Perimeter and Area Class 7 Practice Set 45 Answers Solutions Chapter 12
Question 1
Maharashtra Board Solution
If the side of a square is 12 cm, find its area.
Solution & Step-by-Step Answer:
Area of a square = (side)² = (12)² = 144 sq. cm. ∴ The area of the square is 144 sq. cm.
Question 2
Maharashtra Board Solution
If the length of a rectangle is 15 cm and breadth is 5 cm, find its area.
Solution & Step-by-Step Answer:
Area of a rectangle = length × breadth = 15 × 5 = 75 sq. cm. ∴ The area of the rectangle is 75 sq. cm.
Question 3
Maharashtra Board Solution
The area of a rectangle is 102 sq. cm. If its length is 17 cm, what is its perimeter?
Solution & Step-by-Step Answer:
Area of a rectangle = length × breadth ∴ 102 = 17 × breadth ∴ breadth = = 6 cm Perimeter of rectangle = 2 (length + breadth) = 2 (17 + 6) = 2 × 23 = 46 cm ∴ The perimeter of rectangle is 46 cm.
Question 4
Maharashtra Board Solution
If the side of a square is tripled, how many times will its area be as compared to the area of the original square?
Solution & Step-by-Step Answer:
Let the side of the square be a. ∴ Area of a square = (side)² = a² New side of the square = 3 × a = 3a ∴ New area of the square = (3a)² = 9a² = 9 × area of original square ∴ If the side of a square is tripled, its area will become 9 times the area of the original square.
Maharashtra Board Class 7 Maths Chapter 12 Perimeter and Area Practice Set 45 Intext Questions and Activities
Question 1
Maharashtra Board Solution
A rectangular playground is 65m long and 30m wide. A pathway of 1.5 m width goes all around the ground, outside it. Find the area of the pathway. (Textbook pg. no. 82)
Solution & Step-by-Step Answer:
The playground is rectangular. ₹ABCD is the playground. Around it is a pathway 1.5 m wide. Around ₹ABCD we get the rectangle ₹PQRS Length of new rectangle PQRS = 65 + 1.5 + 1.5 = 68 m Breadth of new rectangle PQRS = 30 + 1.5 + 1.5 = 33m Area of path = Area of rectangle PQRS – Area of rectangle ABCD = 68 x 33 – 65 x 30 = 2244 – 1950 = 294 sq m
Question 2
Maharashtra Board Solution
Is there another way to find the area of the pathway in the problem above? (Textbook pg. no. 82)
Solution & Step-by-Step Answer:
Yes. The area of the pathway can be found by dividing it into rectangles and adding the areas of these rectangles. Length of rectangle 1 = 30 + 1.5 + 1.5 = 33 m Breadth of rectangle 1 = 1.5 m ∴ Area of rectangle 1 = 33 x 1.5 = 49.5 sq. m Area of rectangle 4 = Area of rectangle 1 = 49.5 sq. m. Length of rectangle 2 = 65 m breadth of rectangle 2 = 1.5 m ∴ Area of rectangle 2 = 65 x 1.5 = 97.5 sq. m. Area of rectangle 3 = area of rectangle 2 = 97.5 sq. m. ∴ Area of pathway = Sum of area of the 4 rectangles = 49.5 + 49.5 + 97.5 + 97.5 = 294 sq. m.
Question 3
Maharashtra Board Solution
The length and the width of a mobile phone are 13 cm and 7 cm respectively. It has a screen PQRS as shown in the figure. What is the area of the screen? (Textbook pg. no. 82)
Solution & Step-by-Step Answer:
ABCD is the rectangle formed by the edges of the mobile. PQRS is the rectangle formed by leaving a 1.5 cm wide edge alongside AB, BC, and DC, and a 2 cm edge alongside DA. Length of rectangle PQRS = 9.5 cm Breadth of rectangle PQRS = 4 cm Area of screen = Area of rectangle PQRS = 9.5 x 4 = 38 sq.cm