Maharashtra Board Class 7 Maths Chapter 12 Set 45 Perimeter and Area Solutions

Question 1. If the side of a square is 12 cm, find its area.
Answer: Area of a square = \( (\text{side})^2 = (12)^2 \)
\( = 144 \text{ sq. cm.} \)
\( \therefore \) The area of the square is 144 sq. cm.
In simple words: To find the area of a square, you multiply its side length by itself. For a 12 cm side, the area is 12 multiplied by 12, which is 144 square centimeters.

🎯 Exam Tip: Remember the formula for the area of a square (\( \text{side}^2 \)) and always include the correct units (e.g., sq. cm) in your final answer for full marks.

Question 2. If the length of a rectangle is 15 cm and breadth is 5 cm, find its area.
Answer: Area of a rectangle = length × breadth
\( = 15 \times 5 \)
\( = 75 \text{ sq. cm.} \)
\( \therefore \) The area of the rectangle is 75 sq. cm.
In simple words: To find the area of a rectangle, multiply its length by its breadth. Here, 15 cm multiplied by 5 cm gives an area of 75 square centimeters.

🎯 Exam Tip: Clearly state the formula for the area of a rectangle (\( \text{length} \times \text{breadth} \)) and perform calculations accurately. Ensure units are correctly specified as square units.

Question 3. The area of a rectangle is 102 sq. cm. If its length is 17 cm, what is its perimeter?
Answer: Area of a rectangle = length × breadth
\( \therefore 102 = 17 \times \text{breadth} \)
\( \therefore \text{breadth} = \frac{102}{17} = 6 \text{ cm} \)
Perimeter of rectangle = \( 2 (\text{length} + \text{breadth}) \)
\( = 2 (17 + 6) \)
\( = 2 \times 23 \)
\( = 46 \text{ cm} \)
\( \therefore \) The perimeter of rectangle is 46 cm.
In simple words: First, use the given area and length to find the breadth by dividing the area by the length. Then, use both the length and breadth in the perimeter formula, which is two times the sum of length and breadth, to get the final perimeter.

🎯 Exam Tip: This problem requires a two-step approach: first finding the breadth using the area formula, and then calculating the perimeter. Show both steps clearly, maintaining consistent units for length, breadth, area, and perimeter.

Question 4. If the side of a square is tripled, how many times will its area be as compared to the area of the original square?
Answer: Let the side of the square be a.
\( \therefore \) Area of a square = \( (\text{side})^2 = a^2 \)
New side of the square = \( 3 \times a = 3a \)
\( \therefore \) New area of the square = \( (3a)^2 \)
\( = 9a^2 \)
\( = 9 \times \) area of original square
\( \therefore \) If the side of a square is tripled, its area will become 9 times the area of the original square.
In simple words: When the side of a square is tripled, the area becomes nine times larger because the area calculation involves squaring the side length (3 times 3 equals 9).

🎯 Exam Tip: Demonstrate the relationship between side length and area by using a variable (e.g., 'a') for the original side. Clearly show the squaring of the new side length to derive the multiplicative factor for the area.

Maharashtra Board Class 7 Maths Chapter 12 Perimeter And Area Practice Set 45 Intext Questions And Activities

Question 1. 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)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक आयताकार खेल के मैदान ABCD को दर्शाता है, जिसकी चौड़ाई 30 मीटर है। इस मैदान के चारों ओर 1.5 मीटर चौड़ा एक रास्ता है, जिससे एक बड़ा आयताकार क्षेत्र PQRS बनता है। खेल के मैदान की लंबाई 65 मीटर है।
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 = \( 65 + 1.5 + 1.5 = 68 \text{ m} \)
Breadth of new rectangle PQRS = \( 30 + 1.5 + 1.5 = 33\text{m} \)
Area of path = Area of rectangle PQRS - Area of rectangle ABCD
\( = 68 \times 33 - 65 \times 30 \)
\( = 2244 - 1950 \)
\( = 294 \text{ sq m} \)
In simple words: To find the area of the pathway, first calculate the dimensions of the larger rectangle (playground plus pathway) by adding the pathway width to both length and breadth. Then, find the area of both the large and small (playground) rectangles and subtract the smaller area from the larger one.

🎯 Exam Tip: For problems involving pathways around a field, always calculate the dimensions of the outer and inner rectangles accurately. Clearly show the subtraction of the inner area from the outer area to get the pathway's area, including proper units.

Question 2. Is there another way to find the area of the pathway in the problem above? (Textbook pg. no. 82)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र पिछले प्रश्न में दिए गए आयताकार रास्ते को दर्शाता है, जिसमें रास्ते को चार छोटे आयतों (1, 2, 3, 4) में विभाजित किया गया है। यह विधि रास्ते के क्षेत्रफल को विभाजित करके ज्ञात करने का एक वैकल्पिक तरीका समझाती है।
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 = \( 30 + 1.5 + 1.5 = 33 \text{ m} \)
Breadth of rectangle 1 = \( 1.5 \text{ m} \)
\( \therefore \) Area of rectangle 1 = \( 33 \times 1.5 \)
\( = 49.5 \text{ sq. m.} \)
Area of rectangle 4 = Area of rectangle 1
\( = 49.5 \text{ sq. m.} \)
Length of rectangle 2 = \( 65 \text{ m} \)
breadth of rectangle 2 = \( 1.5 \text{ m} \)
\( \therefore \) Area of rectangle 2 = \( 65 \times 1.5 \)
\( = 97.5 \text{ sq. m.} \)
Area of rectangle 3 = area of rectangle 2
\( = 97.5 \text{ sq. m.} \)
\( \therefore \) Area of pathway = Sum of area of the 4 rectangles
\( = 49.5 + 49.5 + 97.5 + 97.5 \)
\( = 294 \text{ sq. m.} \)
In simple words: Yes, an alternative method is to break down the pathway into smaller, simpler rectangular sections, calculate the area of each section individually, and then add all these individual areas together to get the total area of the pathway.

🎯 Exam Tip: When faced with complex shapes, consider breaking them down into simpler geometric figures like rectangles or squares. Calculate the area of each component and sum them up to find the total area, ensuring all measurements are correctly used for each part.

Question 3. 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)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक मोबाइल फोन (आयताकार ABCD) को दर्शाता है जिसकी लंबाई 13 सेमी और चौड़ाई 7 सेमी है। इसके भीतर एक स्क्रीन (आयताकार PQRS) है। स्क्रीन के चारों ओर मार्जिन दिए गए हैं: ऊपर, दाएं और नीचे 1.5 सेमी, और बाईं ओर 2 सेमी।
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 = \( 9.5 \text{ cm} \)
Breadth of rectangle PQRS = \( 4 \text{ cm} \)
Area of screen = Area of rectangle PQRS
\( = 9.5 \times 4 \)
\( = 38 \text{ sq. cm} \)
In simple words: To find the screen's area, first calculate its exact length and breadth by subtracting the given margins from the total mobile phone dimensions. Then, multiply the calculated length and breadth of the screen to get its area.

🎯 Exam Tip: Carefully subtract the margins from the total length and width of the phone to find the screen's dimensions. A common mistake is miscalculating the effective length or width due to multiple margins. Show clear steps for finding screen dimensions before calculating the area.