NCERT Solutions Class 7 Maths Chapter 11

NCERT Solutions Class 7 Mathematics Chapter 11: Perimeter and Area

Class 7 introduces students to several new Mathematics topics and concepts, which will be further built upon in higher grades. This makes it very important that Class 7 students focus on building a strong foundation in this subject at this stage.

Students can use the NCERT Solutions for Class 7 Mathematics Chapter 11 by Extramarks for their preparations. They can get detailed and accurate solutions to questions given in the NCERT textbook, which will help them understand how to solve the different kinds of problems in a step-by-step manner. If they ever get stuck on a question, they can always refer to the solutions prepared by subject matter experts.

NCERT Solutions Class 7 Mathematics Chapter 11 – 

Access NCERT Solutions for Class 7 Mathematics Chapter 11 – Perimeter and Area

NCERT Solutions for Class 7 Mathematics Chapter 11 – 

NCERT Class 7 Mathematics Chapter 11 teaches students how to calculate the perimeters and areas of different types of basic shapes, including squares, circles, triangles and parallelograms. Students are introduced to a lot of different formulas that they must remember and retain to perform well in the exam. They  are also taught about how to handle the conversion of measurement units.

Some of the core concepts and formulas included in the chapter are as follows:

  • Perimeter is the total length of the boundary of a closed rectilinear figure.
  • The surface circled by a two-dimensional figure is its area.
  • Area of a Parallelogram = Base × Altitude 
  • Area of a Triangle = ½ × Base × Altitude corresponding to the base
  • Circumference of a circle = 2πr.
  • Area of a Circle = πr²
  • 1 cm² = 100 mm²
  • 1m² = 10000 cm²
  • 1 hectare = 10000 m²
  • The perimeter of a regular polygon = Length of a side × Number of the sides 

 Triangle: A triangle is a plane figure surrounded by three lines segments.

Rectilinear Figure: A plane figure encompassed by line segments is called a rectilinear figure. None of the two sides of it intersects at a point other than the vertex.

Quadrilateral: It is a plane figure surrounded by four long segments.

Parallelogram: Parallelogram is a plane figure encircled by four long segments. The opposite sides of a parallelogram are equal and parallel in length. 

Area of a Triangle: The area of a triangle is ½ × Base × Height (Altitude corresponding to the base). 

Example: base = 6 cm, height = 12cm. What is the area of a triangle?

As the formula goes; ½ × Base × Height

= (½ × 6 × 12) cm² = 36 cm²

 Area of a Parallelogram: Base × Altitude (height)

Example: base = 10 cm and height = 5 cm

The formula of parallelogram: base × height 

= (10 × 5) cm²

= 50 cm²

Circumference of a Circle: Perimeter of a circle is similar to the circumference of a circle. The students would draw four circles of different radii and find their circumference with the ratio of the circumferences to the diameter. The ratio will be more than three times its diameter.

 Area of a Circle: The area a circle occupies on a flat plane which has its length and breadth, is known as the area of a surface. The formula to find the area is πr². 

Conversion of Units

Units of Length

Units of Area

1 cm = 10 mm 1 cm² = (10 x 10) mm² = 100 mm²
1 dm = 10 cm 1 dm² = (10 x 10) cm² = 100 cm²
1 m = 10 dm 1 m² = (10 x 10) dm² = 100 dm²
1 dam = 10 m 1 dam² = (10 x 10) m² = 100 m²
1 hm = 10 dam 1 hm² = ( 10 x 10 ) dam² = 100 dam²
1 km = 10 hm 1 km² = ( 10 x 10) hm² = 100 hm²

Note:

1m = 100 cm ∴ 1 m² = 10000 cm²

1 km = 1000 m ∴ 1 km² = 10, 00, 000 m²

1 hectare (ha) = 100 m x 100m = 10000 m²

NCERT Solutions for Class 7 Mathematics

Mathematics is that one subject that many students struggle with. This is because getting good at Mathematics requires one to practise solving many problems and most students don’t do that. With NCERT Solutions of Class 7 Mathematics, Extramarks tries to provide students with a very valuable resource that they can use to get better at the subject and eventually score well in their final exams. Students can use these solutions if they ever get stuck on a problem or to learn the approach to how to solve the different kinds of problems encountered in any NCERT Mathematics Class 7 Chapter 11.

The NCERT Solutions by Extramarks are prepared by subject matter experts who give special attention to providing detailed and accurate solutions in a step-by-step manner so that students can easily grasp the problem-solving approach. Students can use these resources for their preparations, last-minute revisions, and for help with their assignments.

NCERT Solutions for Class 7

Apart from Mathematics, Extramarks provides comprehensive  solutions to textbook questions for all of the other subjects covered in Class 7. All the solutions are prepared by their respective subject experts and extra care is given to providing accurate solutions which students can rely on. Students can download all these solutions from the Extramarks website.

Q.1 Find the area of a square park whose perimeter is 320 m.

Ans

Let the side of the square park be ‘a. Since perimeter of a square=4a So, we get 4a=320 a= 320 4 =80m Therefore, the area of the square park= a 2 = ( 80m ) 2 =6400 m 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@1E02@

Q.2 Find the breadth of a rectangular plot of land, if its area is 440 m2 and the length is 22 m. Also find its perimeter.

Ans

Let the breadth of the rectangular plot be x. Area of the rectangular plot=length×breadth SO, we get 440m 2 =22m×x x= 440 m 2 22m =20m Perimeter of the rectangular plot=2( length+breadth ) =2( 20m+22m ) =2( 42m ) =84mMathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@5ED5@

Q.3 The perimeter of a rectangular sheet is 100 cm. If the length is 35 cm, find its breadth. Also find the area.

Ans

Let the breadth of the rectangular sheet be x. Perimeter of the rectangular sheet=2( length+breadth ) 100 cm=2( 35cm+x ) 100 2 cm=( 35 cm+x ) 50cm=35 cm+x x=50 cm35 cm =15 cm Area of rectangular sheet=length×breadth =35 cm×15 cm = 525 cm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@9138@

Q.4

The area of a square park is the same as of a rectangularpark. If the side of the square park is 60 m and the lengthof the rectangular park is 90 m, find the breadth of therectangular park.

Ans

Let the breadth of rectangular park be x. Since, area of a square park is the same as of a rectangular park. Area of square=Area of rectangular park ( 60m ) 2 =90m×x 3600 m 2 =90m×x x= 3600 m 2 90m =40m MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@44FC@

Q.5

A wire is in the shape of a rectangle. Its length is 40 cmand breadth is 22 cm. If the same wire is rebent in theshape of a square, what will be the measure of each side.Also find which shape encloses more area?

Ans

Perimeter of a rectangle=Perimeter of square 2( length+breadth )=4×side 2( 40m+22m )=4×side side= 124 4 =31cm So, Area of rectangle=length×breadth =40cm×22cm = 880cm 2 Area of square= ( side ) 2 = ( 31cm ) 2 =961 cm 2 Therefore, the square-shaped wire encloses more area. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@AE17@

Q.6

The perimeter of a rectangle is 130 cm. If the breadthof the rectangle is 30 cm, find its length. Also find thearea of the rectangle.

Ans

Let the length of the rectangle be x cm. Perimeter of rectangle = 2 (length+breadth) 130cm=2(30 cm+x) 130 2 cm=30 cm+x x=65 cm30 cm =35 cm Now, area of rectangle=length×breadth =30 cm×35 cm = 1050 cm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@601F@

Q.7 Find the area of each of the following parallelogram :

Ans

(a) Area of Parallelogram=Base×Height Height=4 cm Base=7 cm So, area of parallelogram=7×4 = 28cm 2 (b) Area of Parallelogram=Base×Height Height=3 cm Base=5 cm So, area of parallelogram=5×3 = 15 cm 2 (c) Area of Parallelogram=Base×Height Height=3.5 cm Base=2.5 cm So, area of parallelogram=2.5×3.5 =8 .75 cm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@F5DB@ (d) Area of Parallelogram=Base×Height Height=4.8 cm Base=5 cm So, area of parallelogram=5×4.8 = 24 cm 2 (e) Area of Parallelogram=Base×Height Height=4.4 cm Base=2 cm So, area of parallelogram=2×4.4 =8 .8 cm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaaiabbIcaOiabbsgaKjabbMcaPaqaaiabbgeabjabbkhaYjabbwgaLjabbggaHjabbccaGiabb+gaVjabbAgaMjabbccaGiabbcfaqjabbggaHjabbkhaYjabbggaHjabbYgaSjabbYgaSjabbwgaLjabbYgaSjabb+gaVjabbEgaNjabbkhaYjabbggaHjabb2gaTjabg2da9iabbkeacjabbggaHjabbohaZjabbwgaLjabgEna0kabbIeaijabbwgaLjabbMgaPjabbEgaNjabbIgaOjabbsha0bqaaiaaxMaacaWLjaGaaCzcaiaaysW7caaMe8UaaGPaVlabbIeaijabbwgaLjabbMgaPjabbEgaNjabbIgaOjabbsha0jabg2da9iabbsda0iabb6caUiabbIda4iabbccaGiabbogaJjabb2gaTbqaaiaaxMaacaWLjaGaaCzcaiaaysW7caaMe8UaaGPaVlaaysW7caaMe8UaaGjbVlabbkeacjabbggaHjabbohaZjabbwgaLjabg2da9iabbwda1iabbccaGiabbogaJjabb2gaTbqaaiabbofatjabb+gaVjabbYcaSaqaaiabbggaHjabbkhaYjabbwgaLjabbggaHjabbccaGiabb+gaVjabbAgaMjabbccaGiabbchaWjabbggaHjabbkhaYjabbggaHjabbYgaSjabbYgaSjabbwgaLjabbYgaSjabb+gaVjabbEgaNjabbkhaYjabbggaHjabb2gaTjabg2da9iabbwda1iabgEna0kabbsda0iabb6caUiabbIda4aqaaiaaxMaacaWLjaGaaCzcaiaaxMaacaaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaykW7cqGH9aqpcqqGYaGmcqqG0aancqqGGaaicqqGJbWycqqGTbqBdaahaaWcbeqaaiabikdaYaaaaOqaaiabbIcaOiabbwgaLjabbMcaPaqaaiabbgeabjabbkhaYjabbwgaLjabbggaHjabbccaGiabb+gaVjabbAgaMjabbccaGiabbcfaqjabbggaHjabbkhaYjabbggaHjabbYgaSjabbYgaSjabbwgaLjabbYgaSjabb+gaVjabbEgaNjabbkhaYjabbggaHjabb2gaTjabg2da9iabbkeacjabbggaHjabbohaZjabbwgaLjabgEna0kabbIeaijabbwgaLjabbMgaPjabbEgaNjabbIgaOjabbsha0bqaaiaaxMaacaWLjaGaaCzcaiaaysW7caaMe8UaaGPaVlabbIeaijabbwgaLjabbMgaPjabbEgaNjabbIgaOjabbsha0jabg2da9iabbsda0iabb6caUiabbsda0iabbccaGiabbogaJjabb2gaTbqaaiaaxMaacaWLjaGaaCzcaiaaysW7caaMe8UaaGPaVlaaysW7caaMe8UaaGPaVlabbkeacjabbggaHjabbohaZjabbwgaLjabg2da9iabbkdaYiabbccaGiabbogaJjabb2gaTbqaaiabbofatjabb+gaVjabbYcaSaqaaiabbggaHjabbkhaYjabbwgaLjabbggaHjabbccaGiabb+gaVjabbAgaMjabbccaGiabbchaWjabbggaHjabbkhaYjabbggaHjabbYgaSjabbYgaSjabbwgaLjabbYgaSjabb+gaVjabbEgaNjabbkhaYjabbggaHjabb2gaTjabg2da9iabbkdaYiabgEna0kabbsda0iabb6caUiabbsda0aqaaiaaxMaacaWLjaGaaCzcaiaaxMaacaaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7cqGH9aqpcqqG4aaocqqGUaGlcqqG4aaocqqGGaaicqqGJbWycqqGTbqBdaahaaWcbeqaaiabikdaYaaaaaaa@66E5@

Q.8 Find the area of each of the following triangles :

Ans

(a) Area of a triangle= 1 2 ×Base×Height = 1 2 ×4×3 = 1 2 ×12=6 cm 2 (b) Area of a triangle= 1 2 ×Base×Height = 1 2 ×5×3.2 = 1 2 ×16=8 cm 2 (c) Area of a triangle= 1 2 ×Base×Height = 1 2 ×3×4 = 1 2 ×12=6 cm 2 (d) Area of a triangle= 1 2 ×Base×Height = 1 2 ×3×2 = 1 2 ×6=3 cm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaaiabbIcaOiabbggaHjabbMcaPiabbccaGiabbgeabjabbkhaYjabbwgaLjabbggaHjabbccaGiabb+gaVjabbAgaMjabbccaGiabbggaHjabbccaGiabbsha0jabbkhaYjabbMgaPjabbggaHjabb6gaUjabbEgaNjabbYgaSjabbwgaLjabg2da9maalaaabaGaeGymaedabaGaeGOmaidaaiabgEna0kabbkeacjabbggaHjabbohaZjabbwgaLjabgEna0kabbIeaijabbwgaLjabbMgaPjabbEgaNjabbIgaOjabbsha0bqaaiaaxMaacaWLjaGaaCzcaiaaxMaacqGH9aqpdaWcaaqaaiabigdaXaqaaiabikdaYaaacqGHxdaTcqaI0aancqGHxdaTcqaIZaWmaeaacaWLjaGaaCzcaiaaxMaacaWLjaGaeyypa0ZaaSaaaeaacqaIXaqmaeaacqaIYaGmaaGaey41aqRaeGymaeJaeGOmaiJaeyypa0JaeGOnayJaeeiiaaIaee4yamMaeeyBa02aaWbaaSqabeaacqaIYaGmaaaakeaacqqGOaakcqqGIbGycqqGPaqkcqqGGaaicqqGbbqqcqqGYbGCcqqGLbqzcqqGHbqycqqGGaaicqqGVbWBcqqGMbGzcqqGGaaicqqGHbqycqqGGaaicqqG0baDcqqGYbGCcqqGPbqAcqqGHbqycqqGUbGBcqqGNbWzcqqGSbaBcqqGLbqzcqGH9aqpdaWcaaqaaiabigdaXaqaaiabikdaYaaacqGHxdaTcqqGcbGqcqqGHbqycqqGZbWCcqqGLbqzcqGHxdaTcqqGibascqqGLbqzcqqGPbqAcqqGNbWzcqqGObaAcqqG0baDaeaacaWLjaGaaCzcaiaaxMaacaWLjaGaeyypa0ZaaSaaaeaacqaIXaqmaeaacqaIYaGmaaGaey41aqRaeGynauJaey41aqRaeG4mamJaeiOla4IaeGOmaidabaGaaCzcaiaaxMaacaWLjaGaaCzcaiabg2da9maalaaabaGaeGymaedabaGaeGOmaidaaiabgEna0kabigdaXiabiAda2iabg2da9iabiIda4iabbccaGiabbogaJjabb2gaTnaaCaaaleqabaGaeGOmaidaaaGcbaGaeeikaGIaee4yamMaeeykaKIaeeiiaaIaeeyqaeKaeeOCaiNaeeyzauMaeeyyaeMaeeiiaaIaee4Ba8MaeeOzayMaeeiiaaIaeeyyaeMaeeiiaaIaeeiDaqNaeeOCaiNaeeyAaKMaeeyyaeMaeeOBa4Maee4zaCMaeeiBaWMaeeyzauMaeyypa0ZaaSaaaeaacqaIXaqmaeaacqaIYaGmaaGaey41aqRaeeOqaiKaeeyyaeMaee4CamNaeeyzauMaey41aqRaeeisaGKaeeyzauMaeeyAaKMaee4zaCMaeeiAaGMaeeiDaqhabaGaaCzcaiaaxMaacaWLjaGaaCzcaiabg2da9maalaaabaGaeGymaedabaGaeGOmaidaaiabgEna0kabiodaZiabgEna0kabisda0aqaaiaaxMaacaWLjaGaaCzcaiaaxMaacqGH9aqpdaWcaaqaaiabigdaXaqaaiabikdaYaaacqGHxdaTcqaIXaqmcqaIYaGmcqGH9aqpcqaI2aGncqqGGaaicqqGJbWycqqGTbqBdaahaaWcbeqaaiabikdaYaaaaOqaaiabbIcaOiabbsgaKjabbMcaPiabbccaGiabbgeabjabbkhaYjabbwgaLjabbggaHjabbccaGiabb+gaVjabbAgaMjabbccaGiabbggaHjabbccaGiabbsha0jabbkhaYjabbMgaPjabbggaHjabb6gaUjabbEgaNjabbYgaSjabbwgaLjabg2da9maalaaabaGaeGymaedabaGaeGOmaidaaiabgEna0kabbkeacjabbggaHjabbohaZjabbwgaLjabgEna0kabbIeaijabbwgaLjabbMgaPjabbEgaNjabbIgaOjabbsha0bqaaiaaxMaacaWLjaGaaCzcaiaaxMaacqGH9aqpdaWcaaqaaiabigdaXaqaaiabikdaYaaacqGHxdaTcqaIZaWmcqGHxdaTcqaIYaGmaeaacaWLjaGaaCzcaiaaxMaacaWLjaGaeyypa0ZaaSaaaeaacqaIXaqmaeaacqaIYaGmaaGaey41aqRaeGOnayJaeyypa0JaeG4mamJaeeiiaaIaee4yamMaeeyBa02aaWbaaSqabeaacqaIYaGmaaaaaaa@6A5E@

Q.9 Find the missing values :

S. NO. Base Height Area of the Parallelogram
a. 20 cm 246 cm2
b. 15 cm 154.5 cm2
c. 8.4 cm 78.72 cm2
d. 15.6 cm 16.38 cm2

Ans

Let the height be h and base be b. (a) Area of parallelogram=Base×Height So,we get 246 cm 2 =20 cm×h h= 246 cm 2 20cm =12.3cm (b) Area of parallelogram=Base×Height So,we get 154 .5 cm 2 =b×15 cm b= 154.5 cm 2 15cm =10.3cm MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@3796@ (c) Area of parallelogram=Base×Height So,we get 78 .72 cm 2 =b×8.4 cm b= 48.72 cm 2 8.4cm =5.8cm (d) Area of parallelogram=Base×Height So,we get 16 .38 cm 2 =15.6 cm×h h= 16.38 cm 2 15.6cm =1.05cm So, we get S.No. Base Height Area of the Parallelogram a. 20 cm 12.3cm 246 cm 2 b. 10.3cm 15 cm 154.5 cm 2 c. 5.8cm 8.4 cm 78.72 cm 2 d. 15.6 cm 1.05cm 16.38 cm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@BA07@

Q.10 Find the missing values :

Base Height Area of Triangle
15 cm …………….. 84 cm2
…………….. 31.4 mm 1256 mm2
22 cm …………….. 170.5 cm2

Ans

(a) Let the height be h and base be b. Area of triangle= 1 2 ×base×height 87cm 2 = 1 2 ×15cm×h h= 87 cm 2 ×2 15cm =11.6 cm (b) Let the height be h and base be b. Area of triangle= 1 2 ×base×height 1256 mm 2 = 1 2 ×b×31.4 mm b= 1256 mm 2 ×2 31.4mm =80 mm MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@7243@ (c) Let the height be h and base be b. Area of triangle= 1 2 ×base×height 170 .5 cm 2 = 1 2 ×22cm×h h= 170.5 cm 2 ×2 22cm =15.5 cm So we get Base Height Area of Triangle 15 cm 11.6 cm 84 cm 2 80mm 31.4 mm 1256 mm 2 22 cm 15.5cm 170.5 cm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@4B95@

Q.11

PQRS is a parallelogram. QM is the height from Q to SRand QN is the height from Q to PS. If SR =12 cm andQM = 7.6 cm.Find:a the area of the parallegram PQRSb QN, if PS = 8 cm

Ans

(a) Area of a parallelogram=Base×Height =SR×QM =7.6×12 =91.2 cm 2 (b) Area of a parallelogram=Base×Height =PS×QN =91.2 cm 2 QN×8=91.2 QN= 91.2 8 =11.4cm MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@246A@

Q.12

DL and BM are the heights on sides AB and AD respectivelyof parallelogram ABCD. If the area of the parallelogram is1470 cm2,AB = 35 cm and AD = 49 cm, find the length ofBMand DL.

Ans

Area of parallelogram=Base×Height =AB×DL 1470=35×DL DL= 1470 35 =42cm Also, 1470=AD×BM 1470=49×BM BM= 1470 49 =30cm MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@ECDA@

Q.13

ABC is right angled at A .AD is perpendicular to BC.If AB=5 cm, BC=13 cm and AC=12 cm, Find the area ofABC. Also find the length of AD.

Ans

Since, Area= 1 2 ×Base×Height = 1 2 ×5×12 = 30 cm 2 Also, area of triangle= 1 2 ×AD×BC 30= 1 2 ×AD×13 AD= 30×2 12 =4.6cm MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@DD81@

Q.14

ABC is isosceles with AB=AC=7.5 cm and BC=9 cm. The height AD from A to BC, is 6 cm. Find the area of ABC. What will be the height from C to AB i.e., CE?

Ans

Area of ΔABC= 1 2 ×Base×Height = 1 2 BC×AD= 1 2 ×9×6= 27cm 2 Also, area of ΔABC= 1 2 ×Base×Height = 1 2 AB×CE 27= 1 2 7.5×CE CE= 27×2 7.5 =7.2 cm MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@E877@

Q.15

Find the circumference of the circles with the followingradius: Take π=227a 14 cm b 28 mm c 21 cm

Ans

(a) Circumference of a circle is =2πr =2× 22 7 ×14 =2×22×2 =88cm (b) Circumference of a circle is =2πr =2× 22 7 ×28 =2×22×4 =176mm MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@0482@ (c) Circumference of a circle is =2πr =2× 22 7 ×21 =2×22×3 =132cm MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@A3AF@

Q.16

Find the area of the following circles, given that:a radius = 14 mm Take π=227b diameter = 49 mc radius = 5 cm

Ans

(a) Area of a circle=π r 2 = 22 7 ( 14 ) 2 =616 mm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@7CF8@ (b) Radius= Diameter 2 = 49 2 =24.5mm Area of a circle=π r 2 = 22 7 ( 24.5 ) 2 =1886.5 m 2 (c) Area of a circle=π r 2 = 22 7 ( 5 ) 2 = 550 7 cm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@DCD5@

Q.17 If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet. (Take π =

227

)

Ans

Since, circumference=2πr So, we get MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaaiabbofatjabbMgaPjabb6gaUjabbogaJjabbwgaLjabbYcaSiabbccaGiabbogaJjabbMgaPjabbkhaYjabbogaJjabbwha1jabb2gaTjabbAgaMjabbwgaLjabbkhaYjabbwgaLjabb6gaUjabbogaJjabbwgaLjabg2da9iabbkdaYiabec8aWjabbkhaYbqaaiabbofatjabb+gaVjabbYcaSiabbccaGiabbEha3jabbwgaLjabbccaGiabbEgaNjabbwgaLjabbsha0baaaa@6CE5@ 154=2× 22 7 ×r r= 154×7 2×22 = 49 2 m=4.5m Now, Area=π r 2 = 22 7 × 49 2 × 49 2 =1886.5 m 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@A695@

Q.18 From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet. (Take π = 3.14)

Ans

Outer radius of circular sheet=4 cm Inner radius of circular sheet=3 cm MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@99EC@ Remaining area=( 3.14×3×3 ) =50.2428.26 =21.98 cm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@884A@

Q.19  Find the perimeter of the adjoining figure, which is a semicircle including its diameter.

Ans

Radius=5 cm Length of curved part=πr = 22 7 ×5 =15.71 cm Total permieter=Length of the curved part+Length of diameter =15.71+10 =25.71 cm MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@045B@

Q.20

Shazli took a wire of length 44 cm and bent it into theshape of a circle. Find the radius of that circle. Also findits area. If the same wire is bent into the shape of asquare, what will be the length of each of its sides?Which figure encloses more.

Ans

Circumference=2πr=44 cm 2× 22 7 ×r=44 r=7 cm MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaaiabboeadjabbMgaPjabbkhaYjabbogaJjabbwha1jabb2gaTjabbAgaMjabbwgaLjabbkhaYjabbwgaLjabb6gaUjabbogaJjabbwgaLjabg2da9iabbkdaYiabec8aWjabbkhaYjabg2da9iabbsda0iabbsda0iabbccaGiabbogaJjabb2gaTbqaaiaaxMaacaaMe8UaaGjbVlabbkdaYiabgEna0oaalaaabaGaeGOmaiJaeGOmaidabaGaeG4naCdaaiabgEna0kabdkhaYjabg2da9iabisda0iabisda0aqaaiaaxMaacaWLjaGaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaemOCaiNaeyypa0JaeG4naCJaeeiiaaIaee4yamMaeeyBa0gaaaa@8092@ Area=π r 2 = 22 7 ×7×7 =154 cm 2 If the wire is bent into a square, then the length of each side would be= 44 4 =11 cm Area of square= ( 11 ) 2 =121 cm 2 Therefore, circle encloses more area. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@1AB3@

Q.21

From a circular card sheet of radius 14 cm, two circles ofradius 3.5 cm and a rectangle of length 3 cm and breadth1 cm are removed. (as shown in the adjoiningfigure).Find the area of the remaining sheet. Takeπ=227

Ans

Area of bigger cirle= 22 7 ×14×14 =616 cm 2 Area of 2 small circles=2×π r 2 =2× 22 7 ×3.5×3.5 =77 cm 2 Area of rectangle=Length×Breadth =3×1= 3 cm 2 Remaining area of sheet= 616 cm 2 77 cm 2 3 cm 2 = 536 cm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@3FAA@

Q.22

A circle of radius 2 cm is cut out from a square piece of analuminium sheet of side 6 cm. What is the area of the leftover aluminium sheet?Take π= 3.14

Ans

Area of square-shaped sheet= ( side ) 2 = ( 6 cm 2 ) 2 =36 cm 2 Area of circle=3.14×2×2 =12.56 cm 2 Remaining area= 36 cm 2 12 .56 cm 2 =23 .44 cm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaaiabbgeabjabbkhaYjabbwgaLjabbggaHjabbccaGiabb+gaVjabbAgaMjabbccaGiabbohaZjabbghaXjabbwha1jabbggaHjabbkhaYjabbwgaLjabb2caTiabbohaZjabbIgaOjabbggaHjabbchaWjabbwgaLjabbsgaKjabbccaGiabbohaZjabbIgaOjabbwgaLjabbwgaLjabbsha0jabg2da9maabmaabaGaee4CamNaeeyAaKMaeeizaqMaeeyzaugacaGLOaGaayzkaaWaaWbaaSqabeaacqaIYaGmaaaakeaacaWLjaGaaCzcaiaaxMaacaWLjaGaaCzcaiaaxMaacaaMe8UaaGjbVlaaysW7caaMc8Uaeyypa0ZaaeWaaeaacqaI2aGncqqGGaaicqqGJbWycqqGTbqBdaahaaWcbeqaaiabikdaYaaaaOGaayjkaiaawMcaamaaCaaaleqabaGaeGOmaidaaOGaeyypa0JaeG4mamJaeGOnayJaaGjbVlabbogaJjabb2gaTnaaCaaaleqabaGaeGOmaidaaaGcbaGaaCzcaiaaxMaacaWLjaGaaGjbVlaaysW7caaMe8UaaGPaVlabbgeabjabbkhaYjabbwgaLjabbggaHjabbccaGiabb+gaVjabbAgaMjabbccaGiabbogaJjabbMgaPjabbkhaYjabbogaJjabbYgaSjabbwgaLjabg2da9iabbodaZiabb6caUiabbgdaXiabbsda0iabgEna0kabikdaYiabgEna0kabikdaYaqaaiaaxMaacaWLjaGaaCzcaiaaxMaacaWLjaGaaCzcaiaaysW7caaMe8UaaGjbVlaaysW7caaMc8Uaeyypa0JaeGymaeJaeGOmaiJaeiOla4IaeGynauJaeGOnayJaeeiiaaIaee4yamMaeeyBa02aaWbaaSqabeaacqaIYaGmaaaakeaacaWLjaGaaCzcaiaaxMaacaaMc8UaaGPaVlabbkfasjabbwgaLjabb2gaTjabbggaHjabbMgaPjabb6gaUjabbMgaPjabb6gaUjabbEgaNjabbccaGiabbggaHjabbkhaYjabbwgaLjabbggaHjabg2da9iabbodaZiabbAda2iabbccaGiabbogaJjabb2gaTnaaCaaaleqabaGamGjGbkdaYaaakiabgkHiTiabbgdaXiabbkdaYiabb6caUiabbwda1iabbAda2iabbccaGiabbogaJjabb2gaTnaaCaaaleqabaGaeeOmaidaaaGcbaGaaCzcaiaaxMaacaWLjaGaaCzcaiaaxMaacaWLjaGaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMc8Uaeyypa0JaeeOmaiJaee4mamJaeeOla4IaeeinaqJaeeinaqJaeeiiaaIaee4yamMaeeyBa02aaWbaaSqabeaacqqGYaGmaaaaaaa@0452@

Q.23

The circumference of a circle is 31.4 cm. Find the radius and the area of the circle ? (Take π = 3.14)

Ans

Circumference=2πr=31.4cm 2×3.14×r=31.4 r= 31.4 2×3.14 =5 cm So, Area=3.14×5×5 =78 .50 cm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@AB49@

Q.24 A circular flower bed is surrounded by a path 4 m wide. The diameter of the flower bed is 66 m. What is the area of this path? (π = 3.14)

Ans

Radius of the flower bed= 66 2 =33m So, radius of flower bed and path together=33+4 =37m Area of flower baed and path together=3.14×37×37 =4298.66 m 2 Area of flower bed =3.14×33×33 =3419 .46 m 2 Area of path =4298 .66 m 2 3419 .46m 2 =879 .20m 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@6874@

Q.25

A circular flower garden has an area of 314 m2. A sprinklerat the centre of the garden can cover an area that has aradius of 12 m. Will the sprinkler water the entire garden?Takeπ= 3.14

Ans

Area=π r 2 =314 m 2 3.14× r 2 =314 r 2 =100 r=10 m Yes, the sprinkle will water the whole garden. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@B781@

Q.26 Find the circumference of the inner and the outer circles, shown in the adjoining figure? (Take π = 3.14)

Ans

Radius of outer circle=19 m Circumference=2πr =2×3.14×19 =119.32 m Radius of the inner circle=1910=9 m Circumference=2πr =2×3.14×9 =56.52 m MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaaiabbkfasjabbggaHjabbsgaKjabbMgaPjabbwha1jabbohaZjabbccaGiabb+gaVjabbAgaMjabbccaGiabb+gaVjabbwha1jabbsha0jabbwgaLjabbkhaYjabbccaGiabbogaJjabbMgaPjabbkhaYjabbogaJjabbYgaSjabbwgaLjabg2da9iabbgdaXiabbMda5iabbccaGiabb2gaTbqaaiaaxMaacaaMe8UaaGjbVlaaysW7cqqGdbWqcqqGPbqAcqqGYbGCcqqGJbWycqqG1bqDcqqGTbqBcqqGMbGzcqqGLbqzcqqGYbGCcqqGLbqzcqqGUbGBcqqGJbWycqqGLbqzcqGH9aqpcqqGYaGmcqaHapaCcqqGYbGCaeaacaWLjaGaaCzcaiaaxMaacaWLjaGaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMc8Uaeyypa0JaeeOmaiJaey41aqRaee4mamJaeeOla4IaeeymaeJaeeinaqJaey41aqRaeeymaeJaeeyoaKdabaGaaCzcaiaaxMaacaWLjaGaaCzcaiaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGPaVlabg2da9iabbgdaXiabbgdaXiabbMda5iabb6caUiabbodaZiabbkdaYiabbccaGiabb2gaTbqaaiabbkfasjabbggaHjabbsgaKjabbMgaPjabbwha1jabbohaZjabbccaGiabb+gaVjabbAgaMjabbccaGiabbsha0jabbIgaOjabbwgaLjabbccaGiabbMgaPjabb6gaUjabb6gaUjabbwgaLjabbkhaYjabbccaGiabbogaJjabbMgaPjabbkhaYjabbogaJjabbYgaSjabbwgaLjabg2da9iabbgdaXiabbMda5iabgkHiTiabbgdaXiabbcdaWiabg2da9iabbMda5iabbccaGiabb2gaTbqaaiaaxMaacaWLjaGaaGjbVlaaysW7cqqGdbWqcqqGPbqAcqqGYbGCcqqGJbWycqqG1bqDcqqGTbqBcqqGMbGzcqqGLbqzcqqGYbGCcqqGLbqzcqqGUbGBcqqGJbWycqqGLbqzcqGH9aqpcqqGYaGmcqaHapaCcqqGYbGCaeaacaWLjaGaaCzcaiaaxMaacaWLjaGaaCzcaiaaysW7caaMe8UaaGjbVlaaysW7caaMe8Uaeyypa0JaeeOmaiJaey41aqRaee4mamJaeeOla4IaeeymaeJaeeinaqJaey41aqRaeeyoaKdabaGaaCzcaiaaxMaacaWLjaGaaCzcaiaaxMaacaaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlabg2da9iabbwda1iabbAda2iabb6caUiabbwda1iabbkdaYiabbccaGiabb2gaTbaaaa@18E1@

Q.27 How many times a wheel of radius 28 cm must rotate to go 352 m? (take π =

227

)

Ans

Radius, r=28 cm MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacqqGsbGucqqGHbqycqqGKbazcqqGPbqAcqqG1bqDcqqGZbWCcqqGSaalcqqGGaaicqqGYbGCcqGH9aqpcqqGYaGmcqqG4aaocqqGGaaicqqGJbWycqqGTbqBaaa@5395@ Circumference=2πr =2× 22 7 ×28 =176 cm Number of rotatinos= Total distance to be covered Circumference of the wheel = 352 m 176cm = 35200 176 =200 Therefore, it will rotate 200 times. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@36A6@

Q.28 The minute hand of a circular clock is 15 cm long. How far does the tip of the minute hand move in 1 hour. (Take π = 3.14)

Ans

Since, Distance travelled by the tip of minute hand =Circumference of the clock =2πr =2×3.14×15 =94.2 cm Therfore, minute hand move 94.2 cm in 1 hour. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@F536@

Q.29 A garden is 90 m long and 75 m broad. A path 5 m wide is to be built outside and around it. Find the area of the path. Also find the area of the garden in hectare.

Ans

Length (L) of garden=90 m Breadth (B) of garden=75 m Area of garden=L×B =90 m×75 m = 6750 m 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@C591@ From the figure, it can be observed that the new length and breadth of the garden,when path is also included are 100 m and 85 m respectively Area of the garden including path=100 m ×80 m = 8000 m 2 Area of the path =Area of the garden including pathArea of garden =8000 m 2 6750 m 2 =1750 m 2 1 hectare= 10000 m 2 Therefore, area of the garden in hectare= 6750 10000 =0.675 hectare MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@079B@

Q.30 A 3 m wide path runs outside and around a rectangular park of length 125 m and breadth 65m. Find the area of the path.

Ans

Length (L) of park=125 m Breadth (B) of park=65 m Area of park=L×B =125 m×65 m = 8125 m 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@BF97@ From the figure, it can be observed that the new length and breadth of the park,when path is also included are 131 m and 71 m respectively Area of the park including path=131 m ×71 m = 9301 m 2 Area of the path =Area of the park including pathArea of park =9301 m 2 8125 m 2 =1176 m 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@8DBC@

Q.31 A picture is painted on a cardboard 8 cm long and 5 cm wide such that there is a margin of 1.5 cm along each of its sides. Find the total area of the margin.

Ans

Length (L) of cardboard=125 m Breadth (B) of cardboard=65 m Area of cardboard including margin=L×B =8 cm×5 cm =40 cm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@E873@ From the figure, it can be observed that the new length and breadth of the cardboard,when margin is not included are 5 cm and 2 cm respectively Area of the cardboard not including margin=5 cm ×2 m = 10 cm 2 Area of the margin =Area of the cardboard including margin Area of cardboard not including margin =40 cm 2 10 cm 2 =30 cm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@D639@

Q.32

Two cross roads, each of width 10 m, cut at right anglesthrough the centre of a rectangular park of length 700 mand breadth 300 m and parallel to its sides. Find the areaof the roads. Also find the area of the park excluding crossroads. Give the answer in hectares.

Ans

Length (L) of the park=700 m Breadth (B) of the park=300 m Area of park=700×300 = 210000 m 2 Length of road PQRS=700 m Length of road ABCD=300 m Width of each road=10 m Area of the roads=area(PQRS)+area(ABCD)area(KLMN) =( 700×10 )+( 300×10 )( 10×10 ) =7000+3000100 =9900 m 2 =0.99 hectare MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@AE64@ Area of park excluding roads=210000-9900 = 200100 m 2 =20.01 hectare MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@9E5E@

Q.33 Pragya wrapped a cord around a circular pipe of radius 4 cm (adjoining figure) and cut off the length required of the cord. Then she wrapped it around a square box of side 4 cm (also shown). Did she have any left? (π = 3.14)

Ans

Perimeter of the square=4×side of the square =4×4 =16 cm Perimeter of circular pip=2πr =3×3.14×4 =25.12 cm Therefore, length of chord left with Pragya=25.12 cm16 cm =9.12 cm MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@2F91@

Q.34 The adjoining figure represents a rectangular lawn with a circular flower bed in the middle. Find:
(i) the area of the whole land
(ii) the area of the flower bed
(iii) the area of the lawn excluding the area of the flower bed.

(iv) the circumference of the flower bed.

Ans

(i) Area of whole land=Length×Breadth =10×5 = 50 m 2 (ii) Area of flower bed=π r 2 =3.14×2×2 =12.56 m 2 (iii) Area of the lawn excluding the flower bed =Area of whole landArea of flower bed =50-12.56 =37 .44 m 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@5A62@

Q.35 In the following figures, find the area of the shaded portions:

Ans

(i) Area of EFDC=area(ABCD)area(BCE)area(AFE) =( 18×10 ) 1 2 ( 10×8 ) 1 2 ( 6×10 ) =1804030 =110 cm 2 (ii) area(QTU)=area(PQRS)area(TSU)area(RUQ)area(PQT) =( 20×20 ) 1 2 ( 10×10 ) 1 2 ( 20×10 ) 1 2 ( 20×10 ) =40050100100 =150 cm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@3542@

Q.36

Find the area of the quadrilateral ABCD.Here, AC=22 cm, BM=3 cm,DN=3 cm, and BMAC, DNAC.

Ans

area(ABCD)=area(ABC)+area(ADC) = 1 2 ( 3×22 ) 1 2 ( 3×22 ) =33+33 =66 cm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@9DB2@

Q.37

The length and the breadth of a rectangular piece of landare 500 m and 300 m respectively. Findi its areaii the cost of the land, if 1 m2 of the land costs 10,000.

Ans

(i) Area=Length×breadth =500 m×3000× m = 150000m 2 (ii) 1 m 2 land cost=₹ 10,000 So, cost of 150000 m 2 land = ₹ 10,000×150000 =₹ 1,500,000,000 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@DDC2@

Q.38

A door of length 2 m and breadth 1m is fitted in a wall.The length of the wall is 4.5 m and the breadth is 3.6 m .Find the cost of white washing the wall, if the rate ofwhite washing the wall is 20 per m2.

Ans

Area of the wall=4.5 cm×3.6 cm =16 .2 m 2 Area of the door=2×1 = 2m 2 Area to be white-washed=16 .2 m 2 2 m 2 =14 .2 m 2 Cost of white-washing 1m 2 area= ₹20 So, cost of white-washing 14.2m2 area=14.2×20 =₹ 284 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@517F@

Q.39

A gardener wants to fence a circular garden of diameter21m. Find the length of the rope he needs to purchase, if hemakes 2 rounds of fence. Also find the costs of the rope, if itcost ₹ 4 per meter.Takeπ=227

Ans

Given, d=21 m So, r= 21 2 m Circumference=2πr =2× 22 7 × 21 2 =66m Length of the rope required for fencing=2×66 m =132 m Cost of 1 m rope =₹4 So, cost of 132 m rope=₹ 4×132 =₹528MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@2405@

Q.40

Saima wants to put a lace on the edge of a circular tablecover of diameter 1.5 m. Find the length of the lacerequired and also find its cost if one meter of the lacecosts ₹ 15.Takeπ= 3.14

Ans

Circumference=2πr =2×3.14× d 2 =2×3.14 1.5 2 =4.71m Cost of 1 m of lace=₹ 15 Therefore cost of 4.71 m of lace=₹15×4.71 =₹70.65MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@EFD4@

Q.41

Find the cost of polishing a circular tabletop of diameter1.6 m, if the rate of polishing is Rs 15/m2.Takeπ= 3.14

Ans

Diameter=1.6 m Radius= 1.6 2 =0.8 m Area=3.14×0.8×0.8 =2 .0096 m 2 Cost for polishing 1 m2 area=₹15 Cost for polishing 2.0096 m2 area=₹15×2.0096 =₹30.14 So, it will cost ₹30.14 for polishing such circular table.MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@3FF3@

Q.42

A verandah of width 2.25 m is constructed all alongoutside a room which is 5.5 mlong and 4 m wide. Find:i the area of the verandah.ii the cost of cementing the floor of the verandah at therate of Rs 200 per m2

Ans

(i) Length (L) of room=5.5 m Breadth (B) of room=4 m Area of room including margin=L×B =5.5 m×4 m =22 m 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@D0D8@ From the figure, it can be observed that the new length and breadth of the room,when verandah is also included are 10 m and 8.5 m respectively Area of the verandah including margin=10 m ×8.5 m = 85 m 2 Area of the verandah =Area of the room including verandahArea of the room =85 m 2 22 m 2 =63 m 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@AD42@ (ii)Cost of cementing 1m 2 area of floor of verandah=200 Cost of cementing 63m 2 area of floor of verandah=200×63 =12600 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@D860@

Q.43

A path 1 m wide is built along the border and inside asquare garden of side 30 m. Find:i the area of the pathiithe cost of planting grass in the remaining portion of thegarden at the rate of 40 per m2.

Ans

(i) Side (a) of square garden=30 m Area of square garden including path= a 2 = ( 30 m ) 2 = 900 m 2 From the figure, it can be observed that the side of the square garden, when path is not included, is 28 m. Area of the square garden not including the path= ( 28 ) 2 = 784 m 2 Area of path =Area of the square garden including the path Area of square garden not including the path =900 m 2 784 m 2 =116 m 2 (ii) Cost of planting grass in 1 m 2 area of the garden=₹40 Cost of planting grass in 784 m 2 area of the garden=₹40×784 =₹31360MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@C841@

Q.44

Through a rectangular field of length 90 m and breadth60 m,two roads are constructed which are parallel to thesides and cut each other at right angles through the centreof the fields. If the width of each road is 3 m, findi the area covered by the roads.iithe cost of constructing the roads at the rate of110 per m2.

Ans

Length (L) of the field=90 m Breadth (B) of the field=60 m Area of field=90×60 = 5400 m 2 Length of road PQRS=90 m Length of road ABCD=60 m Width of each road=3 m Area of the roads=area(PQRS)+area(ABCD)area(KLMN) =( 90×3 )+( 60×3 )( 3×3 ) =270+180100 =441 m 2 Cost for constructing 1 m 2 road=₹110 Cost for constructing 4411 m 2 road=₹110×441 =48510 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@0CFE@

For viewing question paper please click here

FAQs (Frequently Asked Questions)
1. Where will I find the class 7 Mathematics chapter-wise solutions?

You will find the class 7 Mathematics chapter-wise solutions on the Extramarks website.

2. Why choose Extramarks NCERT solutions?

The primary reasons to choose Extramarks NCERT solutions are that they are:

  •  drafted by experienced faculties
  • 100% accurate 
  •  made as per CBSE guidelines
  •  self-explanatory

3. What is chapter 11 of class 7 about?

Chapter 11 of Class 7 is about Area and Perimeter. Perimeter is the distance around a close figure whereas area is a metric to quantify the region occupied by a figure. Students will learn how to calculate the perimeters and areas of many different shapes including triangles, circles, squares, and parallelograms in this chapter.

4. What are the area and perimeter?

Perimeter is the distance that surrounds a specific area whereas, an area is the part of a plane that is encircled by the perimeter. An area is a particular shape that includes a square, rectangle, triangle, etc.

5. What is the formula for the area of a circle?

The formula of the area of a circle is πr2.