Maharashtra Board Class 7 Maths Chapter 12 Set 46 Perimeter and Area Solutions

Question 1. A page of a calendar is 45 cm long and 26 cm wide. What is its area?
Answer:
Solution:
Area of page of a calendar = length × breadth
= 45 × 26
= 1170 sq. cm.
∴ The area of the page of the calendar is 1170 sq. cm.
In simple words: To find the area of a rectangular object like a calendar page, multiply its length by its breadth.

🎯 Exam Tip: Remember to always include the correct square units (e.g., sq. cm, sq. m) when stating the area.

Question 2. What is the area of a triangle with base 4.8 cm and height 3.6 cm?
Answer:
Solution:
Area of triangle = \( \frac{1}{2} \) × base × height
= \( \frac{1}{2} \) x 4.8 x 3.6
= \( \frac{1}{2} \) × 17.28
= 8.64 sq. cm.
∴ The area of the triangle is 8.64 sq. cm.
In simple words: The area of a triangle is half the product of its base and height.

🎯 Exam Tip: Ensure you use consistent units for base and height before calculating the area.

Question 3. What is the value of a rectangular plot of land 75.5 m long and 30.5 m broad at the rate of Rs. 1000 per square metre?
Answer:
Solution:
Area of the rectangular plot = length × breadth
= 75.5 x 30.5
= 2302.75 sq. m.
Value of the plot = area of the plot × rate per square metre = 2302.75 × 1000
= Rs. 230275
∴ The value of the plot is Rs. 23,02,750.
In simple words: First, calculate the total area of the plot by multiplying its length and breadth. Then, multiply this area by the given rate per square metre to find the total value.

🎯 Exam Tip: Pay close attention to unit conversions and large number multiplications for accuracy in value calculations.

Question 4. A rectangular hall is 12 m long and 6 m broad. Its flooring is to be made of square tiles of side 30 cm. How many tiles will fit in the entire hall? How many would be required if tiles of side 15 cm were used?
Answer:
Solution:
Area of the rectangular hall = length x breadth
= 12 x 6
= 72 sq. m.
Side of the square shaped tile = 30 cm
= \( \frac{30}{100} \) m ...[1cm = \( \frac{1}{100} \)m]
= \( \frac{3}{10} \) m
Area of the tile = (side)²
= \( (\frac{3}{10})^2 \)
= \( \frac{9}{100} \) sq.m
Number of tiles required = \( \frac{\text{Area of the hall}}{\text{Area of each tile}} \)
= \( 72 \div \frac{9}{100} \)
\( \implies 72 \times \frac{100}{9} \)
= 800
∴ 800 square shaped tiles of 30 cm side will be required.
If the side of the square is reduced to half, its area will become \( \frac{1}{4} \) times the original.
i. e. number of tiles required will become 4 times the original tiles.
∴ Number of tiles required = 4 × number of tiles of side 30 cm
= 4 × 800
= 3200
∴ 3200 square shaped tiles of 15 cm side will be required.
In simple words: Calculate the hall's area. Convert the tile side length to the same unit as the hall, then find the area of one tile. Divide the hall's area by one tile's area to find the number of tiles. If tile side is halved, its area becomes one-fourth, so four times as many tiles are needed.

🎯 Exam Tip: Crucially, ensure all measurements are in the same units (e.g., meters) before calculating areas and tile counts.

Question 5. Find the perimeter and area of a garden with measures as shown in the figure alongside.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक 'प्लस' (+) चिह्न के आकार के बगीचे को दर्शाता है। यह आकृति पाँच समान वर्गों से मिलकर बनी है, जिसमें एक वर्ग केंद्र में है और चार अन्य वर्ग उसकी चारों भुजाओं से जुड़े हुए हैं। प्रत्येक वर्ग की भुजा 13 मीटर लंबी है।
Answer:
Solution:
The boundary of the garden is made of 12 sides each of length 13 m.
Perimeter of the garden = sum of the lengths of all sides
= 12 x 13
= 156 m
The garden in the given figure can be divided into 5 squares each of side 13 m.
∴ Area of the garden = 5 × area of each square part
= 5 × (side)²
= 5 × (13)²
= 5 × 169
= 845 sq. m.
∴ The perimeter and area of a garden are 156 m and 845 sq. m. respectively.
In simple words: For a composite shape, find the perimeter by summing all outer boundary lengths. Calculate the area by dividing the shape into simpler, known figures (like squares) and summing their individual areas.

🎯 Exam Tip: For complex shapes, break them down into basic geometric figures to simplify area calculation, and carefully trace the outer boundary for the perimeter.