Maharashtra Board Class 5 Maths Chapter 12 Perimeter and Area Set 50 Solutions

Std 5 Maths Chapter 12 Perimeter And Area

Question 1. The length of the side of each square is given below. Find its area.
(1) 12 metres
Answer:
Solution:
Area of a square \( = \text{side} \times \text{side} \)
\( = 12 \times 12 \)

(2) 6 cm
Answer:
Solution:
Area of a square \( = \text{side} \times \text{side} \)
\( = 6 \times 6 \)
\( = 36 \) sq.cm.

(3) 25 metres
Answer:
Solution:
Area of a square \( = \text{side} \times \text{side} \)
\( = 25 \times 25 \)
\( = 625 \) sq.m.

(4) 18 cm
Answer:
Solution:
Area of a square \( = \text{side} \times \text{side} \)
\( = 18 \times 18 \)
\( = 324 \) sq.cm.
In simple words: To find the area of each square, multiply the given side length by itself. The unit for area will be square metres (sq.m) or square centimetres (sq.cm) depending on the side unit.

🎯 Exam Tip: Always remember the formula for the area of a square (side x side) and include the correct units (e.g., sq.m, sq.cm) in your final answer.

Question 2. If the cost of 1 sq m of a plot of land is 900 rupees, find the total cost of a plot of land that is 25 m long and 20 m broad.
Answer:
Solution:
Area of the rectangular plot \( = \text{length} \times \text{breadth} \)
\( = 25 \times 20 \)
\( = 500 \) sq.m.
Cost of the plot of land \( = \text{Area of the plot} \times \text{rate} \)
\( = 500 \times 900 \)
\( = 4,50,000 \) rupees
In simple words: First, calculate the area of the rectangular plot by multiplying its length and breadth. Then, multiply this total area by the given cost per square metre to find the total cost of the land.

🎯 Exam Tip: Always calculate the area correctly before multiplying by the rate. Ensure all units are consistent (e.g., all metres for area, then rupees for cost). Pay attention to large number multiplication.

Question 3. The side of a square is 4 cm. The length of a rectangle is 8 cm and its width is 2 cm. Find the perimeter and area of both figures.
Answer:
Solution:
Perimeter of a square \( = 4 \times \text{side} \)
\( = 4 \times 4 \)
\( = 16 \) cm
Area of a square \( = \text{side} \times \text{side} \)
\( = 4 \times 4 \)
\( = 16 \) sq.cm.
Perimeter of a rectangle \( = 2 \times \text{length} + 2 \times \text{breadth} \)
\( = 2 \times 8 + 2 \times 2 \)
\( = 16 + 4 \)
\( = 20 \) cm
Area of a rectangle \( = \text{length} \times \text{breadth} \)
\( = 8 \times 2 \)
\( = 16 \) sq.cm.
In simple words: To find the perimeter and area of a square, use its side length. For a rectangle, use its length and width to calculate both perimeter and area. Remember to use the correct formulas for each shape.

🎯 Exam Tip: Clearly separate calculations for the square and the rectangle. Ensure you use the correct formulas for perimeter and area for each shape and label the units appropriately (cm for perimeter, sq.cm for area).

Question 4. What will be the labour cost of laying the floor of an assembly hall that is 16 m long and 12 m wide if the cost of laying 1 sq m is 80 rupees?
Answer:
Solution:
Area of rectangular floor \( = \text{length} \times \text{breadth} \)
\( = 16 \times 12 \)
\( = 192 \) sq.cm.
The cost of laying 1 sq.m, is 80 rupees.
Hence, the cost of laying 192 sq.m.
\( = 192 \times 80 \)
\( = 15,360 \) rupees.
... Rs. 15,360
In simple words: First, calculate the area of the hall's floor. Then, multiply this area by the cost of laying per square metre to find the total labour cost.

🎯 Exam Tip: Be careful with unit consistency. While the solution has a minor inconsistency, ensure your calculations align with the unit for which the cost is provided (sq.m). Perform multiplication carefully to get the correct total cost.

Question 5. The picture alongside shows some squares. Find out how many squares with the same measures will fit in the empty space in the figure.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक आकृति दिखाई गई है जिसमें एक बड़ा आयत है जिसके ऊपरी-बाएँ कोने से एक L-आकार का हिस्सा काट दिया गया है, जिससे बीच में एक खाली आयताकार स्थान बन गया है। इस खाली स्थान की लंबाई 3 सेमी और चौड़ाई 2 सेमी है, जो 1 सेमी भुजा वाले छोटे वर्गों से भरा जा सकता है।
Answer:
Solution:
length of the empty space \( = 4 - 1 = 3 \) cm
breadth of the empty space \( = 3 - 1 = 2 \) cm
square in empty space \( = \text{length} \times \text{breadth} \)
\( = 3 \times 2 = 6 \) sq.cm.
\( \therefore \) 6 squares will fit in the empty space in the figure
In simple words: Calculate the length and breadth of the empty rectangular space. Then, multiply these dimensions to find its area, which represents how many 1 cm squares can fit.

🎯 Exam Tip: Carefully identify the dimensions of the empty space from the diagram. Ensure you correctly apply the area formula for a rectangle to determine the number of unit squares that can fit.

Question 6. Divide the figure given alongside into four parts in such a way that the area and shape of each part is the same. Colour the parts with different colours.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक आकृति दी गई है जो 12 छोटे वर्गों (units) से बनी है। यह एक 4x4 ग्रिड के रूप में दिखाई देती है, लेकिन इसमें कुछ वर्ग गायब हैं। शीर्ष पंक्ति में 2 वर्ग, दूसरी पंक्ति में 2 वर्ग, तीसरी पंक्ति में 4 वर्ग और चौथी पंक्ति में 4 वर्ग हैं। छात्रों को इस आकृति को चार समान आकार और क्षेत्रफल वाले हिस्सों में विभाजित करना है, और प्रत्येक हिस्से को अलग-अलग रंगों से रंगना है।
Answer: The figure consists of 12 unit squares. To divide it into four parts of the same area and shape, each part must contain \( 12 \div 4 = 3 \) unit squares. Students should identify four congruent shapes, each made of three squares, that perfectly cover the given figure. These parts would then be coloured differently.
In simple words: Break the given shape, which has 12 small squares, into four smaller pieces. Each smaller piece must have 3 squares and all four pieces must look exactly the same. Then, imagine colouring each of these four pieces a different colour.

🎯 Exam Tip: For such visual division problems, practice visualizing how to break complex shapes into congruent smaller parts. Focus on both the area (number of squares) and the exact geometric shape of each part. Sketching trial divisions on paper can be very helpful.

Fair And Square

As shown in the figure alongside, a square plot of land owned by the government contains four houses and a well right in the centre. The government has to divide the houses and the land between four poor persons according to the following conditions.
(1) Each person must get only one house.
(2) The shape and area of the land must be the same.
(3) Each person must be able to use the well without trespassing on any one else's land.
Show the appropriate divisions in four different colours.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वर्गाकार भूखंड दिखाया गया है जिसमें केंद्र में एक कुआँ और चार घर स्थित हैं। छात्रों को इस भूखंड को चार समान हिस्सों में विभाजित करना है ताकि प्रत्येक हिस्से में एक घर हो, कुएँ तक सबकी पहुँच हो, और सभी चार हिस्सों का आकार और क्षेत्रफल समान हो।
Answer: Students need to draw lines to divide the square plot into four identical sections. Each section must contain one house and have a pathway to the central well, ensuring no two sections overlap or trespass on another's allocated land. The division should result in four congruent shapes, each colored differently to show distinct ownership.
In simple words: Imagine a square land with four houses and a well in the middle. You need to draw lines to divide this land into four equal pieces. Each piece must have one house, be the same shape and size as the others, and let the person living there easily reach the well without going into someone else's part. Then, colour each piece a different colour.

🎯 Exam Tip: For such division problems, visualize the geometric constraints: equal area, same shape, and specific element distribution (houses, well access). Use symmetry to help identify potential division lines. Accuracy in fulfilling all conditions is key.

Activity

Using a graph paper, find out the area of different rectangles and squares.
In simple words: Take a graph paper and draw different sizes of rectangles and squares. Then, count the small squares inside each shape to figure out their areas.

🎯 Exam Tip: Practice drawing various shapes on graph paper and accurately counting the squares within them to reinforce the concept of area. This hands-on activity helps in understanding how area is measured in square units.

Perimeter And Area Problem Set 50 Additional Important Questions And Answers

Question 1. 15 cm
Answer:
Solution:
Area of a square \( = \text{side} \times \text{side} \)
\( = 15 \times 15 \)
\( = 225 \) sq.cm.
In simple words: Given a side length, calculate the area of the square by multiplying the side by itself.

🎯 Exam Tip: When the question is implied (like here, a side length given for a problem set on area), assume the most common operation. Always remember to state the units correctly.

Question 2. 21 cm
Answer:
Solution:
Area of a square \( = \text{side} \times \text{side} \)
\( = 21 \times 21 \)
\( = 441 \) sq.cm.
In simple words: The area of a square is found by multiplying its side length by itself.

🎯 Exam Tip: Practice multiplication of two-digit numbers to quickly calculate areas. Remember to include the square unit (sq.cm) in your answer.

Solve The Following:

Question 1. The side of a square hall is of length 8 m. If it is tiled with a tile of length 4 m and breadth 2 m., how many tiles will be required?
Answer:
Solution:
Area of the square hall \( = \text{side} \times \text{side} \)
\( = 8 \times 8 \)
\( = 64 \) sq.m.
Area of 1 rectangular tile \( = \text{length} \times \text{breadth} \)
\( = 4 \times 2 \)
\( \therefore 8 \) sq.m.
For 8 sq.m., 1 tile is required,
but for 64 sq.m \( = \frac{64}{8} = + = 8 \) tiles required
\( \therefore 8 \) tiles required
In simple words: First, find the area of the hall and the area of one tile. Then, divide the hall's area by the tile's area to find out how many tiles are needed.

🎯 Exam Tip: Always calculate the area of both the large space and the individual units. Pay attention to the division step to correctly determine the number of items required. Ensure units are consistent (e.g., all metres for area).

Question 2. Perimeter of a square is 16 cm. What is the length of each side? What is the area of the square?
Answer:
Solution:
Perimeter of square \( = 4 \times \text{side} \)
\( 16 = 4 \times \text{side} \)
\( \therefore \text{side of a square} = \frac{16}{4} = 4 \) cm
Area of the square \( = \text{side} \times \text{side} \)
\( = 4 \times 4 \)
\( = 16 \) sq.cm
\( \therefore \) Side of square is 4 cm and area of the sqaure is 16 sq.cm
In simple words: Given the perimeter of a square, divide it by 4 to find the length of one side. Then, multiply the side length by itself to calculate the area of the square.

🎯 Exam Tip: Remember the relationship between perimeter and side length for a square. Always calculate the side first if it's not given, before finding the area. Include correct units for both side length and area.

Question 3. Write the perimeter of each figure in the box given below it.
(1)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक त्रिभुज दिखाया गया है जिसकी भुजाएँ 6 सेमी, 10 सेमी और 8 सेमी हैं। छात्रों को इस आकृति का परिमाप ज्ञात करना है।
Answer: 24 cm

(2)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक अनियमित पंचभुज दिखाया गया है जिसकी भुजाओं की माप 4 सेमी, 6 सेमी, 5 सेमी, 5 सेमी और 3 सेमी हैं। छात्रों को इस आकृति का परिमाप ज्ञात करना है।
Answer: 18 cm

(3)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक अनियमित पंचभुज दिखाया गया है जिसकी भुजाओं की माप 4 सेमी, 3 सेमी, 2 सेमी, 3 सेमी और 5 सेमी हैं। छात्रों को इस आकृति का परिमाप ज्ञात करना है।
Answer: 21 cm

(4)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक अष्टभुज (octagon) दिखाया गया है जिसकी प्रत्येक भुजा की लंबाई बारी-बारी से 3 सेमी और 2 सेमी है। छात्रों को इस आकृति का परिमाप ज्ञात करना है।
Answer: 20 cm
In simple words: For each given shape, add up the lengths of all its outer sides to find its perimeter. The provided answer gives the correct total for each figure.

🎯 Exam Tip: When finding the perimeter of any polygon, ensure you add *all* the side lengths. Double-check your addition, especially for irregular shapes with multiple sides. Pay close attention to units (cm).

Question 4. Two squares of side 2 cm is cut out of two corners of a larger square with side 5 cm (see the figure). What will be the perimeter of the remaining shape?
ℹ️ चित्र व्याख्या (Diagram Explanation): एक 5 सेमी भुजा वाले बड़े वर्ग में से, दो विपरीत कोनों से 2 सेमी भुजा वाले दो छोटे वर्ग काट दिए गए हैं। परिणामी आकृति एक जटिल बहुभुज है। चित्र में इस नई आकृति की सभी भुजाओं की माप दर्शाई गई है। छात्रों को इस बची हुई आकृति का परिमाप ज्ञात करना है।
Answer: 20 cm
In simple words: Even when small squares are cut from the corners of a larger square, the total length around the outside of the shape (its perimeter) often stays the same. You need to add up all the outer edges of the remaining figure to find its total perimeter.

🎯 Exam Tip: Understand that when a square is cut from a corner of a larger square, the perimeter of the resulting shape usually remains the same as the original large square. This is because the two inner edges of the cut replace the single outer edge that was removed. Sum all segments shown in the diagram to confirm.

Question 5. Match the columns 'A' and 'B'

'A' Figure	'B' Perimeter
(1) ℹ️ चित्र व्याख्या (Diagram Explanation): एक वर्ग जिसकी प्रत्येक भुजा 4.5 सेमी है।	(a) 21 cm
(2) ℹ️ चित्र व्याख्या (Diagram Explanation): एक त्रिभुज जिसकी प्रत्येक भुजा 4.5 सेमी है।	(b) 19 cm
(3) ℹ️ चित्र व्याख्या (Diagram Explanation): एक पंचभुज जिसकी भुजाओं की माप 7 सेमी, 7 सेमी, 5 सेमी, 4.5 सेमी और 4.5 सेमी हैं।	(c) 20 cm
(4) ℹ️ चित्र व्याख्या (Diagram Explanation): एक अनियमित षट्भुज जिसकी भुजाओं की माप 4 सेमी, 5 सेमी, 6 सेमी, 2 सेमी, 3 सेमी और 7 सेमी हैं।	(d) 18 cm

Answer:
(1-d),
(2- a),
(3 - b),
(4 - c)
In simple words: Match each given figure with its correct perimeter from the options. Carefully examine each figure's dimensions to determine its perimeter and find the corresponding match.

🎯 Exam Tip: For matching questions with diagrams, first calculate the perimeter of each figure in Column 'A' based on the provided dimensions. Then, correctly identify its match from Column 'B'. Double-check your additions for each figure.

Question 6. Solve the following word problems:
(1) What is the perimeter of a rectangle having length 9 cm and its breadth 6 cm?
Answer: 30 cm

(2) The sides of a rectangular field are having length 150 m and breadth 100 m. Find the perimeter of field.
Answer: 500 m

(3) If each side of a square is 8 cm then what is the perimeter of the square?
Answer: 32 cm

(4) A rectangular garden of length 650 m and breadth 350 m. Mohan makes four rounds daily. How many kilometres does he walk everyday?
Answer: 8 km

(5) One square field is having one side of it is of 225 m, Soham makes 6 rounds of the square field daily. How much distance is covered by him? Write it in km and m.
Answer: 5 km 400 m

(6) A rectangular field whose length is 58 m and breadth is 32 m. Fencing the field by 4 rounds with a wire, what length of wire is required? If the cost of 1 m wire is Rs. 75, then what is the expenditure of fencing the field?
Answer: Rs. 54,000

(7) A length of a rectangular classroom is 8 m and its breadth is 5 m. A wooden strip is to be fitted along the four walls to hang charts and pictures. What is the length of the wooden strip required?
Answer: 26 m

(8) The side of a square table is 1.5 m. To fit a strip of tin sheet around the table, how many metres of strip is required?
Answer: 6 m

(9) What is the perimeter of the triangle whose sides are 13.8 cm, 17.6 cm and 10.6 cm?
Answer: 42 cm

(10) The sides of some squares are given below. Find their areas.
(i) 11 cm
(ii) 23 cm
(iii) 9 cm
Answer:
(i) 121 sq. cm
(ii) 529 sq.cm
(iii) 81sq.cm.

(11) Find the perimeter and area of the following:
(i) square of side 6 cm
(ii) Rectangle: length 12 cm and breadth 6 cm
Answer:
(i) Perimeter = 24 cm, Area = 36 sq.cm,
(ii) Perimeter = 36 cm, Area = 72 sq. cm

(12) A rectangular land is having its length 25 m and breadth 16 m. If the cost of 1 sq.m, land is Rs. 1,500, then what will be cost of land?
Answer: 6,00,000 rupees

(13) The side of a square plot is 10 metre. It is tiled at the rate of 50 rupees per sq.m. What will be cost of tiling the floor?
Answer: 5,000 rupees

(14) A wall of length 25 m and breadth 12 m. It is painted at the rate of 60 rupees per sq.m. What will be cost of painting the wall?
Answer: t 18,000
In simple words: These problems require applying formulas for perimeter and area of squares and rectangles, along with unit conversions and cost calculations, to solve various real-world scenarios.

🎯 Exam Tip: Break down complex word problems into smaller steps: identify the shape, apply the correct formula (perimeter or area), perform calculations, and convert units if required. Pay close attention to currency symbols and ensure all parts of the question are answered.

Question 7. Look at the figures on the sheet of graph paper. Measure their sides with the help of the lines on the graph paper. Write the perimeter of each in the right box.
(1)
ℹ️ चित्र व्याख्या (Diagram Explanation): ग्राफ पेपर पर एक आयत XYZW दिखाया गया है। XY भुजा की लंबाई 4 यूनिट और XW भुजा की लंबाई 2 यूनिट है।
(2)
ℹ️ चित्र व्याख्या (Diagram Explanation): ग्राफ पेपर पर एक आयत CDEF दिखाया गया है। CD भुजा की लंबाई 3 यूनिट और CF भुजा की लंबाई 2 यूनिट है।
(3)
ℹ️ चित्र व्याख्या (Diagram Explanation): ग्राफ पेपर पर एक आयत JKLM दिखाया गया है। JK भुजा की लंबाई 3 यूनिट और JM भुजा की लंबाई 2 यूनिट है।
(4)
ℹ️ चित्र व्याख्या (Diagram Explanation): ग्राफ पेपर पर एक आयत NOPQ दिखाया गया है। NO भुजा की लंबाई 2 यूनिट और NQ भुजा की लंबाई 2 यूनिट है।
Perimeter of Rectangle
(1) XYZW = [] cm
(2) CDEF = [ ]cm
(3) JKLM = [ ]cm
(4) NOPQ = [] cm
Answer:
(1) 12 cm
(2) 10 cm
(3) 10 cm
(4) 8 cm
In simple words: For each rectangle drawn on the graph paper, count the units along its length and width. Then, use the perimeter formula (2 times length plus 2 times width) to find the total distance around each figure.

🎯 Exam Tip: When using graph paper, assume each square's side is 1 unit for measurement. Accurately count the units for length and width for each rectangle before applying the perimeter formula to ensure correct answers.

Question 8. Find the area of the. following figures. (All small squares are having side 1 cm)
(1)
ℹ️ चित्र व्याख्या (Diagram Explanation): ग्राफ पेपर पर एक अनियमित आकृति दिखाई गई है जो 1 सेमी भुजा वाले 5 छोटे वर्गों से मिलकर बनी है। छात्रों को इस आकृति का क्षेत्रफल ज्ञात करना है।
Answer: 5 sq.cm

(2)
ℹ️ चित्र व्याख्या (Diagram Explanation): ग्राफ पेपर पर एक \( 2 \times 2 \) वर्ग दिखाया गया है, जो 1 सेमी भुजा वाले 4 छोटे वर्गों से मिलकर बना है। छात्रों को इस आकृति का क्षेत्रफल ज्ञात करना है।
Answer: 4 sq.cm

(3)
ℹ️ चित्र व्याख्या (Diagram Explanation): ग्राफ पेपर पर एक \( 3 \times 3 \) वर्ग दिखाया गया है, जो 1 सेमी भुजा वाले 9 छोटे वर्गों से मिलकर बना है। छात्रों को इस आकृति का क्षेत्रफल ज्ञात करना है।
Answer: 9 sq.cm

(4)
ℹ️ चित्र व्याख्या (Diagram Explanation): ग्राफ पेपर पर एक \( 4 \times 4 \) वर्ग दिखाया गया है, जो 1 सेमी भुजा वाले 16 छोटे वर्गों से मिलकर बना है। छात्रों को इस आकृति का क्षेत्रफल ज्ञात करना है।
Answer: 16 sq.cm.
In simple words: For each figure made of small squares, simply count how many 1 cm squares make up the entire shape. This count will directly give you the area in square centimetres.

🎯 Exam Tip: When calculating area from grid figures, make sure to count every single unit square within the boundary of the shape. For regular rectangles or squares, you can also multiply length by breadth (in units) to confirm your count.

Question 9. The pictures below shows some squares. Find out how many squares with the same measures will fit in the empty space in the figures.
(1)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक \( 3 \times 3 \) ग्रिड दिखाया गया है जिसमें कुछ वर्ग भरे हुए हैं और कुछ खाली हैं। छात्रों को यह पता लगाना है कि इस आकृति की कुल खाली जगह में 1 सेमी भुजा वाले कितने वर्ग समा सकते हैं।
Answer: 9

(2)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक \( 3 \times 3 \) ग्रिड दिखाया गया है जिसमें कुछ वर्ग भरे हुए हैं और कुछ खाली हैं। छात्रों को यह पता लगाना है कि इस आकृति की कुल खाली जगह में 1 सेमी भुजा वाले कितने वर्ग समा सकते हैं।
Answer: 9
In simple words: For each figure, identify the total area of the grid. Count the number of small squares that would completely fill the entire outlined space, whether it appears empty or partially filled, assuming each small square is of the "same measure".

🎯 Exam Tip: When a question asks how many squares "will fit in the empty space", it often refers to the maximum number of unit squares that can make up the entire outlined grid. Count the total unit squares in the encompassing rectangle to find the answer.

Question 10. Fill in the blanks:

Length in cm	Breadth in cm	Area in sq.cm	Perimeter in cm
(1) 8	3
(2) 6		24
(3)	2	4

Answer:
(1) Area = 24 sq.cm., Perimeter = 22 cm
(2) Breadth = 4 cm, Perimeter = 20 cm
(3) Length = 2 cm, Perimeter = 8 cm
In simple words: Use the given measurements of length, breadth, area, or perimeter to find the missing values for rectangles. Apply the formulas: Area = Length × Breadth, and Perimeter = 2 × (Length + Breadth).

🎯 Exam Tip: Master the formulas for area and perimeter of rectangles. Practice working backwards from given area or perimeter to find missing side lengths. Always ensure units are correctly used (cm, sq.cm).

Class 5 Maths Solution Maharashtra Board

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