Solution :

Volume of each box

= 6 × 3.5 × 2.5

= 52.5

Hence, Volume of 12 such boxes

= 12 × 52.5

= 630 cm³

Thus, volume of a packet containing 12 such match boxes is 630 cm³

(3) The length, breadth and height of a chocolate box are in the ratio 5 : 4 : 3. If its volume is 7500 cm3, then find its dimensions.

Solution :

Let, Length = 5x

Height = 3x

Therefore, Volume

= 5x × 4x × 3x

= 60x³

Now, 60x³ = 7500

= x³ = 125

= x = 5

Therefore, the dimensions of the box will be, 25 × 20 × 15

(4) The length, breadth and depth of a pond are 20.5 m, 16 m and 8 m respectively. Find the capacity of the pond in litres.

Solution :

Volume = 20.5 × 16 × 8 = 2624

We know, 1m³ = 10000 litres

Hence, 2624 m³ = 26240000 litres

Therefore, the capacity of the pond 26240000 litres.

(5) The dimensions of a brick are 24 cm × 12 cm × 8 cm. How many such bricks will be required to build a wall of 20 m length, 48 cm breadth and 6 m height?

Solution :

Volume of the brick,

= 24 × 12 × 8

= 2304 cm³

Wall dimensions are,

Length = 20m – 2000 cm

Height = 6m = 600 cm

Hence, Volume of the wall,

= 2000 × 48 × 600

Volume of the brick = 2309

Therefore, No of the bricks required,

= 2000 × 48 × 600/2304

= 25000

(6) The volume of a container is 1440 m3. The length and breadth of the container are 15 m and 8 m respectively. Find its height.

Solution :

Volume = 1440 m³

Let, the height = h

15 × 8 × h = 1440

= h = 1440/15 × 8

= h = 12

Therefore, its height will be 12

(7) Find the volume of a cube each of whose side is (i) 5 cm (ii) 3.5 m (iii) 21 cm

Solution :

(i) For a = 5 cm

Volume of a cube,

= 5³

= 125

So, the volume of a cube is 125

(ii) For a = 3.5 cm

Volume of a cube,

= (3.5)³

= 42.875 cm³

∴ The volume of a cube is 42.875 cm³

(iii) For a = 21 cm

Volume of a cube,

= (21)³

= 9261

Therefore, the volume of a cube is 9261.

(8) A cubical milk tank can hold 125000 litres of milk. Find the length of its side in metres.

Solution :

Volume = 125000 = 125 m³

∴ a³ = 125

= a = 5

Hence, the length of its side will be 5 m.

(9) A metallic cube with side 15 cm is melted and formed into a cuboid. If the length and height of the cuboid is 25 cm and 9 cm respectively then find the breadth of the cuboid.

Solution :

Volume of the cub melted = 25 × 3 × h

∴ The volume of the cube melted

= 5 × 15 × 15

= 25 × 9 × h

= 5 × 15 × 15

= h = 15 × 15 × 5/25 × 3

= h = 15 cm

Therefore, the breadth of the cuboid is 15 cm.

#### Mensuration Exercise 7.4 Solutions :

(1) The semi-perimeter of a triangle having sides 15 cm, 20 cm and 25 cm is

(1) 60 cm

(2) 45 cm

(3) 30 cm

(4) 15 cm

Solution :

Correct option – (3)

30 cm is the correct answer.

(2) If the sides of a triangle are 3 cm, 4 cm and 5 cm, then the area is

(1) 3 cm²

(2) 6 cm²

(3) 9 cm²

(4) 12 cm²

Solution :

Correct option – (2)

If the sides of a triangle are 3 cm, 4 cm and 5 cm, then the area is 6 cm².

(3) The perimeter of an equilateral triangle is 30 cm. the area is

(1) 10√3 cm²

(2) 12√3 cm²

(3) 15√3 cm²

(4) 25√3 cm²

Solution :

Correct option – (4)

Therefore, the are will be 25√3 cm²

(4) The lateral surface area of a cube of side 12 cm is

(1) 144 cm²

(2) 196 cm²

(3) 576 cm²

(4) 664 cm²

Solution :

Correct option – (3)

The lateral surface area of a cube will be 576 cm²

(5) If the lateral surface area of a cube is 600 cm², then the total surface area is

(1) 150 cm²

(2) 400 cm²

(3) 900 cm²

(4) 1350 cm²

Solution :

Correct option – (3)

If the lateral surface area of a cube is 600 cm², then the total surface area is 900 cm².

(6) The total surface area of a cuboid with dimension 10 cm × 6 cm × 5 cm is

(1) 280 cm²

(2) 300 cm²

(3) 360 cm²

(4) 600 cm²

Solution :

Correct option – (1)

The total surface area of a cuboid will be 280 cm²

(7) If the ratio of the sides of two cubes are 2:3, then ratio of their surface areas will be

(1) 4 : 6

(2) 4 : 9

(3) 6 : 9

(4) 16 : 36

Solution :

Correct option – (2)

If the ratio of the sides of two cubes are 2 : 3, then ratio of their surface areas will be 4 : 9

(8) The volume of a cuboid is 660 cm³ and the area of the base is 33 cm². its height is

(1) 10 cm

(2) 12 cm

(3) 20 cm

(4) 22 cm

Solution :

Correct option – (3)

The volume of a cuboid is 660 cm³ and the area of the base is 33 cm². its height is 20 cm.

(9) The capacity of a water tank of dimensions 10 m × 5 m × 1.5 m is

(1) 75 litres

(2) 750 litres

(3) 7500 litres

(4) 75000 litres

Solution :

Correct option – (4)

The capacity of the water tank will be 75000 litres.

(10) The number of bricks each measuring 50 cm × 30 cm × 20 cm that will be required to build a wall whose dimensions are 5 m × 3 m × 2 m is

(1) 1000

(2) 2000

(3) 3000

(4) 5000

Solution :

Correct option – (1)

1000 brick will be required to build the wall.

Next Chapter Solution :

👉 Statistics

Updated: August 1, 2023 — 4:08 pm