Maharashtra Board Class 7 Maths Chapter 12 Set 44 Perimeter and Area Solutions

Question 1. If the length and breadth of a rectangle are doubled, how many times the perimeter of the old rectangle will that of the new rectangle be?
Answer: Let the length of the old rectangle be l and breadth be b.
∴ Perimeter of old rectangle = \(2(l + b)\)
Length of new rectangle = \(2l\) and breadth = \(2b\)
∴ Perimeter of new rectangle = \(2(2l + 2b)\)
= \(2 \times 2 (l + b)\)
= \(2 \times\) perimeter of old rectangle
∴ The perimeter of new rectangle will be twice the perimeter of old rectangle.
In simple words: If a rectangle's length and breadth are doubled, its perimeter also doubles because the formula for perimeter involves adding length and breadth.

🎯 Exam Tip: Remember to clearly define variables for the original and new dimensions. Showing the perimeter calculation for both is crucial for full marks.

Question 2. If the side of a square is tripled, how many times the perimeter of the first square will that of the new square be?
Answer: Let the length of the square be a.
Perimeter of square = \(4 \times\) side
= \(4 \times a = 4a\)
Side of new square = \(3 \times a = 3a\)
Perimeter of new square = \(4 \times\) side
= \(4 \times 3a = 3 \times 4a = 3 \times\) perimeter of original square.
∴ The perimeter of new square will be three times the perimeter of original square.
In simple words: When the side of a square is tripled, its perimeter also triples, as the perimeter is directly proportional to the side length (Perimeter = 4 * side).

🎯 Exam Tip: Clearly show the relationship between the original perimeter and the new perimeter by factoring out the original perimeter expression. This demonstrates a strong understanding of the concept.

Question 3. Given alongside is the diagram of a playground. It shows the length of its sides. Find the perimeter of the playground.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक 'L' आकार का खेल का मैदान है जिसके किनारे की मापें दी गई हैं। ऊपर की क्षैतिज भुजा 10 मीटर है, बाईं ऊर्ध्वाधर भुजा 15 मीटर है, निचली क्षैतिज भुजा 15 मीटर है, दाहिनी ऊर्ध्वाधर भुजा 15 मीटर है, और आंतरिक क्षैतिज व ऊर्ध्वाधर भुजाएँ क्रमशः 5 मीटर हैं।
Answer: Side AF = side BC + side DE
∴ Side AF = \(15 + 15 = 30\) m
Side FE = side AB + side CD
∴ Side FE = \(10 + 5 = 15\) m
∴ Perimeter of the playground = side AB + side BC + side CD + side DE + side FE + side AF
= \(10 + 15 + 5 + 15 + 15 + 30\)
= \(90\) m.
∴ The perimeter of the playground is \(90\) m.
In simple words: To find the perimeter of the L-shaped playground, we first calculate the lengths of the unknown outer sides by adding the lengths of the corresponding inner segments, and then sum up all the outer boundary lengths.

🎯 Exam Tip: For irregular shapes, break down complex sides into known segments. Always ensure all outer boundaries are accounted for in the final sum. Clearly label derived side lengths.

Question 4. As shown in the figure, four napkins all of the same size were made from a square piece of cloth of length 1 m. What length of lace will be required to trim all four sides of all the napkins?
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वर्ग का चित्र है जिसे चार समान छोटे वर्गों में विभाजित किया गया है। यह दर्शाता है कि 1 मीटर भुजा वाले एक वर्गाकार कपड़े से चार समान वर्गाकार नैपकिन बनाए गए हैं।
Answer: Side of the square piece of cloth = \(1\) m
∴ Side of each napkin = \(0.5\) m
Length of lace that will be required for 1 napkin = perimeter of the napkin
= \(4 \times\) side = \(4 \times 0.5 = 2\) m
∴ Perimeter of 4 napkins = \(4 \times 2 = 8\) m
∴ \(8\) metre long lace will be required to trim all four napkins.
In simple words: Since a 1m square cloth is divided into four equal napkins, each napkin is a 0.5m square. The lace required for one napkin is its perimeter (2m), so for four napkins, it's 8m.

🎯 Exam Tip: Clearly state the dimensions of the individual napkins derived from the main cloth. Remember to multiply the perimeter of one napkin by the total number of napkins to get the final lace length.