# Samacheer Kalvi Class 7 Maths Term 1 Chapter 2 Solutions

## Samacheer Kalvi Class 7 Maths Term 1 Chapter 2 Measurements Solutions

Welcome to NCTB Solutions. Here with this post we are going to help 7th class students by providing Solutions for Samacheer Kalvi Class 7 Maths Term 1 chapter 2 Measurements. Here students can easily find all the solutions for Measurements Exercise 2.1, 2.2, 2.3 and 2.4. 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 2 solutions. Here all the solutions are based on Tamil Nadu State Board latest syllabus.

#### Measurements Exercise 2.1 Solutions :

(1) Find the area and perimeter of the following parallelograms

Solution :

Figure – (i)

Given, ABCD is a parallelogram

Since, Base of a parallelogram (b) = 11 cm and Height of a parallelogram (h) = 3 cm

We know,

The area of parallelogram

= b × h

= 11 × 3

= 33 sq.cm

Also know,

The perimeter of parallelogram = sum of the length of the four sides.

= 11 + 4 + 11 + 4

= 30 cm

Figure – (ii)

Given, PQRS is a parallelogram

From fig, PS = RQ

Since, Base of a parallelogram (b) = 7 cm and Height of a parallelogram (h) = 10 cm

We know,

The area of parallelogram

= b × h

= 7 × 10

= 70 sq.cm

Also know,

The perimeter of parallelogram = sum of the length of the four sides.

= 7 + 13 + 7 + 13

= 40 cm

(2) Find the missing values.

 Sr. No. Base Height Area (i) 18 cm 5 cm (ii) 8 m 56 sq.m (iii) 17mm 221 sq.mm

Solution :

(i) Base – 18 cm

Height – 5 cm

Area – 90 sq.cm

(ii) Base – 8 m

Height – 7 m

Area – 56 sq.m

(iii) Base – 13 mm

Height – 17 mm

Area – 221 sq.mm

(3) Suresh won a parallelogram-shaped trophy in a state level Chess tournament. He knows that the area of the trophy is 735 sq. cm and its base is 21 cm. What is the height of that trophy?

Solution :

Given, Suresh won a parallelogram-shaped trophy in a state level Chess tournament.

Since, base of trophy = 21 cm

We know that , area of of trophy = 735 sq.cm

b × h = 735

21 × h = 735

h = 735 /21

h = 35 cm

Hence, 35 cm is the height of that trophy.

(4) Janaki has a piece of fabric in the shape of a parallelogram. Its height is 12 m and its base is 18 m. She cuts the fabric into four equal parallelograms by cutting the parallel sides through its mid-points. Find the area of each new parallelogram.

Solution :

Janaki has a piece of fabric in the shape of a parallelogram.

Since, it’s height = 12 m and base = 18 m

Thus, area of this fabric

= h × b

= 12 × 18

= 216 Sq.cm

Now, She cuts the fabric into four equal parallelograms by cutting the parallel sides
through its mid-points.

Therefore,

= 216 ÷ 4

= 54 sq.cm

Hence, the area of each new parallelogram is 54 sq.cm.

(5) A ground is in the shape of parallelogram. The height of the parallelogram is 14 metres and the corresponding base is 8 metres longer than its height. Find the cost of levelling the ground at the rate of Rs 15 per sq. m.

Solution :

Given, a ground is in the shape of parallelogram.

Since, it’s height = 14 m

Thus, base

= 8 + 14

= 22 m

We know that, area of parallelogram

= b × h

= 22 × 14

= 308 sq.m

Since, the rate of 1 sq.m = ₹15

Thus, the rate of 308 sq.m

= 308 × 15

= ₹4620

Hence, the cost of levelling the ground is ₹4620

Objective Type Question Solutions :

(6) The perimeter of a parallelogram whose adjacent sides are 6 cm and 5 cm is

(i) 12 cm

(ii) 10 cm

(iii) 24 cm

(iv) 22 cm

Solution :

Correct Option → (iv)

The perimeter of a parallelogram will be 22 cm

(7) The area of a parallelogram whose base 10 m and height 7 m is

(i) 70 sq. m

(ii) 35 sq. m

(iii) 7 sq. m

(iv) 10 sq. m

Solution :

Correct Option → (i)

The area of a parallelogram will be 70 sq. m.

(8) The base of the parallelogram with area is 52 sq. cm and height 4 cm is

(i) 48 cm

(ii) 104 cm

(iii) 13 cm

(iv) 26 cm

Solution :

Correct Option → (iii)

The base of the parallelogram  13 cm

(9) What happens to the area of the parallelogram, if the base is increased 2 times and the height is halved

(i) Decreases to half

(ii) Remains the same

(iii) Increases by two times

(iv) none

Solution :

Correct Option → (ii)

If the base is increased 2 times and the height is halved then the area of the parallelogram Remains the same.

(10) In a parallelogram the base is three times its height. if the height is 8 cm then the area is

(i) 64 sq. cm

(ii) 192 sq. cm

(iii) 32 sq. cm

(iv) 72 sq. cm

Solution :

Correct Option → (ii)

Then the area will be 192 sq. cm.

#### Measurements Exercise 2.2 Solutions :

(1) Find the area of rhombus PQRS shown in the following figures.

Solution :

Figure – (i)

From fig, d1 = 8 cm and d2 = 16 cm

We know,

Area of the rhombus = 1/2 × d1 × d2

= 1/2 × 8 × 16

= 4 × 16

= 64 sq.cm

Therefore, area of rhombus PQRS is 64 sq.cm

Figure – (ii)

From fig, Base = 15 cm and height = 11 cm

We know, Area of the rhombus

= b × h = 15 × 11

= 165 sq.cm

Therefore, area of rhombus PQRS is 165 sq.cm

(2) Find the area of a rhombus whose base is 14 cm and height is 9 cm.

Solution :

Given, Base = 14 cm and height = 9 cm

We know, Area of the rhombus

= b × h = 14 × 9

= 126 sq.cm

Therefore, area of rhombus is 126 sq.cm

(3) Find the missing value.

 Sr. No Diagonal (d1) Diagonal (d2) Area (i) 19 cm 16 cm (ii) 26 m 468 sq.m (iii) 12 mm 180 mm

Solution :

(i) Diagonal (d1) – 19 cm

Diagonal (d2) – 16 cm

Area – 152 sq.cm

(ii) Diagonal (d1) – 26 m

Diagonal (d2) – 36 m

Area – 468 sq.m

(iii) Diagonal (d1) – 30 mm

Diagonal (d2) – 12 mm

Area – 180 mm

(4) The area of a rhombus is 100 sq. cm and length of one of its diagonals is 8 cm. Find the length of the other diagonal.

Solution :

Given, the length of one diagonal (d1) = 8 cm

Let, the length of the other diagonal be d2 cm

Area of the rhombus = 100 sq. cm (given) 1/2 x (d1× d2) = 100

1/2 x (8× d2) = 100

8 x d2 = 100 x 2

d2 = 200/8 = 25

Therefore, length of the other diagonal is 25 cm.

(5) A sweet is in the shape of rhombus whose diagonals are given as 4 cm and 5 cm. The surface of the sweet should be covered by an aluminum foil. Find the cost of aluminum foil used for 400 such sweets at the rate of ₹ 7 per 100 sq. cm.

Solution :

Given, a sweet is in the shape of rhombus whose diagonals are given as 4 cm and 5 cm.

Thus, d1 = 4 cm and d2 = 5 cm

We know,

Area of the sweet

= 1/2 x (d1× d2)

= 1/2 × 4× 5

= 2 × 5

= 10 sq.cm

There are 400 such sweet

Therefore, total area

= 10 × 400

= 4000 sq.cm

Since, the rate of 100 sq.cm = ₹7

So, the rate of 4000 sq.cm

= 4000×7/100

= 40 × 7

= ₹280

Hence, the cost of aluminum foil used for 400 such sweets is ₹280.

Objective Type Question Solution :

(6) The area of the rhombus with side 4 cm and height 3 cm is

(i) 7 sq. cm

(ii) 24 sq. cm

(iii) 12 sq. cm

(iv) 10 sq. cm

Solution :

Correct Option → (iii)

The area of the rhombus will be 12 sq. cm

(7) The area of the rhombus when both diagonals measuring 8 cm is July 6, 2021 by Arup Sarkar 1 Comment (Edit)

(i) 64 sq. cm

(ii) 32 sq. cm

(iii) 30 sq. cm

(iv) 16 sq. cm

Solution :

Correct Option → (ii)

The area of the rhombus  will be 32 sq. cm

(8) The area of the rhombus is 128 sq. cm and the length of one diagonal is 32 cm. the length of the other diagonal is

(i) 12 cm

(ii) 8 cm

(iii) 4 cm

(iv) 20 cm

Solution :

Correct Option → (ii)

The length of the other diagonal will be 8 cm.

(9) The height of the rhombus whose area 96 sq. m and side 24 m is

(i) 8 m

(ii) 10 m

(iii) 2 m

(iv) 4 m

Solution :

Correct Option → (iv)

The height of the rhombus will be 4 m

(10) The angle between the diagonals of a rhombus is

(i) 120º

(ii) 180º

(iii) 90º

(iv) 100º

Solution :

Correct Option → (iii)

The angle between the diagonals of a rhombus is 90º

#### Measurements Exercise 2.3 Solutions :

(1) Find the missing values.

 Sr. No Height  ‘h’ Parallel  side  ‘a’ Parallel  side  ‘b Area (i) 10 m 12 m 20 m (ii) 13 cm 28 cm 492 sq.cm (iii) 19 m 16 m 323 sq.m (iv) 16 cm 15 cm 360 sq.cm

Solution :

(i) Height ‘h’ – 10 m

Parallel side ‘a’ – 12 m

Parallel side ‘b – 20 m

Area – 160 sq.cm

(ii) Height ‘h’ – 24 cm

Parallel side ‘a’ – 13 cm

Parallel side ‘b – 28 cm

Area – 492 sq.cm

(iii) Height ‘h’ – 19 m

Parallel side ‘a’ – 18 m

Parallel side ‘b – 16 m

Area – 323 sq.m

(iv) Height ‘h’ – 16 cm

Parallel side ‘a’ – 15 cm

Parallel side ‘b – 30 cm

Area – 360 sq.cm

(2) Find the area of a trapezium whose parallel sides are 24 cm and 20 cm and the distance between them is 15 cm.

Solution :

Given height (h) = 15 cm

Parallel sides are (a) = 24 cm and (b) = 20 cm

Area of the trapezium = 1/2 x h (a + b) sq. units

= 1/2 × 15 × (24 + 20)

= 1/2 × 15 × 44

= 15 × 22

= 330 sq.cm

Therefore, the area of the trapezium is 330 sq.cm

(3) The area of a trapezium is 1586 sq. cm. The distance between its parallel sides is 26 cm. If one of the parallel sides is 84 cm then, find the other side.

Solution :

Given, the area of a trapezium = 1586 sq. cm.

Parallel sides are (a) = 84 cm and height (h) = 26 cm

To find Parallel sides are (b) = ?

We know,

Area of the trapezium = 1/2 × h (a + b) sq. units

1586 = 1/2 × 26 × (84 + b)

1586 = 13 × (84 + b)

(84 + b) = 1586/13

(84 + b) = 122

b = 122 – 84

b = 38 cm

Hence, other parallel side of trapezium is 38 cm

(4) The area of a trapezium is 1080 sq. cm. If the lengths of its parallel sides are 55.6 cm and 34.4 cm, find the distance between them.

Solution :

Given, the area of a trapezium = 1080 sq. cm.

Parallel sides are (a)

= 55.6 cm and Parallel sides are (b)

= 34.4 cm

To find height (h) = ?

We know,

Area of the trapezium = 1/2 × h (a+b) sq. units

1080 = 1/2 × h × ( 55.6 + 34.4)

1080 = 1/2 × h × 90

1080 = h × 45

h = 1080/45

h = 1080/45

h = 24 cm

Hence, distance between parallel side of trapezium is 24 cm.

(5) The area of a trapezium is 180 sq. cm and its height is 9 cm. If one of the parallel sides is longer than the other by 6 cm, find the length of the parallel sides.

Solution :

Given, the area of a trapezium = 180 sq. cm.

Let parallel sides are (a) = x cm and Parallel sides are (b) = (x + 6) cm

Height (h) = 9 cm

To find x = ?

We know,

Area of the trapezium = 1/2 × h (a + b) sq. units

180 = 1/2 × 9 ×(x + x + 6)

180/9 = 1/2 × (2x + 6)

20 = 1/2 × (2x + 6)

(2x + 6) = 20 × 2

(2x + 6) = 40

2x = 40 – 6

2x = 34

x = 34/2 = 17

Thus, a = 17 cm and b = x + 6

= 17 + 6

= 23 cm

Hence, parallel sides of trapezium is 23 cm and 17 cm.

(6) The sunshade of a window is in the form of isosceles trapezium whose parallel sides are 81 cm and 64 cm and the distance between them is 6 cm. Find the cost of painting the surface at the rate of Rs 2 per sq. cm.

Solution :

Given, height (h) = 6 cm

Parallel sides are (a) = 81 cm and (b) = 64 cm

Area of the trapezium

= 1/2 × h (a + b) sq. units

= 1 /2 × 6 × (81 + 64)

= 3 × (145)

= 435 sq.cm

Since, cost of painting for 1 sq.cm = ₹2

Therefore, cost of painting for 435 sq.cm

= ₹(2 × 435)

= ₹ 870

Hence, the cost of painting the surface is ₹870

(7) A window is in the form of trapezium whose parallel sides are 105 cm and 50 cm respectively and the distance between the parallel sides is 60 cm. Find the cost of the glass used to cover the window at the rate of ` 15 per 100 sq. cm.

Solution :

Given height (h) = 60 cm

Parallel sides are (a) = 105 cm and (b) = 50 cm

Area of the trapezium = 1/2 × h (a + b) sq. units

= 1/2 × 60 × (105 + 50)

= 30 × (155)

= 4650 sq.cm

Since, the cost of the glass used to cover the window at the rate of ₹ 15 per 100 sq. cm.

Thus, for 100 sq.cm = ₹15

So, for 4650 sq.cm = ₹x

Therefore, ₹x = 4650×15/100

= 46.5 × 15

= ₹697.5

Hence, the cost of the glass used to cover the window is ₹697.5

Objective Type Question Solutions :

(8) The area of the trapezium, if the parallel sides are measuring 8 cm and 10 cm and the height 5 cm is

(i) 45 sq. cm

(ii) 40 sq. cm

(iii) 18 sq. cm

(iv) 50 sq. cm

Solution :

Correct Option → (i)

45 sq. cm is the correct answer to this question.

(9) In a trapezium if the sum of the parallel sides is 10 m and the area is 140 sq. m, then the height is

(i) 7 cm

(ii) 40 cm

(iii) 14 cm

(iv) 28 cm

Solution :

Correct Option → (iv)

28 cm is the correct answer.

(10) When the non-parallel sides of a trapezium are equal then it is known as

(i) a square

(ii) a rectangle

(iii) an isosceles trapezium

(iv) a parallelogram

Solution :

Correct Option → (iii)

When the non-parallel sides of a trapezium are equal then it is known as an isosceles trapezium.

#### Measurements Exercise 2.4 Solutions :

(1) The base of the parallelogram is 16 cm and the height is 7 cm less than its base. Find the area of the parallelogram.

Solution :

Given, base of the parallelogram = 16 cm

Thus height of the parallelogram = 16 – 7 = 9 cm

We know,

Area of the parallelogram,

= b × h

= 16 × 9

= 144 sq.cm

Therefore, the area of parallelogram is 144 sq.cm

(2) An agricultural field is in the form of a parallelogram, whose area is 68.75 sq. hm. The distance between the parallel sides is 6.25 hm. Find the length of the base.

Solution :

Given, an agricultural field is in the form of a parallelogram, whose area is 68.75 sq. hm.

Also, The distance between the parallel sides is 6.25 hm.

That is, height of parallelogram = 6.25 hm

To find length of base = ?

Area of the parallelogram = b × h

68.75 = b × 6.25

b = 68.75/6.25

b = 11 hm

Hence, the length of base is 11 hm.

(3) A square and a parallelogram have the same area. If the side of the square is 48 m and the height of the parallelogram is 18 m, find the length of the base of the parallelogram.

Solution :

Given, a square and a parallelogram have the same area.

Thus, side of square = 48 m

Area of square = side × side

= 48 × 48

= 2304 sq m

Therefore, area of parallelogram = 2304 sq.m

Since, height of parallelogram = 18 m

We know, Area of the parallelogram = b × h

2304 = b × 18

b = 2304/18

b = 128 m

Hence, the length of the base of the parallelogram is 128 m.

(4) The height of the parallelogram is one fourth of its base. If the area of the parallelogram is 676 sq. cm, find the height and the base.

Solution :

Given, the area of the parallelogram is 676 sq. cm

Let b cm be base of parallelogram

Thus, height of parallelogram = b/4 cm

Area of the parallelogram = b × h

676 = b × b/4

b × b = 676 × 4

b² = 2704

taking square root on both sides

b = 52 cm

Thus, h = 52/4

h = 13 cm

Hence, base and height of parallelogram is 52 cm and 13 cm respectively.

(5) The area of the rhombus is 576 sq. cm and the length of one of its diagonal is half of the length of the other diagonal then find the length of the diagonals.

Solution :

Given, the area of the rhombus is 576 sq. cm

Let one of the diagonal is d cm

Thus, other diagonal is d/2 cm

We know,

Area of rhombus = 1/2 × d1 × d2

576 = 1/2 × d × d/2

576 = d × d/4

d2 = 576 × 4

d2 = 2304

d = 48 cm

Thus, d/2 = 48/2

= 24 cm

Hence, the diagonal of the rhombus is 48 cm and 24 cm

(6) A ground is in the form of isosceles trapezium with parallel sides measuring 42 m and 36 m long. The distance between the parallel sides is 30 m. Find the cost of levelling it at the rate of ` 135 per sq.m.

Solution :

Given height (h) = 30 cm

Parallel sides are (a) = 42 cm and (b) = 36 cm

Area of the trapezium,

= 1/2 × h (a + b) sq. units

= 1/2 × 30 × (42 + 36)

= 15 × 78

= 1170 sq m

Since, the cost of levelling it at the rate of ₹135 per sq.m

Thus for 1 sq.m = ₹135

Therefore, for 1170 sq.m

= 1170 × 135

= ₹1,57 950

Challenge Problem Solutions :

(7) In a parallelogram PQRS (see the diagram) PM and PN are the heights corresponding to the sides QR and RS respectively. If the area of the parallelogram is 900 sq. cm and the length of PM and PN are 20 cm and 36 cm respectively, find the length of the sides QR and SR.

Solution :

Given, the area of the parallelogram is 900 sq. cm

To find the length of the sides QR and SR.

Suppose we consider QR is base of parallelogram then corresponding height is PM = 20 cm

We know,

Area of the parallelogram = b ×h

900 = b × 20

b = 900/20 =45 cm

That is QR = 45 cm

Also, Suppose we consider RS is base of parallelogram then corresponding height is PN = 36 cm

Area of the parallelogram = b ×h

900 = b × 36

b = 900/36

b = 25 cm

That is SR = 25 cm

Hence, the length of the sides QR and SR is 45 cm and 25 cm respectively.

(8) If the base and height of a parallelogram are in the ratio 7:3 and the height is 45 cm then, find the area of the parallelogram.

Solution :

Given, base : height = 7 : 3 and height = 45 cm

Thus, base/45 = 7/3

base = 7 × 45/3

= 7 × 15

= 105 cm

We know, the area of the parallelogram = b × h

= 105 × 45

= 4725 sq.cm

Hence, the area of the parallelogram is 4725 sq.cm.

(9) Find the area of the parallelogram ABCD, if AC is 24 cm and BE = DF = 8 cm.

Solution :

Given, AC = 24 cm and BE = DF= 8 cm

From fig,

area of the parallelogram ABCD = area of ΔADC + area of ΔABC

= 1/2 × DF × AC + 1/2 × EB × AC

= 1/2 × AC (DF + EB)

= 1/2 × 24 (8 + 8)

= 12 × 16

= 192 sq.cm

Hence, area of the parallelogram ABCD is 192 sq.cm

(10) The area of the parallelogram ABCD is 1470 sq cm. If AB = 49 cm and AD = 35 cm then, find the heights DF and BE.

Solution :

Given, area of the parallelogram ABCD = 1470 sq cm

AB = 49 cm and AD = 35 cm

To find the length of the heights DF and BE.

Suppose we consider DF is height of parallelogram then corresponding base is AB = 49 cm

We know,

Area of the parallelogram = b × h

1470 = 49 × h

h = 1470/49

h = 30 cm

That is , DF = 30 cm

Also, Suppose we consider BE is height of parallelogram then corresponding base is AD = 35 cm

We know,

Area of the parallelogram = b × h

1470 = 35 × h

h = 1470/35

h = 42 cm

That is, BE = 42 cm

Hence, the length of the heights DF and BE is 30 cm and 42 cm respectively.

(12) A man has to build a rhombus shaped swimming pool. One of the diagonal is 13 m and the other is twice the first one. Then find the area of the swimming pool and also find the cost of cementing the floor at the rate of Rs 15 per sq.cm.

Solution :

Given, a man has to build a rhombus shaped swimming pool. One of the diagonal is 13 m and the other is twice the first one.

Thus, d1 = 13 m and d2

= 2 × 13

= 26 cm

We know, area of rhombus = 1/2 × d1 ×d2

= 1/2 × 13 × 26

= 13 × 13

= 169 sq.cm

Since, the cost of cementing the floor at the rate of ₹15 per sq.cm.

Thus, for 1 sq cm = ₹ 15

Therefore, for 169 sq cm,

= 15 × 169

= ₹2535

Hence, the area of the swimming pool is 169 sq.cm and the cost of cementing the floor is ₹2535.

(13) Find the height of the parallelogram whose base is four times the height and whose area is 576 sq. cm.

Solution :

Given area of parallelogram = 576 sq.cm

Since the height of the parallelogram whose base is four times the height

Let height of parallelogram is h

Thus, base (b) = 4 h

We know,

area of parallelogram = b × h

576 = 4h × h

h² = 576/4

h² = 144

h = 12 cm

Hence, the height of the parallelogram is 12 cm.

(14) The table top is in the shape of trapezium with measurements given in the figure. Find the cost of the glass used to cover the table at the rate of Rs 6 per 10 sq. cm.

Solution :

From fig, height (h) = 50 cm

Parallel sides are (a) = 200 cm and (b) = 150 cm

Area of the trapezium,

= 1/2 × h (a+b) sq. units

= 1/2 × 50 × (200+150)

= 25 × 350

= 8750 sq.cm

Since, the cost of the glass used to cover the table at the rate of ₹ 6 per 10 sq. cm.

Thus, for 10 sq.cm = ₹ 6

So, for 8750 sq.cm

= 8750 × 6/10

= 875 × 6

= ₹5250

Hence, the cost of the glass used to cover the table is ₹ 5250

(15) Arivu has a land ABCD with the measurements given in the figure. If a portion ABED is used for cultivation (where E is the mid-point of DC), find the cultivated area.

Solution :

From fig, land ABCD is rectangle shape

Thus, AB = DC

As AB = 24 m so DC = 24 m

Since, E is the mid point of DC

Thus, DE = DC/2

= 24/2

= 12 m

Also from fig we can say that a portion ABED is trapezium shape

From fig, height (h) = 18 m

Parallel sides are (a) = 24 m and (b) = 12 m

Area of the trapezium

= 1/2 × h (a + b) sq. units

= 1/2 × 18 × (24 + 12)

= 9 × 36

= 324 sq.m

Hence, the cultivated area will be 324 sq.m

Next Chapter Solution :

👉 Algebra

Updated: July 28, 2023 — 3:52 pm