GSEB Class 6 Maths Solutions Chapter 10 માપન Exercise 10.1

Question 1. Find the perimeter of each of the following figures:
(i) A trapezoid with sides 5 cm, 1 cm, 2 cm, and 4 cm.
(ii) A trapezoid with sides 40 cm, 35 cm, 23 cm, and 35 cm.
(iii) A rhombus with all sides measuring 15 cm.
(iv) A regular pentagon with all sides measuring 4 cm.
(v) An arrow-shaped polygon with sides 1 cm, 4 cm, 0.5 cm, 2.5 cm, 2.5 cm, 0.5 cm, and 4 cm.
(vi) A complex polygon with sides 4 cm, 3 cm, 2 cm, 3 cm, 1 cm, 4 cm, 3 cm, 2 cm, 3 cm, 1 cm, 4 cm, 3 cm, 2 cm, 3 cm, and 1 cm.

Answer:
(i) The perimeter is the total of all side lengths.
Perimeter \( = 5 \text{ cm} + 1 \text{ cm} + 2 \text{ cm} + 4 \text{ cm} = 12 \text{ cm} \).
(ii) The perimeter is the total of all side lengths.
Perimeter \( = 40 \text{ cm} + 35 \text{ cm} + 23 \text{ cm} + 35 \text{ cm} = 133 \text{ cm} \).
(iii) The perimeter is the total of all side lengths.
Perimeter \( = 15 \text{ cm} + 15 \text{ cm} + 15 \text{ cm} + 15 \text{ cm} = 60 \text{ cm} \).
(iv) The perimeter is the total of all side lengths.
Perimeter \( = 4 \text{ cm} + 4 \text{ cm} + 4 \text{ cm} + 4 \text{ cm} + 4 \text{ cm} = 20 \text{ cm} \).
(v) The perimeter is the total of all side lengths.
Perimeter \( = 1 \text{ cm} + 4 \text{ cm} + 0.5 \text{ cm} + 2.5 \text{ cm} + 2.5 \text{ cm} + 0.5 \text{ cm} + 4 \text{ cm} = 15 \text{ cm} \).
(vi) The perimeter is the total of all side lengths.
Perimeter \( = 4 \text{ cm} + 3 \text{ cm} + 2 \text{ cm} + 3 \text{ cm} + 1 \text{ cm} + 4 \text{ cm} + 3 \text{ cm} + 2 \text{ cm} + 3 \text{ cm} + 1 \text{ cm} + 4 \text{ cm} + 3 \text{ cm} + 2 \text{ cm} + 3 \text{ cm} + 1 \text{ cm} = 52 \text{ cm} \).
In simple words: To find the perimeter, you simply add up the lengths of all the outside edges of each shape.

Exam Tip: Always make sure to include all sides when calculating the perimeter of a polygon. Double-check your addition for accuracy.

Question 2. A rectangular box lid has a length of 40 cm and a width of 10 cm. It is taped all around with tape. What length of tape is needed?
Answer: The length of tape needed equals the perimeter of the rectangular lid. The formula for the perimeter of a rectangle is \( 2 \times (\text{length} + \text{width}) \).
Perimeter \( = 2 \times (40 \text{ cm} + 10 \text{ cm}) \)
\( = 2 \times 50 \text{ cm} \)
\( = 100 \text{ cm} \)
Since 100 cm makes 1 meter, the needed tape length is 1 meter.
In simple words: We need to find the total distance around the lid. Add the length and width, then multiply by two. The answer is 100 cm, which is 1 meter.

Exam Tip: Remember that perimeter is the total distance around an object. For rectangles, always use the formula \( 2 \times (L+W) \) and be careful with unit conversions (e.g., cm to m).

Question 3. A tabletop has a length of 2 meters 25 cm and a width of 1 meter 50 cm. What is the perimeter of this surface?
Answer: First, convert all measurements to meters for consistency.
Length \( = 2 \text{ m } 25 \text{ cm} = 2.25 \text{ m} \)
Width \( = 1 \text{ m } 50 \text{ cm} = 1.50 \text{ m} \)
The perimeter of a rectangle is found using the formula \( 2 \times (\text{length} + \text{width}) \).
Perimeter \( = 2 \times (2.25 \text{ m} + 1.50 \text{ m}) \)
\( = 2 \times 3.75 \text{ m} \)
\( = 7.50 \text{ m} \)
So, the perimeter of the tabletop is 7 meters 50 cm.
In simple words: Change everything to meters first. Add the length and width, then multiply by two to get the total distance around the table.

Exam Tip: When working with mixed units (meters and centimeters), it's always best to convert everything to a single unit before performing calculations to avoid errors.

Question 4. To make a frame for a photograph with a length of 32 cm and a width of 21 cm, what length of wooden strip is needed?
Answer: The length of the wooden strip required will be the perimeter of the photograph frame.
Given length of the frame \( = 32 \text{ cm} \)
Given width of the frame \( = 21 \text{ cm} \)
Using the perimeter formula for a rectangle: Perimeter \( = 2 \times (\text{length} + \text{width}) \).
Perimeter \( = 2 \times (32 \text{ cm} + 21 \text{ cm}) \)
\( = 2 \times 53 \text{ cm} \)
\( = 106 \text{ cm} \)
Therefore, 106 cm of wooden strip is needed to make the frame.
In simple words: To know how much wood is needed for the frame, we find the perimeter of the photo. Add its length and width, then multiply by two.

Exam Tip: Frame questions are essentially perimeter questions. Be sure to use the correct formula for the shape (rectangle in this case) and provide the answer in the specified units.

Question 5. A rectangular plot of land has a length of 0.7 km and a width of 0.5 km. How much wire is needed to fence it with four rows of wire on all sides?
Answer: First, calculate the perimeter for one row of fencing.
Given length of the land \( = 0.7 \text{ km} \)
Given width of the land \( = 0.5 \text{ km} \)
Perimeter for one row \( = 2 \times (\text{length} + \text{width}) \)
\( = 2 \times (0.7 \text{ km} + 0.5 \text{ km}) \)
\( = 2 \times 1.2 \text{ km} \)
\( = 2.4 \text{ km} \)
Since the land needs to be fenced with four rows of wire, multiply the perimeter of one row by 4.
Total wire length \( = 4 \times 2.4 \text{ km} = 9.6 \text{ km} \)
Thus, 9.6 km of wire will be needed.
In simple words: First, find the distance around the land for one fence line. Since four fence lines are needed, multiply that distance by four to get the total wire length.

Exam Tip: For multiple rows of fencing, calculate the perimeter for one row first, then multiply by the number of rows. Always include units in your final answer.

Question 6. Find the perimeter of each of the following shapes:
(a) A triangle with side lengths of 3 cm, 4 cm, and 5 cm.
(b) An equilateral triangle with a side length of 9 cm.
(c) An isosceles triangle whose equal sides measure 8 cm each, and the third side measures 6 cm.
Answer:
(a) For a triangle, the perimeter is the sum of its three side lengths.
Perimeter \( = 3 \text{ cm} + 4 \text{ cm} + 5 \text{ cm} = 12 \text{ cm} \).
(b) An equilateral triangle has all three sides of equal length. So, the perimeter is 3 times the side length.
Perimeter \( = 3 \times 9 \text{ cm} = 27 \text{ cm} \).
(c) An isosceles triangle has two equal sides. Here, two sides are 8 cm each, and the third side is 6 cm.
Perimeter \( = 8 \text{ cm} + 8 \text{ cm} + 6 \text{ cm} = 22 \text{ cm} \).
In simple words: To get the perimeter of any triangle, just add up all its side lengths. For special triangles like equilateral (all sides same) or isosceles (two sides same), use the properties to simplify the sum.

Exam Tip: Understand the properties of different types of triangles (equilateral, isosceles, scalene) to efficiently calculate their perimeters.

Question 7. Find the perimeter of a triangle whose sides measure 10 cm, 14 cm, and 15 cm.
Answer: The perimeter of any triangle is calculated by adding the lengths of all its sides.
Given side lengths are 10 cm, 14 cm, and 15 cm.
Perimeter \( = 10 \text{ cm} + 14 \text{ cm} + 15 \text{ cm} \)
\( = 39 \text{ cm} \)
The perimeter of the given triangle is 39 cm.
In simple words: Just add the lengths of the three sides together to find the triangle's perimeter.

Exam Tip: For any polygon, the perimeter is simply the sum of the lengths of all its boundary sides. This is a fundamental concept.

Question 8. Find the perimeter of a regular hexagon whose each side measures 8 meters.
Answer: A regular hexagon has six sides of equal length. To find its perimeter, multiply the length of one side by 6.
Length of each side \( = 8 \text{ meters} \)
Perimeter of a regular hexagon \( = 6 \times \text{side length} \)
\( = 6 \times 8 \text{ meters} \)
\( = 48 \text{ meters} \)
So, the perimeter of the regular hexagon is 48 meters.
In simple words: A hexagon has six equal sides. Multiply the length of one side by six to find its total perimeter.

Exam Tip: For any regular polygon (where all sides are equal), the perimeter is simply the number of sides multiplied by the length of one side.

Question 9. Find the measure of one side of a square with a perimeter of 20 meters.
Answer: A square has four equal sides. The perimeter of a square is 4 times the length of one side. So, to find the side length, divide the perimeter by 4.
Given perimeter of the square \( = 20 \text{ meters} \)
Perimeter \( = 4 \times \text{side length} \)
Side length \( = \frac{\text{Perimeter}}{4} \)
Side length \( = \frac{20 \text{ meters}}{4} \)
Side length \( = 5 \text{ meters} \)
Therefore, one side of the square measures 5 meters.
In simple words: A square has four sides of the same length. If you know the total distance around it (perimeter), just divide that total by four to find how long one side is.

Exam Tip: For squares, remembering the relationship "perimeter = \( 4 \times \) side" is key. When solving for an unknown side, simply rearrange this formula.

Question 10. The perimeter of a regular pentagon is 100 cm. What is the length of each of its sides?
Answer: A regular pentagon has five sides of equal length. To find the length of each side, divide its total perimeter by 5.
Given perimeter of the regular pentagon \( = 100 \text{ cm} \)
Perimeter \( = 5 \times \text{side length} \)
Side length \( = \frac{\text{Perimeter}}{5} \)
Side length \( = \frac{100 \text{ cm}}{5} \)
Side length \( = 20 \text{ cm} \)
Thus, each side of the regular pentagon is 20 cm long.
In simple words: A regular pentagon has five equal sides. If you know its total perimeter, divide it by five to get the length of just one side.

Exam Tip: The term "regular" implies all sides are equal. Always count the number of sides for the specific polygon (penta- means five, hexa- means six, etc.) when finding side length from perimeter.

Question 11. The length of a piece of string is 30 cm. If this string is used to form (a) a square (b) an equilateral triangle (c) a regular hexagon, what will be the length of one side in each figure?
Answer: The total length of the string (30 cm) will represent the perimeter of each shape formed.
(a) If a square is formed, it has 4 equal sides.
Side length of the square \( = \frac{\text{Perimeter}}{4} = \frac{30 \text{ cm}}{4} = 7.5 \text{ cm} \).
(b) If an equilateral triangle is formed, it has 3 equal sides.
Side length of the equilateral triangle \( = \frac{\text{Perimeter}}{3} = \frac{30 \text{ cm}}{3} = 10 \text{ cm} \).
(c) If a regular hexagon is formed, it has 6 equal sides.
Side length of the regular hexagon \( = \frac{\text{Perimeter}}{6} = \frac{30 \text{ cm}}{6} = 5 \text{ cm} \).
In simple words: The string's length is the total distance around any shape you make. Just divide the string's length by the number of equal sides each shape has to find one side's length.

Exam Tip: This question tests your understanding of regular polygons and how perimeter is distributed among equal sides. Always divide the total perimeter by the number of sides.

Question 12. Two sides of a triangle measure 12 cm and 14 cm. If the perimeter of this triangle is 36 cm, what is the measure of its third side?
Answer: The perimeter of a triangle is the sum of the lengths of all its three sides.
Let the three sides be \( s_1, s_2, \) and \( s_3 \).
Given \( s_1 = 12 \text{ cm} \)
Given \( s_2 = 14 \text{ cm} \)
Given Perimeter \( = 36 \text{ cm} \)
Perimeter \( = s_1 + s_2 + s_3 \)
\( 36 \text{ cm} = 12 \text{ cm} + 14 \text{ cm} + s_3 \)
\( 36 \text{ cm} = 26 \text{ cm} + s_3 \)
\( s_3 = 36 \text{ cm} - 26 \text{ cm} \)
\( s_3 = 10 \text{ cm} \)
Thus, the measure of the third side of the triangle is 10 cm.
In simple words: To find the missing side of a triangle, subtract the sum of the two known sides from the total perimeter.

Exam Tip: Always remember that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. This is a good way to check your answer.

Question 13. A square garden has a side length of 250 meters. What will be the cost of fencing it at Rs 20 per meter?
Answer: First, calculate the perimeter of the square garden to find the total length of fencing required.
Side length of the square garden \( = 250 \text{ meters} \)
Perimeter of a square \( = 4 \times \text{side length} \)
Perimeter \( = 4 \times 250 \text{ meters} \)
Perimeter \( = 1000 \text{ meters} \)
Cost of fencing per meter \( = \text{Rs } 20 \)
Total cost of fencing \( = \text{Perimeter} \times \text{cost per meter} \)
Total cost \( = 1000 \text{ meters} \times \text{Rs } 20/\text{meter} \)
Total cost \( = \text{Rs } 20,000 \)
Therefore, the total cost of fencing the square garden will be Rs 20,000.
In simple words: First, find the total length around the garden by multiplying its side by four. Then, multiply this total length by the cost per meter to find the total fencing cost.

Exam Tip: Remember to calculate the total length (perimeter) first, then multiply by the given rate to find the total cost. Always include the currency symbol in your final answer for cost-related problems.

Question 14. A rectangular garden has a length of 175 meters and a width of 125 meters. What will be the cost of fencing it at Rs 12 per meter?
Answer: First, calculate the perimeter of the rectangular garden to find the total length of fencing required.
Length of the rectangular garden \( = 175 \text{ meters} \)
Width of the rectangular garden \( = 125 \text{ meters} \)
Perimeter of a rectangle \( = 2 \times (\text{length} + \text{width}) \)
Perimeter \( = 2 \times (175 \text{ meters} + 125 \text{ meters}) \)
Perimeter \( = 2 \times 300 \text{ meters} \)
Perimeter \( = 600 \text{ meters} \)
Cost of fencing per meter \( = \text{Rs } 12 \)
Total cost of fencing \( = \text{Perimeter} \times \text{cost per meter} \)
Total cost \( = 600 \text{ meters} \times \text{Rs } 12/\text{meter} \)
Total cost \( = \text{Rs } 7200 \)
Therefore, the total cost of fencing the rectangular garden will be Rs 7200.
In simple words: Find the total distance around the rectangular garden by adding its length and width, then multiplying by two. Then, multiply this total distance by the given cost for each meter to get the final fencing cost.

Exam Tip: Ensure you use the correct perimeter formula for a rectangle (\( 2(L+W) \)) and pay attention to unit consistency and correct multiplication for the total cost.

Question 15. Sweetie runs around a square garden with a side of 75 meters. Bulbul runs around a rectangular garden with a length of 60 meters and a width of 45 meters. Who runs a shorter distance?
Answer: First, calculate the distance covered by Sweetie.
Sweetie runs around a square garden.
Side of the square garden \( = 75 \text{ meters} \)
Distance covered by Sweetie \( = \text{Perimeter of square} = 4 \times \text{side} \)
Distance covered by Sweetie \( = 4 \times 75 \text{ meters} = 300 \text{ meters} \).
Next, calculate the distance covered by Bulbul.
Bulbul runs around a rectangular garden.
Length of the rectangular garden \( = 60 \text{ meters} \)
Width of the rectangular garden \( = 45 \text{ meters} \)
Distance covered by Bulbul \( = \text{Perimeter of rectangle} = 2 \times (\text{length} + \text{width}) \)
Distance covered by Bulbul \( = 2 \times (60 \text{ meters} + 45 \text{ meters}) \)
Distance covered by Bulbul \( = 2 \times 105 \text{ meters} = 210 \text{ meters} \).
Comparing the distances: 210 meters (Bulbul) \( < \) 300 meters (Sweetie).
Therefore, Bulbul runs a shorter distance.
In simple words: Calculate the total distance Sweetie runs around her square garden. Then, calculate the total distance Bulbul runs around her rectangular garden. Compare the two numbers to see who ran less.

Exam Tip: This question requires two separate perimeter calculations followed by a comparison. Clearly state the perimeter for each person before making the final comparison.

Question 16. What is the perimeter of each of the following figures? What do you infer from your answers?
(a) A square with a side of 25 cm.
(b) A rectangle with a length of 40 cm and a width of 10 cm.
(c) A rectangle with a length of 30 cm and a width of 20 cm.
(d) A triangle with sides of 40 cm, 30 cm, and 30 cm.
Answer:
(a) For the square: Perimeter \( = 4 \times \text{side length} = 4 \times 25 \text{ cm} = 100 \text{ cm} \).
(b) For the rectangle: Perimeter \( = 2 \times (\text{length} + \text{width}) = 2 \times (40 \text{ cm} + 10 \text{ cm}) = 2 \times 50 \text{ cm} = 100 \text{ cm} \).
(c) For the rectangle: Perimeter \( = 2 \times (\text{length} + \text{width}) = 2 \times (30 \text{ cm} + 20 \text{ cm}) = 2 \times 50 \text{ cm} = 100 \text{ cm} \).
(d) For the triangle: Perimeter \( = \text{sum of all sides} = 40 \text{ cm} + 30 \text{ cm} + 30 \text{ cm} = 100 \text{ cm} \).
Inference: From the calculations, we can observe that the perimeter of all four figures is the same, which is 100 cm. This shows that different shapes can have the same perimeter.
In simple words: Calculate the distance around each shape. You'll notice that even though the shapes look different, the total distance around them is the same. This means shapes with different forms can have the same perimeter.

Exam Tip: This question highlights that perimeter is a measure of total boundary length, and distinct geometric shapes can share the same perimeter. Always perform calculations for each figure before drawing conclusions.

Question 17. Avaneet buys nine square tiles, each with a side length of \( \frac{1}{2} \) meter. She arranges these pieces in a square shape.
(a) What is the perimeter of the arrangement in figure (i)?
(b) Shaarin does not like the arrangement in figure (ii). She arranges the tiles in a cross shape. What is the perimeter of her arrangement?
(c) Which arrangement has a greater perimeter?
(d) Avaneet wonders if any arrangement is possible that yields a greater perimeter? Can you find such a way? (Tiles' sides must meet perfectly, i.e., tiles cannot be broken.)
Answer:
(a) Avaneet arranges 9 square tiles in a square shape (3x3 grid).
Each tile has a side length of \( \frac{1}{2} \) meter.
The side length of the large square formed by the 3x3 arrangement will be \( 3 \times \frac{1}{2} \text{ meter} = \frac{3}{2} \text{ meters} \).
The perimeter of this square arrangement \( = 4 \times \text{side length} \)
\( = 4 \times \frac{3}{2} \text{ meters} \)
\( = 2 \times 3 \text{ meters} \)
\( = 6 \text{ meters} \).
(b) Shaarin arranges the 9 tiles in a cross shape as shown in figure (ii).
In this cross arrangement, there are 16 exposed segments, each with a length of \( \frac{1}{2} \) meter.
Perimeter of the cross arrangement \( = 16 \times \frac{1}{2} \text{ meters} \)
\( = 8 \text{ meters} \).
(c) Comparing the perimeters: Arrangement (a) has a perimeter of 6 meters, and arrangement (b) has a perimeter of 8 meters.
Since \( 8 \text{ meters} > 6 \text{ meters} \), the cross arrangement (b) has a greater perimeter.
(d) Yes, it is possible to achieve an even greater perimeter. To maximize the perimeter for a fixed number of tiles, arrange them in a single long row.
If all 9 tiles are arranged in a single straight row, it forms a rectangle.
The length of this rectangle would be \( 9 \times \frac{1}{2} \text{ meter} = \frac{9}{2} \text{ meters} \).
The width of this rectangle would be \( \frac{1}{2} \text{ meter} \).
Perimeter of this new arrangement \( = 2 \times (\text{length} + \text{width}) \)
\( = 2 \times (\frac{9}{2} \text{ meters} + \frac{1}{2} \text{ meters}) \)
\( = 2 \times (\frac{10}{2} \text{ meters}) \)
\( = 2 \times 5 \text{ meters} \)
\( = 10 \text{ meters} \).
This arrangement gives a perimeter of 10 meters, which is greater than both the square (6m) and the cross (8m) arrangements.
In simple words: For part (a), arranging 9 tiles in a 3x3 square means the large square's side is \( 3 \times \frac{1}{2} \) meter, so multiply that by 4 for the perimeter. For part (b), count all the exposed edges of the cross shape, each being \( \frac{1}{2} \) meter, and add them up. For part (c), compare the two perimeters. For part (d), to get the largest perimeter, arrange all the tiles in one long line. Calculate its new length and width, then find its perimeter.

Exam Tip: This question illustrates that for a fixed area (number of tiles), different arrangements can lead to different perimeters. Long, thin arrangements generally have larger perimeters than compact, square-like ones. Carefully count all exposed sides for perimeter calculation.