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Exercise 11.1 - Chapter 11 Mensuration class 8 ncert solutions Maths - SaraNextGen [2024]


Question 1:

A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area?

https://img-nm.mnimgs.com/img/study_content/curr/1/8/5/74/2459/Chapter%2011_html_72b63cb0.jpg

Answer:

Perimeter of square = 4 (Side of the square) = 4 (60 m) = 240 m

Perimeter of rectangle = 2 (Length + Breadth)

= 2 (80 m + Breadth)

= 160 m + 2 × Breadth

It is given that the perimeter of the square and the rectangle are the same.

160 m + 2 × Breadth = 240 m

Breadth of the rectangle = https://img-nm.mnimgs.com/img/study_content/curr/1/8/5/74/2459/Chapter%2011_html_m28ceb42a.gif  = 40 m

Area of square = (Side)2 = (60 m)2 = 3600 m2

Area of rectangle = Length × Breadth = (80 × 40) m2 = 3200 m2

Thus, the area of the square field is larger than the area of the rectangular field.

Question 2:

Mrs. Kaushik has a square plot with the measurement as shown in the following figure. She wants to construct a house in the middle of the plot. A garden is developed around the house. Find the total cost of developing a garden around the house at the rate of Rs 55 per m2.

https://img-nm.mnimgs.com/img/study_content/curr/1/8/5/74/2472/Chapter%2011_html_m7162e91a.jpg

Answer:

Area of the square plot = (25 m)2 = 625 m2

Area of the house = (15 m) × (20 m) =300 m2

Area of the remaining portion = Area of square plot − Area of the house

= 625 m2 − 300 m2 = 325 m2

The cost of developing the garden around the house is Rs 55 per m2.

Total cost of developing the garden of area 325 m2 = Rs (55 × 325)

= Rs 17,875

Question 3:

The shape of a garden is rectangular in the middle and semi circular at the ends as shown in the diagram. Find the area and the perimeter of the garden [Length of rectangle is 20 − (3.5 + 3.5) metres]

https://img-nm.mnimgs.com/img/study_content/curr/1/8/5/74/2474/Chapter%2011_html_3c8f839b.jpg

Answer:

Length of the rectangle = [20 − (3.5 + 3.5)] metres = 13 m

Circumference of 1 semi-circular part = πhttps://img-nm.mnimgs.com/img/study_content/curr/1/8/5/74/2474/Chapter%2011_html_m4e6268ec.gif

Circumference of both semi-circular parts = (2 × 11) m = 22 m

https://img-nm.mnimgs.com/img/study_content/curr/1/8/5/74/2474/Chapter%2011_html_m55a8dff6.jpg

Perimeter of the garden = AB + Length of both semi-circular regions BC and

DA + CD

= 13 m + 22 m + 13 m = 48 m­

Area of the garden = Area of rectangle + 2 × Area of two semi-circular regions

https://img-nm.mnimgs.com/img/study_content/curr/1/8/5/74/2474/Chapter%2011_html_m91fa687.gif

Question 4:

A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m2? (If required you can split the tiles in whatever way you want to fill up the corners).

Answer:

Area of parallelogram = Base × Height

Hence, area of one tile = 24 cm × 10 cm = 240 cm2

Required number of tiles = https://img-nm.mnimgs.com/img/study_content/curr/1/8/5/74/2475/Chapter%2011_html_4f048c62.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/8/5/74/2475/Chapter%2011_html_m74c9d94d.gif = 45000 tiles

Thus, 45000 tiles are required to cover a floor of area 1080 m2.

Question 5:

An ant is moving around a few food pieces of different shapes scattered on the floor. For which food − piece would the ant have to take a longer round? Remember, circumference of a circle can be obtained by using the expression c = 2πr, where r is the radius of the circle.

https://img-nm.mnimgs.com/img/study_content/curr/1/8/5/74/2476/Chapter%2011_html_43059154.jpg  https://img-nm.mnimgs.com/img/study_content/curr/1/8/5/74/2476/Chapter%2011_html_1deef71e.jpg  https://img-nm.mnimgs.com/img/study_content/curr/1/8/5/74/2476/Chapter%2011_html_m123b4981.jpg

Answer:

(a)Radius (r) of semi-circular part = https://img-nm.mnimgs.com/img/study_content/curr/1/8/5/74/2476/Chapter%2011_html_5cfa9fc8.gif

Perimeter of the given figure = 2.8 cm + πr

https://img-nm.mnimgs.com/img/study_content/curr/1/8/5/74/2476/Chapter%2011_html_m2885b6d2.gif

(b)Radius (r) of semi-circular part = https://img-nm.mnimgs.com/img/study_content/curr/1/8/5/74/2476/Chapter%2011_html_5cfa9fc8.gif

Perimeter of the given figure = 1.5 cm + 2.8 cm + 1.5 cm +π (1.4 cm)

https://img-nm.mnimgs.com/img/study_content/curr/1/8/5/74/2476/Chapter%2011_html_3b36d38d.gif

(c)Radius (r) of semi-circular part = https://img-nm.mnimgs.com/img/study_content/curr/1/8/5/74/2476/Chapter%2011_html_5cfa9fc8.gif

Perimeter of the figure(c) = 2 cm + πr + 2 cm

https://img-nm.mnimgs.com/img/study_content/curr/1/8/5/74/2476/Chapter%2011_html_m44f5c44d.gif

Thus, the ant will have to take a longer round for the food-piece (b), because the perimeter of the figure given in alternative (b) is the greatest among all.

Also Read : Exercise-11.2-Chapter-11-Mensuration-class-8-ncert-solutions-Maths

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