Samacheer Kalvi Class 6 Maths Solutions Term 3 Chapter 3 Perimeter and Area Exercise 3.2

Miscellaneous Practice Problems

Question 1. A piece of wire is 36 cm long. What will be the length of each side if we form
(i) a square
(ii) an equilateral triangle
Answer:
(i) When the wire is used to form a square, its total length becomes the perimeter of the square. A square has 4 equal sides.
\( \text{Perimeter of the square} = 36 \text{ cm} \)
\( 4 \times \text{side} = 36 \text{ cm} \)
\( \text{side} = \frac{36}{4} = 9 \text{ cm} \)
So, each side of the square will be 9 cm long.
(ii) When the wire is used to form an equilateral triangle, its total length becomes the perimeter of the triangle. An equilateral triangle has 3 equal sides.
\( \text{Perimeter of the equilateral triangle} = 36 \text{ cm} \)
\( 3 \times \text{side} = 36 \text{ cm} \)
\( \text{side} = \frac{36}{3} = 12 \text{ cm} \)
Therefore, each side of the equilateral triangle will be 12 cm long. An equilateral triangle has all angles equal to 60 degrees.
In simple words: The wire's total length is the outside edge of the shape. For a square, divide the length by 4. For an equilateral triangle, divide it by 3.

🎯 Exam Tip: Remember that perimeter is the total length of the boundary of a closed shape. For regular polygons (like squares or equilateral triangles), you can find the side length by dividing the perimeter by the number of equal sides.

Question 2. From one vertex of an equilateral triangle with a side of 40 cm, an equilateral triangle with 6 cm side is removed. What is the perimeter of the remaining portion?
Answer:Let the original equilateral triangle have side \( A = 40 \text{ cm} \). A smaller equilateral triangle with side \( B = 6 \text{ cm} \) is removed from one vertex. When a small equilateral triangle is removed from a vertex, two sides of the original triangle become shorter, and one new side (the base of the removed triangle) is added to the perimeter. Each of the two original sides connected to the vertex loses 6 cm in length. So, the length of each of these two modified sides becomes \( 40 - 6 = 34 \text{ cm} \). The third side of the original triangle remains unchanged at 40 cm. The base of the removed triangle, which is 6 cm, now forms part of the new perimeter. So, the perimeter of the remaining portion consists of:
1. Two altered sides: \( 34 \text{ cm} + 34 \text{ cm} \)
2. One original side: \( 40 \text{ cm} \)
3. The new inner side (base of removed triangle): \( 6 \text{ cm} \) Perimeter of the remaining portion \( = (40 - 6) + (40 - 6) + 40 + 6 \)
\( = 34 + 34 + 40 + 6 \)
\( = 114 \text{ cm} \) The total perimeter becomes 114 cm. This technique is often used in calculating areas of complex shapes by subtracting or adding simple ones.
In simple words: When a small triangle is cut from a corner of a big triangle, two sides of the big triangle get shorter by the size of the small triangle. But, the edge where the small triangle was cut adds a new line to the outline. So, we add the two shorter sides, the one full side, and the new inside line to find the total length around.

🎯 Exam Tip: When a shape is removed from a corner, carefully trace the new outer boundary. The lengths of the original sides change, and new internal lines might become part of the perimeter. Always visualize how the shape's outline changes.

Question 3. Rahim and Peter go for a morning walk, Rahim walks around a square path of side 50 m and Peter walks around a rectangular path with a length of 40 m and a breadth of 30 m. If both of them walk 2 rounds each, who covers more distance and by how much?
Answer:First, we need to find the distance covered in one round by Rahim and Peter.
**For Rahim (Square Path):**
Side of the square path \( = 50 \text{ m} \)
Perimeter of the square \( = 4 \times \text{side} \)
\( = 4 \times 50 \text{ m} = 200 \text{ m} \)
Distance covered by Rahim in 1 round \( = 200 \text{ m} \)
Distance covered by Rahim in 2 rounds \( = 2 \times 200 \text{ m} = 400 \text{ m} \)
**For Peter (Rectangular Path):**
Length of the rectangular path \( = 40 \text{ m} \)
Breadth of the rectangular path \( = 30 \text{ m} \)
Perimeter of the rectangle \( = 2 \times (\text{length} + \text{breadth}) \)
\( = 2 \times (40 \text{ m} + 30 \text{ m}) \)
\( = 2 \times 70 \text{ m} = 140 \text{ m} \)
Distance covered by Peter in 1 round \( = 140 \text{ m} \)
Distance covered by Peter in 2 rounds \( = 2 \times 140 \text{ m} = 280 \text{ m} \)
**Comparing Distances:**
Distance covered by Rahim \( = 400 \text{ m} \)
Distance covered by Peter \( = 280 \text{ m} \) Since \( 400 \text{ m} > 280 \text{ m} \), Rahim covers more distance.
Difference in distance \( = 400 \text{ m} - 280 \text{ m} = 120 \text{ m} \) Rahim covers 120 m more distance than Peter. This shows how different shapes can lead to different distances even if they appear similar in size.
In simple words: First, find the distance around one path for Rahim (square) and Peter (rectangle). Then, multiply each by 2 for two rounds. Compare the final distances. Rahim walked more distance, and the difference was 120 meters.

🎯 Exam Tip: Always clearly calculate the perimeter for each shape first, then multiply by the number of rounds to get the total distance. Pay close attention to the units (meters in this case).

Question 4. The length of a rectangular park is 14 m more than its breadth. If the perimeter of the park is 200 m, what is its length? Find the area of the park.
Answer:Let the breadth of the rectangular park be \( b \text{ m} \). According to the problem, the length \( l \) is 14 m more than its breadth. So, \( l = b + 14 \text{ m} \). The perimeter of the park is given as 200 m. The formula for the perimeter of a rectangle is \( P = 2 \times (l + b) \).
Substituting the given values:
\( 200 = 2 \times ((b + 14) + b) \)
\( 200 = 2 \times (2b + 14) \)
Now, divide both sides by 2:
\( \frac{200}{2} = 2b + 14 \)
\( 100 = 2b + 14 \)
Subtract 14 from both sides:
\( 100 - 14 = 2b \)
\( 86 = 2b \)
Divide by 2 to find \( b \):
\( b = \frac{86}{2} = 43 \text{ m} \) Now we have the breadth, we can find the length:
\( l = b + 14 = 43 + 14 = 57 \text{ m} \) Finally, we need to find the area of the park. The formula for the area of a rectangle is \( A = l \times b \).
\( A = 57 \text{ m} \times 43 \text{ m} \)
\( A = 2451 \text{ m}^2 \) The area of the park is 2451 square meters. This problem illustrates how algebraic equations help solve geometry problems.
In simple words: First, use the perimeter formula and the clue about length and breadth to find the breadth. Once you have the breadth, find the length. Finally, multiply the length and breadth to get the area of the park.

🎯 Exam Tip: When given a relationship between length and breadth, express one in terms of the other (e.g., \( l = b + 14 \)). Then, substitute this into the perimeter formula to solve for a single variable. Remember to use correct units for both length and area.

Question 5. Your garden is in the shape of a square of side 5 m. Each side is to be fenced with 2 rows of wire. Find how much amount is needed to fence the garden at Rs 10 per metre.
Answer:First, we need to find the perimeter of the square garden to know the total length of wire needed for one row.
Side of the square garden \( a = 5 \text{ m} \)
Perimeter of the square \( = 4 \times a \)
\( = 4 \times 5 \text{ m} = 20 \text{ m} \) This 20 m is the length of wire needed for 1 row of fencing. The problem states that the garden needs to be fenced with 2 rows of wire.
Total length of wire needed \( = 2 \times \text{Perimeter for 1 row} \)
\( = 2 \times 20 \text{ m} = 40 \text{ m} \) The cost of fencing is Rs 10 per metre.
Total amount needed to fence the garden \( = \text{Total length of wire} \times \text{Cost per metre} \)
\( = 40 \text{ m} \times \text{Rs } 10/\text{m} \)
\( = \text{Rs } 400 \) So, Rs 400 is needed to fence the garden with two rows of wire. This is a common practical application of perimeter calculations.
In simple words: Find the distance around the garden (perimeter). Since you need two rows of wire, double that distance. Then, multiply the total wire length by the cost for each meter to get the final amount.

🎯 Exam Tip: Always read carefully to see if the fencing requires more than one row. Calculate the total length of wire first, then multiply by the cost per unit length. Don't forget to include the currency symbol for the final answer.

Challenge Problems

Question 6. A closed shape has 20 equal sides and one of its sides is 3 cm. Find its perimeter.
Answer:A closed shape with equal sides is called a regular polygon.
Number of equal sides \( = 20 \)
Length of one side \( = 3 \text{ cm} \) The perimeter of a polygon is found by adding the lengths of all its sides. Since all sides are equal:
Perimeter \( = \text{Number of sides} \times \text{Length of one side} \)
\( = 20 \times 3 \text{ cm} \)
\( = 60 \text{ cm} \) The perimeter of the shape is 60 cm. This applies to any regular polygon, whether it's a pentagon, hexagon, or a 20-sided Icosagon.
In simple words: To find the perimeter of a shape with many equal sides, just multiply the number of sides by the length of one side.

🎯 Exam Tip: For any regular polygon (a shape with all sides and angles equal), its perimeter is simply the number of sides multiplied by the length of one side. Remember to use the correct units in your final answer.

Question 7. A rectangle has length 40 cm and breadth 20 cm. How many squares with side 10 cm can be formed from it.
Answer:First, calculate the area of the rectangle.
Length of the rectangle \( = 40 \text{ cm} \)
Breadth of the rectangle \( = 20 \text{ cm} \)
Area of the rectangle \( = \text{length} \times \text{breadth} \)
\( = 40 \text{ cm} \times 20 \text{ cm} \)
\( = 800 \text{ cm}^2 \) Next, calculate the area of one small square.
Side of one square \( = 10 \text{ cm} \)
Area of one square \( = \text{side} \times \text{side} \)
\( = 10 \text{ cm} \times 10 \text{ cm} \)
\( = 100 \text{ cm}^2 \) To find how many small squares can be formed from the rectangle, divide the area of the rectangle by the area of one square.
Number of squares \( = \frac{\text{Area of Rectangle}}{\text{Area of 1 Square}} \)
\( = \frac{800 \text{ cm}^2}{100 \text{ cm}^2} \)
\( = 8 \) So, 8 squares of side 10 cm can be formed from the given rectangle. This process is useful for tiling or cutting materials efficiently.
In simple words: Find the total space inside the big rectangle. Then, find the space inside one small square. Divide the big space by the small space to see how many small squares fit.

🎯 Exam Tip: When asked to find how many smaller shapes fit into a larger shape, calculate the area of both and then divide the larger area by the smaller area. Ensure both areas are in the same units.

Question 8. The length of a rectangle is three times its breadth. If its perimeter is 64 cm, find the sides of the rectangle.
Answer:Let the breadth of the rectangle be \( b \text{ cm} \). According to the problem, the length \( l \) is three times its breadth. So, \( l = 3 \times b = 3b \text{ cm} \). The perimeter of the rectangle is given as 64 cm. The formula for the perimeter of a rectangle is \( P = 2 \times (l + b) \).
Substituting the given values:
\( 64 = 2 \times (3b + b) \)
\( 64 = 2 \times (4b) \)
\( 64 = 8b \)
Now, divide both sides by 8 to find \( b \):
\( b = \frac{64}{8} = 8 \text{ cm} \) Now that we have the breadth, we can find the length:
\( l = 3b = 3 \times 8 = 24 \text{ cm} \) So, the length of the rectangle is 24 cm and the breadth is 8 cm. This demonstrates a common way to use a perimeter to find unknown side lengths.
In simple words: Use the perimeter formula and the clue that length is three times the breadth. This helps you find the breadth first. Then, multiply the breadth by three to find the length.

🎯 Exam Tip: When solving for unknown dimensions, always set up equations using variables (like \( l \) and \( b \)). Write down the given relationships and the perimeter formula, then substitute and solve systematically. Double-check your calculations.

Question 9. How many different rectangles can be made with a 48 cm long string? Find the possible pairs of length and breadth of the rectangles.
Answer:The length of the string, 48 cm, represents the perimeter of the rectangle.
Perimeter of the rectangle \( = 48 \text{ cm} \) The formula for the perimeter of a rectangle is \( P = 2 \times (l + b) \).
So, \( 48 = 2 \times (l + b) \)
Divide both sides by 2:
\( \frac{48}{2} = l + b \)
\( 24 = l + b \) Now we need to find pairs of positive integers for length \( l \) and breadth \( b \) such that their sum is 24. Since length is usually considered greater than or equal to breadth \( (l \ge b) \), we list the possible pairs:
If \( b = 1 \), then \( l = 24 - 1 = 23 \) (Pair: (23, 1))
If \( b = 2 \), then \( l = 24 - 2 = 22 \) (Pair: (22, 2))
If \( b = 3 \), then \( l = 24 - 3 = 21 \) (Pair: (21, 3))
If \( b = 4 \), then \( l = 24 - 4 = 20 \) (Pair: (20, 4))
If \( b = 5 \), then \( l = 24 - 5 = 19 \) (Pair: (19, 5))
If \( b = 6 \), then \( l = 24 - 6 = 18 \) (Pair: (18, 6))
If \( b = 7 \), then \( l = 24 - 7 = 17 \) (Pair: (17, 7))
If \( b = 8 \), then \( l = 24 - 8 = 16 \) (Pair: (16, 8))
If \( b = 9 \), then \( l = 24 - 9 = 15 \) (Pair: (15, 9))
If \( b = 10 \), then \( l = 24 - 10 = 14 \) (Pair: (14, 10))
If \( b = 11 \), then \( l = 24 - 11 = 13 \) (Pair: (13, 11))
If \( b = 12 \), then \( l = 24 - 12 = 12 \) (Pair: (12, 12)) - This is a square, which is a special type of rectangle. Counting these pairs, there are 12 different rectangles that can be made. This illustrates how many geometric variations can exist with a fixed perimeter.
In simple words: The string length is the perimeter. Divide the perimeter by 2 to find what length plus breadth must add up to. Then, list all the possible pairs of whole numbers that add up to that sum, making sure the length is not smaller than the breadth. Each pair makes a different rectangle.

🎯 Exam Tip: When finding pairs of length and breadth for a given perimeter, start by finding \( l + b \). Then systematically list integer pairs, ensuring \( l \ge b \). Remember that a square is also a type of rectangle (when \( l = b \)).

Question 10. Draw a square B whose side is twice of the square A. Calculate the perimeters of the squares A and B.
Answer:Let's assume Square A has a side of 2 m based on the visual.
**Square A:**
Side \( s_A = 2 \text{ m} \)
Perimeter of Square A \( P_A = 4 \times s_A = 4 \times 2 \text{ m} = 8 \text{ m} \)
**Square B:**
The side of Square B is twice the side of Square A.
Side \( s_B = 2 \times s_A = 2 \times 2 \text{ m} = 4 \text{ m} \)
Perimeter of Square B \( P_B = 4 \times s_B = 4 \times 4 \text{ m} = 16 \text{ m} \)
Here is a representation of the squares:

We can observe that \( P_B = 16 \text{ m} \) and \( P_A = 8 \text{ m} \). This means \( P_B = 2 \times P_A \). So, the perimeter of B is twice the perimeter of A. This relationship holds true when the side length is doubled.
In simple words: First, find the side of square A (let's say 2m from the drawing). Then, double that to find the side of square B. For each square, multiply its side by 4 to get its perimeter. You will see that the perimeter of the bigger square is twice the perimeter of the smaller one.

🎯 Exam Tip: Remember that if the side length of a square is doubled, its perimeter also doubles. However, its area would increase by a factor of four. Be careful not to confuse perimeter and area scaling.

Question 11. What will be the area of a new square formed if the side of a square is made one fourth?
Answer:Let the side of the original square be \( s \) units. The area of the original square \( A_{original} = s \times s = s^2 \) square units. If the side of the new square is made one-fourth of the original side:
Side of the new square \( s_{new} = \frac{1}{4} \times s = \frac{s}{4} \) units. Now, calculate the area of the new square:
Area of the new square \( A_{new} = s_{new} \times s_{new} \)
\( A_{new} = \left(\frac{s}{4}\right) \times \left(\frac{s}{4}\right) \)
\( A_{new} = \frac{s \times s}{4 \times 4} = \frac{s^2}{16} \)
\( A_{new} = \frac{1}{16} \times s^2 \) Since \( s^2 \) is the area of the original square, we can write:
\( A_{new} = \frac{1}{16} \times A_{original} \) So, the area of the new square will be reduced to \( \frac{1}{16} \) times that of the original area. This highlights how changing dimensions significantly impacts area.
In simple words: If you make the side of a square four times smaller, its new area will become 16 times smaller than the old area. This is because you divide by 4 twice (once for length, once for width).

🎯 Exam Tip: When the side of a square is scaled by a factor \( k \), its perimeter scales by \( k \) and its area scales by \( k^2 \). If the side is made \( \frac{1}{4} \), then \( k = \frac{1}{4} \), so the area scales by \( \left(\frac{1}{4}\right)^2 = \frac{1}{16} \).

Question 12. Two plots have the same perimeter. One is a square of side 10 m and another is a rectangle of breadth 8 m. Which plot has the greater area and by how much?
Answer:First, calculate the perimeter and area for the square plot.
**For the Square Plot:**
Side of the square \( a = 10 \text{ m} \)
Perimeter of the square \( P_{square} = 4 \times a = 4 \times 10 \text{ m} = 40 \text{ m} \)
Area of the square \( A_{square} = a \times a = 10 \text{ m} \times 10 \text{ m} = 100 \text{ m}^2 \) Next, calculate the dimensions and area for the rectangular plot.
**For the Rectangular Plot:**
Breadth of the rectangle \( b = 8 \text{ m} \)
The problem states that both plots have the same perimeter.
So, Perimeter of the rectangle \( P_{rectangle} = 40 \text{ m} \) The formula for the perimeter of a rectangle is \( P = 2 \times (l + b) \).
\( 40 = 2 \times (l + 8) \)
Divide both sides by 2:
\( \frac{40}{2} = l + 8 \)
\( 20 = l + 8 \)
Subtract 8 from both sides to find \( l \):
\( l = 20 - 8 = 12 \text{ m} \) Now, calculate the area of the rectangle:
Area of the rectangle \( A_{rectangle} = l \times b = 12 \text{ m} \times 8 \text{ m} = 96 \text{ m}^2 \)
**Comparing Areas:**
Area of the square plot \( = 100 \text{ m}^2 \)
Area of the rectangular plot \( = 96 \text{ m}^2 \) Since \( 100 \text{ m}^2 > 96 \text{ m}^2 \), the square plot has the greater area.
Difference in area \( = 100 \text{ m}^2 - 96 \text{ m}^2 = 4 \text{ m}^2 \) The square plot has a greater area by 4 m². This is a general principle: among all rectangles with the same perimeter, the square has the largest area.
In simple words: First, find the perimeter and area of the square. Since the rectangle has the same perimeter, use that to find its length, then its area. Compare the areas; the square has more area, and you can find how much more.

🎯 Exam Tip: This problem illustrates a key concept: for a fixed perimeter, a square will always enclose the largest area compared to any non-square rectangle. If the question asks to maximize area for a fixed perimeter, think square!

Question 13. Look at the picture of the house given and find the total area of the shaded portion.
Answer:The shaded portion of the house can be divided into two basic shapes: a rectangle at the bottom and a triangle on top.
**Area of the Rectangle (bottom part):**
From the picture, the length of the rectangle is 9 cm and the breadth (height) is 6 cm.
Area of rectangle \( = \text{length} \times \text{breadth} \)
\( = 9 \text{ cm} \times 6 \text{ cm} \)
\( = 54 \text{ cm}^2 \)
**Area of the Triangle (top part):**
From the picture, the base of the triangle is 3 cm and the height is 4 cm.
Area of triangle \( = \frac{1}{2} \times \text{base} \times \text{height} \)
\( = \frac{1}{2} \times 3 \text{ cm} \times 4 \text{ cm} \)
\( = 6 \text{ cm}^2 \)
**Total Area of the Shaded Portion:**
Total Area \( = \text{Area of Rectangle} + \text{Area of Triangle} \)
\( = 54 \text{ cm}^2 + 6 \text{ cm}^2 \)
\( = 60 \text{ cm}^2 \) The total area of the shaded portion is 60 square centimeters. This method of breaking down complex shapes into simpler ones is fundamental in geometry.
In simple words: To find the area of the whole house shape, split it into a rectangle and a triangle. Calculate the area of the rectangle (length times breadth) and the area of the triangle (half times base times height). Then, add these two areas together.

🎯 Exam Tip: For complex shapes, always break them down into simpler, identifiable geometric figures (like rectangles, squares, triangles). Calculate the area of each component and then add them up to find the total area. Ensure correct formulas and units are used.

Question 14. Find the approximate area of the flower in the given square grid.
Answer:To find the approximate area, we count the number of full squares and half squares covered by the flower. We assume each square in the grid has an area of \( 1 \text{ cm}^2 \).
**1. Count Full Squares:**
By carefully looking at the grid, the number of full squares completely inside the flower is 11.
Area of full squares \( = 11 \times 1 \text{ cm}^2 = 11 \text{ cm}^2 \)
**2. Count Half Squares (or more than half):**
Next, count the squares that are partially covered. For estimation, squares that are half or more than half covered are counted as a full square. Squares less than half covered are usually ignored. However, the given solution counts 9 half squares. Let's follow that.
Number of (approximately) half squares \( = 9 \)
Area of half squares \( = 9 \times \frac{1}{2} \text{ cm}^2 = 4.5 \text{ cm}^2 \)
**3. Total Approximate Area:**
Add the area of full squares and half squares:
Total approximate area of the flower \( = 11 \text{ cm}^2 + 4.5 \text{ cm}^2 \)
\( = 15.5 \text{ cm}^2 \) The approximate area of the flower is 15.5 cm². This method is useful for estimating areas of irregular shapes when precise measurements are difficult.
In simple words: To guess the area of the flower, count all the squares that are fully covered. Then, count the squares that are about half covered and add half their area. Add these two amounts together for the total guess.

🎯 Exam Tip: For approximate area calculations using a grid, count full squares accurately. For partial squares, a common rule is to count squares more than half-filled as full squares, and ignore those less than half-filled. Sometimes, as in this solution, half-filled squares are specifically counted as half a unit.