Naresh Sawlani

Sector 34, Faridabad, India - 121003

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Sector 34, Faridabad, India - 121003.

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Dynamic, versatile, hands-on Learning & Development, Project Management, Educator and Training Professional with a passion of imparting and sharing knowledge and experience with future and current management professional and business leaders.

Have trained / mentored over 2000 professionals in various technical and project management disciplines.

Lectured into reputable management institutes as visiting faculty. Impart knowledge and corporate leanings of Marketing, Strategy, HRM and Project Management to the future business professionals.

Credentials hold: PGDBM (Marketing), PMP, B.Com, ITIL, Facet 5.

Have trained / mentored over 2000 professionals in various technical and project management disciplines.

Lectured into reputable management institutes as visiting faculty. Impart knowledge and corporate leanings of Marketing, Strategy, HRM and Project Management to the future business professionals.

Credentials hold: PGDBM (Marketing), PMP, B.Com, ITIL, Facet 5.

Sindhi

English

Hindi

Tolani Institute of Management Studies, Gujarat 1999

Master of Business Administration (M.B.A.)

Project Management Institute (PMI) 2008

Project Management Professional (PMP)

Sector 34, Faridabad, India - 121003

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Class I-V Tuition

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class I-V Tuition

7

Experience in School or College

Regularly take classes in a private institute in Lajpat Nagar

Teaching Experience in detail in Class I-V Tuition

I love to teach and see the progress my students make. Classes Teach: Grade 4 - Grade 9 (Maths, Science, English) Location: Student's Home (in the vicinity of 5-6 KM of Ashoka Enclave), At Home, Online. When you schedule a trial lesson with me, we’ll go over your goals and make a plan. I am very flexible to your needs. I’ll focus a lesson on exactly what you want to learn. I tend to make the learning process go smoothly. Can start the session the same day on agreeing to the contract. 1st Demo class will be free. Thanks for your time. Please contact if suits your requirement.

Taking a MBA class

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1. Which classes do you teach?

I teach Class 8 Tuition and Class I-V Tuition Classes.

2. Do you provide a demo class?

Yes, I provide a paid demo class.

3. How many years of experience do you have?

I have been teaching for 7 years.

Answered on 06 Aug CBSE/Class 9/Mathematics/Unit 5-Mensuration/Heron's formula/NCERT Solutions/Exercise 12.2

Radha made a picture of an aeroplane with coloured paper as shown in Figure. Find the total area of the paper used. ...more

Radha made a picture of an aeroplane with coloured paper as shown in Figure. Find the total area of the paper used.

For finding the area of the paper used, determine the area of each part separately and then find the sum of the areas to get the area of used paper. For region I (Triangle) Length of the sides of the triangle section I = a=5cm, b=1cm and c=5cm Semi Perimeter of the triangle, s =( a+b+c)/2s=(5 + 5 + 1)/2=... ...more

For finding the area of the paper used, determine the area of each part separately and then find the sum of the areas to get the area of used paper.

**For region I (Triangle)**

Length of the sides of the triangle section I = a=5cm, b=1cm and c=5cm

Semi Perimeter of the triangle,

s =( a+b+c)/2

s=(5 + 5 + 1)/2= 11/2cm

Semi perimeter = 11/2 cm = 5.5cm

Using heron’s formula,

Area of section I = √s (s-a) (s-b) (s-c)

= √5.5(5.5 – 5) (5.5 – 5) (5.5 – 1) cm²

= √5.5 × 0.5 × 0.5 × 4.5 cm²

= √5.5 × 0.5 × 0.5 × 4.5 cm²

= 0.75√11 cm²= 0.75 ×3.32 cm²

= 2.49 cm² (approx)

**Section II( rectangle)**

Length of the sides of the rectangle of section II = 6.5cm and 1cm

Area of section II = l ×b= 6.5 × 1

= 6.5cm²

**Section III is an isosceles trapezium**

In ? AMD

AD = 1cm (given)

AM + NB = AB – MN = 1cm

Therefore, AM = 0.5cm

Now,AD² =AM² +MD²

MD²= 1² – 0.5²

MD²= 1- 0.25= 0.75

MD = √0.75= √75/100=√3/4cm

Now, area of trapezium = ½(sum of parallel sides)×height

=½×(AB+DC)×MD

=½×(2+1)×√3/4

= ½(3)×√(3/4)= ½×3×√3×2=(3/4)√3

= (3/4)×1.73= 1.30cm²(approx)

[√3=1.73....]

Hence, area of trapezium = 1.30cm²

**Section IV and V** are 2 congruent right angled triangles with base 6cm and height 1.5cm

Area of region IV and V = 2 (½ × 6 × 1.5)cm² = 9cm²

Total area of the paper used = Area I + Area II + Area III + (Area IV + Area V) = (2.49+ 6.5 + 1.30 + 9)

= 19.3 cm² (approx)

Like 0

Answers 1 Comments Answered on 06 Aug CBSE/Class 9/Mathematics/Unit 5-Mensuration/Heron's formula/NCERT Solutions/Exercise 12.1

Heron's Formula for the area of a triangle(Hero's Formula) A method for calculating the area of a triangle when you know the lengths of all three sides. Let a=15m,b=11m,c=6m be the lengths of the sides of a triangle. The area is given by: Area = √p(p−a)(p−b)(p−c) where p is half... ...more

Heron's Formula for the area of a triangle(Hero's Formula)

A method for calculating the area of a triangle when you know the lengths of all three sides.

Let a=15m,b=11m,c=6m be the lengths of the sides of a triangle. The area is given by:

Area = √p(p−a)(p−b)(p−c)

where p is half the perimeter, or (a+b+c)/2 = (15+11+6)/2 =16m

Therefore, Area = √16(16-15)(16-11)(16-6) = √16(1)(5)(10) =20√2 = 20 × 1.414 = 28.28 m²

Like 0

Answers 3 Comments Naresh SawlaniDirections

x Class I-V Tuition 5.0

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class I-V Tuition

7

Experience in School or College

Regularly take classes in a private institute in Lajpat Nagar

Teaching Experience in detail in Class I-V Tuition

I love to teach and see the progress my students make. Classes Teach: Grade 4 - Grade 9 (Maths, Science, English) Location: Student's Home (in the vicinity of 5-6 KM of Ashoka Enclave), At Home, Online. When you schedule a trial lesson with me, we’ll go over your goals and make a plan. I am very flexible to your needs. I’ll focus a lesson on exactly what you want to learn. I tend to make the learning process go smoothly. Can start the session the same day on agreeing to the contract. 1st Demo class will be free. Thanks for your time. Please contact if suits your requirement.

Class 8 Tuition 5.0

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class 8 Tuition

7

Experience in School or College

Teaching experience of over 3 years

Teaching Experience in detail in Class 8 Tuition

3 years tuition and teaching experience to students from VI-IX Subjects taught: English, Mathematics, Science

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Answered on 06 Aug CBSE/Class 9/Mathematics/Unit 5-Mensuration/Heron's formula/NCERT Solutions/Exercise 12.2

Radha made a picture of an aeroplane with coloured paper as shown in Figure. Find the total area of the paper used. ...more

Radha made a picture of an aeroplane with coloured paper as shown in Figure. Find the total area of the paper used.

For finding the area of the paper used, determine the area of each part separately and then find the sum of the areas to get the area of used paper. For region I (Triangle) Length of the sides of the triangle section I = a=5cm, b=1cm and c=5cm Semi Perimeter of the triangle, s =( a+b+c)/2s=(5 + 5 + 1)/2=... ...more

For finding the area of the paper used, determine the area of each part separately and then find the sum of the areas to get the area of used paper.

**For region I (Triangle)**

Length of the sides of the triangle section I = a=5cm, b=1cm and c=5cm

Semi Perimeter of the triangle,

s =( a+b+c)/2

s=(5 + 5 + 1)/2= 11/2cm

Semi perimeter = 11/2 cm = 5.5cm

Using heron’s formula,

Area of section I = √s (s-a) (s-b) (s-c)

= √5.5(5.5 – 5) (5.5 – 5) (5.5 – 1) cm²

= √5.5 × 0.5 × 0.5 × 4.5 cm²

= √5.5 × 0.5 × 0.5 × 4.5 cm²

= 0.75√11 cm²= 0.75 ×3.32 cm²

= 2.49 cm² (approx)

**Section II( rectangle)**

Length of the sides of the rectangle of section II = 6.5cm and 1cm

Area of section II = l ×b= 6.5 × 1

= 6.5cm²

**Section III is an isosceles trapezium**

In ? AMD

AD = 1cm (given)

AM + NB = AB – MN = 1cm

Therefore, AM = 0.5cm

Now,AD² =AM² +MD²

MD²= 1² – 0.5²

MD²= 1- 0.25= 0.75

MD = √0.75= √75/100=√3/4cm

Now, area of trapezium = ½(sum of parallel sides)×height

=½×(AB+DC)×MD

=½×(2+1)×√3/4

= ½(3)×√(3/4)= ½×3×√3×2=(3/4)√3

= (3/4)×1.73= 1.30cm²(approx)

[√3=1.73....]

Hence, area of trapezium = 1.30cm²

**Section IV and V** are 2 congruent right angled triangles with base 6cm and height 1.5cm

Area of region IV and V = 2 (½ × 6 × 1.5)cm² = 9cm²

Total area of the paper used = Area I + Area II + Area III + (Area IV + Area V) = (2.49+ 6.5 + 1.30 + 9)

= 19.3 cm² (approx)

Like 0

Answers 1 Comments Answered on 06 Aug CBSE/Class 9/Mathematics/Unit 5-Mensuration/Heron's formula/NCERT Solutions/Exercise 12.1

Heron's Formula for the area of a triangle(Hero's Formula) A method for calculating the area of a triangle when you know the lengths of all three sides. Let a=15m,b=11m,c=6m be the lengths of the sides of a triangle. The area is given by: Area = √p(p−a)(p−b)(p−c) where p is half... ...more

Heron's Formula for the area of a triangle(Hero's Formula)

A method for calculating the area of a triangle when you know the lengths of all three sides.

Let a=15m,b=11m,c=6m be the lengths of the sides of a triangle. The area is given by:

Area = √p(p−a)(p−b)(p−c)

where p is half the perimeter, or (a+b+c)/2 = (15+11+6)/2 =16m

Therefore, Area = √16(16-15)(16-11)(16-6) = √16(1)(5)(10) =20√2 = 20 × 1.414 = 28.28 m²

Like 0

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