ML Aggarwal Solutions for Class 7 Maths Chapter 16- Perimeter and Area

Class 7: Maths Chapter 16 solutions. Complete Class 7 Maths Chapter 16 Notes.

ML Aggarwal Solutions for Class 7 Maths Chapter 16- Perimeter and Area

ML Aggarwal 7th Maths Chapter 16, Class 7 Maths Chapter 16 solutions

1. ABCD is a square of side 24 cm. EF is parallel to BC and AE = 15 cm. By how much does

(i) the perimeter of AEFD exceed the perimeter of EBCF?

(ii) the area of AEFD exceed the area of EBCF?

Solution:

Given

The side of the square ABCD = 24 cm

EF || BC || AD

AE is of length 15 cm

EB = 24 – 15

EB = 9 cm

(i) Perimeter of AEFD = 2 (15 + 24) cm

= 2 (39) cm

We get,

= 78 cm

Perimeter of EBCF = 2 (9 + 24) cm

= 2 (33) cm

We get,

= 66 cm

Now,

Difference between the two perimeters = 78 – 66

= 12 cm

(ii) Area of AEFD = l × b

= 15 × 24

We get,

= 360 sq. cm

Area of EBCF = 9 × 24

= 216 sq. cm

Now, difference between the two areas = 360 – 216

= 144 sq. cm

2. Nagma runs around a rectangular park 180 m long and 120 m wide at the rate of 7.5 km/ hour. In how much time will she complete five rounds?

Solution:

Length of rectangular plot (l) = 180 m

Breadth of rectangular (b) = 120 m

Perimeter = 2 (l + b)

= 2 (180 + 120) m

= 2 (300)

We get,

= 600 m

Distance travelled in 5 rounds = 600 × 5

= 3000 m i.e

= 3 km

Given speed = 7.5 km/hr

Time taken = (3 / 7.5) h

= {(3 × 10) / 75} h

= (30 / 75) h

We get,

= (2 / 5) h

= (2 / 5) × 60

= 2 × 12

= 24 minutes

3. The area of a rectangular plot is 540 m², if its length is 27 m, find its breadth and perimeter.

Solution:

Given

Area of a rectangular plot = 540 m²

Length (l) = 27 m

Therefore,

Breadth = (Area) / (Length)

= (540 / 27) m

We get,

= 20 m

And Perimeter = 2 (l + b)

= 2 (27 + 20) m

= 2 (47) m

We get,

= 94 m

Therefore, the breadth of the rectangular plot is 20 m and its perimeter is 94 m

4. Find the area of each of the following parallelogram:

Solution:

(i) Base of the parallelogram (b) = 8 cm and

Height of the parallelogram (h) = 4.5 cm

Area = Base × Height

= 8 × 4.5

= 36 cm²

(ii) Base of the parallelogram (b) = 2 cm

Height of the parallelogram (h) = 4.4 cm

Area = Base × Height

= 2 × 4.4

= 8.8 cm²

(iii) Base of the parallelogram (b) = 2.5 cm

Height of the parallelogram (h) = 3.5 cm

Area = Base × Height

= 2.5 × 3.5

= 8.75 cm²

5. Find the area of each of the following triangles:

Solution:

(i) Base of the triangle (b) = 6.4 cm

Height of the triangle (h) = 6 cm

Area of the triangle = (1 / 2) × base × height

= (1 / 2) × 6.4 × 6 cm²

We get,

= 19.2 cm²

(ii) Base of the triangle (b) = 5 cm

Height of the triangle (h) = 6 cm

Area of the triangle = (1 / 2) × base × height

= (1 / 2) × 5 × 6

We get,

= 15 cm²

(iii) Base of the triangle (b) = 4.5 cm

Height of the triangle (h) = 6 cm

Area of the triangle = (1 / 2) × base × height

= (1 / 2) × 4.5 × 6

We get,

= 13.5 cm²

6. Find the circumference of the circles with the following radius:

(i) 7 cm

(ii) 21 cm

(iii) 28 mm

(iv) 3.5 cm

Solution:

(i) Radius of the circle (r) = 7 cm

We know that,

Circumference of the circle = 2πr

= 2 × (22 / 7) × 7

We get,

= 44 cm

Hence, the circumference of the circle is 44 cm

(ii) Radius of the circle (r) = 21 cm

We know that,

Circumference of the circle = 2πr

= 2 × (22 / 7) × 21

We get,

= 132 cm

Hence, the circumference of the circle is 132 cm

(iii) Radius of the circle (r) = 28 mm

We know that,

Circumference of the circle = 2πr

= 2 × (22 / 7) × 28

We get,

= 176 mm

Hence, the circumference of the circle is 176 mm

(iv) Radius of the circle = 3.5 cm

We know that,

Circumference of the circle = 2πr

= 2 × (22 / 7) × 3.5

We get,

= 22 cm

Hence, the circumference of the circle is 22 cm

7. Find the area of the circles, given that:

(i) radius = 14 mm

(ii) diameter = 49 m

(iii) diameter = 9.8 m

(iv) radius = 5 cm

Solution:

(i) Radius (r) = 14 mm

Hence,

Area = πr²

= (22 / 7) × 14 × 14

We get,

= 616 cm²

Therefore, the area of the circle is 616 cm²

(ii) Diameter = 49 m

Hence,

Radius = (49 / 2) m

Area = πr²

= (22 / 7) × (49 / 2) × (49 / 2)

We get,

= 3773 m²

Therefore, the area of the circle is 3773 m²

(iii) Diameter = 9.8 m

Hence,

Radius = (9.8 / 2)

= 4.9 m

Area = πr²

= (22 / 7) × 4.9 × 4.9

We get,

= 75. 46 m²

Therefore, the area of the circle is 75.46 m²

(iv) Radius = 5 cm

Area = πr²

= (22 / 7) × 5 × 5

We get,

= (550 / 7)

=
cm²

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ML Aggarwal Solutions for Class 7 Maths Chapter 16- Perimeter and Area

Download PDF: ML Aggarwal Solutions for Class 7 Maths Chapter 16- Perimeter and Area PDF

Chapterwise ML Aggarwal Solutions for Class 7 Maths :

• Chapter 1- Integers
• Chapter 2- Fractions and Decimals
• Chapter 3- Rational Numbers
• Chapter 4- Exponents and Powers
• Chapter 5- Sets
• Chapter 6- Ratio and Proportion
• Chapter 7- Percentage and Its Applications
• Chapter 8- Algebraic Expressions
• Chapter 9- Linear Equations and Inequalitie
• Chapter 10- Lines and Angles
• Chapter 11- Triangles and its Properties
• Chapter 12- Congruence of Triangles
• Chapter 13- Practical Geometry
• Chapter 14- Symmetry
• Chapter 15- Visualising Solid Shapes
• Chapter 16- Perimeter and Area
• Chapter 17- Data Handling

About ML Aggarwal

M. L. Aggarwal, is an Indian mechanical engineer, educator. His achievements include research in solutions of industrial problems related to fatigue design. Recipient Best Paper award, Manipal Institute of Technology, 2004. Member of TSTE.