GSEB Class 6 Maths Solutions Chapter 10 Mensuration Exercise 10.1

Gujarat Board Textbook Solutions Class 6 Maths Chapter 10 Mensuration Ex 10.1

Question 1. Find the perimeter of each of the following figures:
(a) Figure (a)
(b) Figure (b)
(c) Figure (c)
(d) Figure (d)
(e) Figure (e)
(f) Figure (f)
Answer:
(a) Perimeter of the given figure \( = \text{sum of the sides} = 4 \text{ cm} + 5 \text{ cm} + 1 \text{ cm} + 2 \text{ cm} = 12 \text{ cm} \)
(b) Perimeter of the given figure \( = \text{sum of the sides} = 23 \text{ cm} + 35 \text{ cm} + 40 \text{ cm} + 35 \text{ cm} = 133 \text{ cm} \)
(c) Perimeter of the given figure \( = \text{sum of the sides} = 15 \text{ cm} + 15 \text{ cm} + 15 \text{ cm} + 15 \text{ cm} = 60 \text{ cm} \)
(d) Perimeter of the given figure \( = \text{sum of the sides} = 4 \text{ cm} + 4 \text{ cm} + 4 \text{ cm} + 4 \text{ cm} + 4 \text{ cm} = 20 \text{ cm} \)
(e) Perimeter of the given figure \( = \text{sum of the sides} = 1 \text{ cm} + 4 \text{ cm} + 0.5 \text{ cm} + 2.5 \text{ cm} + 2.5 \text{ cm} + 0.5 \text{ cm} + 4 \text{ cm} = 15 \text{ cm} \)
(f) Perimeter of the given figure \( = \text{sum of the sides} = 4 \text{ cm} + 3 \text{ cm} + 2 \text{ cm} + 3 \text{ cm} + 1 \text{ cm} + 4 \text{ cm} + 3 \text{ cm} + 2 \text{ cm} + 3 \text{ cm} + 1 \text{ cm} + 4 \text{ cm} + 3 \text{ cm} + 2 \text{ cm} + 3 \text{ cm} + 1 \text{ cm} + 4 \text{ cm} + 3 \text{ cm} + 2 \text{ cm} + 3 \text{ cm} + 1 \text{ cm} = 52 \text{ cm} \)
In simple words: The perimeter of any figure is the total length of its outer boundary. Just add up the lengths of all the sides to find it.

Exam Tip: Carefully identify all sides of a figure, especially complex ones, and ensure each side is included in the sum to avoid errors in perimeter calculation.

Question 2. The lid of a rectangular box of sides 40 cm by 10 cm is sealed all round with tape. What is the length of the tape required?
Answer: The length of the tape required is the perimeter of the rectangular box lid.
Length of the tape required \( = \text{perimeter of the lid of a rectangular box} \)
\( = 2(\text{length} + \text{breadth}) = 2 \times (40 \text{ cm} + 10 \text{ cm}) \)
\( = 2 \times 50 \text{ cm} = 100 \text{ cm or } 1 \text{ m} \)
In simple words: The tape needed is equal to the perimeter of the box lid. You get this by adding the length and width, then multiplying by two. The result is 100 cm, or 1 meter.

Exam Tip: Remember that "sealed all round" implies finding the perimeter. Convert all units to be consistent before calculation, or state units clearly in your final answer.

Question 3. A table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table-top?
Answer:
Length of the table-top \( = 2 \text{ m } 25 \text{ cm} = 225 \text{ cm} \)
Breadth of the table-top \( = 1 \text{ m } 50 \text{ cm} = 150 \text{ cm} \)
Perimeter of the table-top \( = 2(\text{length} + \text{breadth}) \)
\( = 2(225 \text{ cm} + 150 \text{ cm}) \)
\( = 2(375 \text{ cm}) \)
\( = 750 \text{ cm or } 7.5 \text{ m} \)
In simple words: First, convert all measurements to the same unit, like centimeters. Then, add the length and width and multiply the sum by two to find the total distance around the table.

Exam Tip: Always convert mixed units (like meters and centimeters) into a single unit before performing calculations to prevent errors.

Question 4. What is the length of the wooden strip required to frame a photograph of length and breadth 32 cm and 21 cm respectively?
Answer:
Length of the frame \( = 32 \text{ cm} \)
Breadth of the frame \( = 21 \text{ cm} \)
Perimeter of the frame \( = 2 [\text{length} + \text{breadth}] \)
\( = 2(32 \text{ cm} + 21 \text{ cm}) \)
\( = 2(53 \text{ cm}) = 106 \text{ cm} \)
Length of the wooden strip required \( = 106 \text{ cm} \)
In simple words: The total length of the wooden strip needed is the perimeter of the photograph frame. Add the length and breadth, then multiply by two to get the answer.

Exam Tip: Framing a picture involves covering its outer boundary, which means you need to find the perimeter. Ensure to correctly apply the perimeter formula for a rectangle.

Question 5. A rectangular piece of land measures 0.7 km by 0.5 km. Each side is to be fenced with 4 rows of wires. What is the length of the wire needed?
Answer:
Length of the land \( = 0.7 \text{ km} \)
Breadth of the land \( = 0.5 \text{ km} \)
Perimeter of the rectangular piece of land \( = 2 (\text{length} + \text{breadth}) \)
\( = 2 (0.7 \text{ km} + 0.5 \text{ km}) \)
\( = 2 (1.2 \text{ km}) = 2.4 \text{ km} \)
Length of wire fencing for 1 row \( = 2.4 \text{ km} \)
Length of wire fencing for 4 rows \( = 4 \times 2.4 \text{ km} = 9.6 \text{ km} \)
In simple words: First, find the perimeter of the land by adding length and breadth then multiplying by two. Since there are four rows of wire, multiply this perimeter by four to get the total wire length.

Exam Tip: For problems involving multiple rows of fencing, always calculate the perimeter for one row first, then multiply by the number of rows to find the total material needed.

Question 6. Find the perimeter of each of the following shapes:
(a) A triangle of sides 3 cm, 4 cm, and 5 cm.
(b) An equilateral triangle of side 9 cm.
(c) An isosceles triangle with equal sides 8 cm each and third side 6 cm.
Answer:
(a) Sides of the triangle are: \( 3 \text{ cm}, 4 \text{ cm, and } 5 \text{ cm} \)
Perimeter \( = \text{sum of the sides} = 3 \text{ cm} + 4 \text{ cm} + 5 \text{ cm} = 12 \text{ cm} \)
(b) Length of each side \( = 9 \text{ cm} \)
The perimeter of an equilateral triangle \( = 3 \times \text{length of one side} \)
The perimeter of the given equilateral triangle \( = 3 \times 9 \text{ cm} = 27 \text{ cm} \)
(c) Sides of the given triangle are: \( 8 \text{ cm}, 6 \text{ cm and } 8 \text{ cm} \)
Perimeter of the triangle \( = \text{sum of the sides} = 8 \text{ cm} + 6 \text{ cm} + 8 \text{ cm} = 22 \text{ cm} \)
In simple words: To find the perimeter of any triangle, just add the lengths of all three of its sides. For an equilateral triangle, multiply the side length by three.

Exam Tip: Remember specific properties of triangles: an equilateral triangle has three equal sides, and an isosceles triangle has two equal sides. Use these properties to simplify perimeter calculations.

Question 7. Find the perimeter of a triangle with sides measuring 10 cm, 14 cm, and 15 cm.
Answer:
Sides of the triangle are: \( 10 \text{ cm}, 14 \text{ cm and } 15 \text{ cm} \)
Perimeter \( = \text{sum of the sides} = 10 \text{ cm} + 14 \text{ cm} + 15 \text{ cm} = 39 \text{ cm} \)
In simple words: To get the perimeter of any triangle, simply add up the measurements of all three of its sides.

Exam Tip: Always ensure you add all three side lengths of a triangle to find its perimeter. A common mistake is to miss one side.

Question 8. Find the perimeter of a regular hexagon with each side measuring 8 m.
Answer:
The perimeter of a regular hexagon \( = 6 \times \text{one side of the hexagon} \)
Perimeter of the given regular hexagon \( = 6 \times 8 \text{ m} = 48 \text{ m} \)
In simple words: A regular hexagon has six sides that are all the same length. So, to find its perimeter, just multiply the length of one side by six.

Exam Tip: For any regular polygon, the perimeter is simply the number of sides multiplied by the length of one side. Know the number of sides for common polygons like hexagons (6 sides).

Question 9. Find the side of the square whose perimeter is 20 m.
Answer:
The perimeter of the square \( = 20 \text{ m} \).
Perimeter of square \( = 4 \times \text{length of one side} \)
\( 4 \times \text{Length of a side} = 20 \text{ m} \)
Length of a side \( = \frac{20 \text{ m}}{4} = 5 \text{ m} \)
In simple words: Since a square has four equal sides, to find the length of one side, divide the total perimeter by four.

Exam Tip: Remember the relationship between the perimeter of a square and its side length: Perimeter = 4 × side. You can use this to find either value if the other is known.

Question 10. The perimeter of a regular pentagon is 100 cm. How long is each side?
Answer:
Perimeter of a regular pentagon \( = 100 \text{ cm} \)
Perimeter \( = 5 \times \text{Length of one side} \)
\( 5 \times \text{Length of one side} = 100 \text{ cm} \)
Length of one side \( = \frac{100 \text{ cm}}{5} = 20 \text{ cm} \)
In simple words: A regular pentagon has five equal sides. To find the length of one side, divide the total perimeter by five.

Exam Tip: For any regular polygon, you can find the length of one side by dividing its perimeter by the number of sides it has.

Question 11. A piece of string is 30 cm long. What will be the length of each side of the string is used to form:
(a) a square?
(b) an equilateral triangle?
(c) a regular hexagon?
Answer:
Length of the given string \( = 30 \text{ cm} \)
(a) The string is used to form a square. The perimeter of a square \( = 4 \times \text{Side} \)
Side \( = \frac{30 \text{ cm}}{4} = 7.5 \text{ cm} \)
The length of each side \( = 7.5 \text{ cm} \)
(b) The string is in the form of an equilateral triangle. The perimeter \( = 3 \times \text{length of one side} \)
\( 3 \times \text{length of one side} = 30 \text{ cm} \)
Length of one side \( = \frac{30 \text{ cm}}{3} = 10 \text{ cm} \)
(c) The string is in the form of a regular hexagon. The perimeter of a regular hexagon \( = 6 \times \text{length of one side} \)
\( 6 \times \text{Length of one side} = 30 \text{ cm} \)
Length of one side \( = \frac{30 \text{ cm}}{6} = 5 \text{ cm} \)
In simple words: To find the length of each side, divide the total string length (the perimeter) by the number of sides of the shape being formed.

Exam Tip: For problems involving reshaping a fixed length of material, the length of the material represents the perimeter of the new shape. Divide by the number of sides to find each side's length.

Question 12. Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is its third side?
Answer:
Perimeter of the triangle \( = \text{sum of its sides} \)
\( 36 \text{ cm} = 12 \text{ cm} + 14 \text{ cm} + \text{third side} \)
\( 36 \text{ cm} = 26 \text{ cm} + \text{third side} \)
Third side \( = 36 \text{ cm} - 26 \text{ cm} \)
Third side \( = 10 \text{ cm} \)
Hence, the length of the third side of the triangle \( = 10 \text{ cm} \)
In simple words: Add the lengths of the two sides you know. Then, subtract that sum from the total perimeter to find the length of the missing third side.

Exam Tip: The perimeter of any polygon is the sum of all its sides. If you know the total perimeter and all but one side, you can find the missing side by subtracting the known sides from the total.

Question 13. Find the cost of fencing a square park of side 250 m at the rate of Rs. 20 per meter.
Answer:
Side of the park \( = 250 \text{ m} \)
The park is in the form of a square.
Perimeter \( = 4 \times \text{length of one side} = 4 \times 250 \text{ m} = 1000 \text{ m} \)
Rate of fencing \( = \text{Rs. } 20 \text{ per meter} \)
Cost of fencing \( = \text{Rs. } 20 \times 1000 = \text{Rs. } 20,000 \)
In simple words: First, calculate the total length around the park (its perimeter). Since it's a square, multiply the side length by four. Then, multiply this total length by the cost for each meter to get the final fencing price.

Exam Tip: Remember that "cost of fencing" problems require two steps: first calculate the perimeter of the area to be fenced, and then multiply by the given rate per unit length.

Question 14. Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of Rs. 12 per meter.
Answer:
Length of the park \( = 175 \text{ m} \)
The breadth of the park \( = 125 \text{ m} \)
Park is in the form of a rectangle.
Perimeter of a rectangle \( = 2 \times [\text{length} + \text{breadth}] \)
Perimeter of the given park \( = 2 \times [175 \text{ m} + 125 \text{ m}] \)
\( = 2 \times 300 \text{ m} = 600 \text{ m} \)
Rate of fencing \( = \text{Rs. } 12 \text{ per metre} \)
Cost of fencing \( = \text{Rs. } 12 \times 600 = \text{Rs. } 7200 \)
In simple words: Calculate the park's total boundary length by adding its length and breadth, then doubling that sum. Finally, multiply this total length by the given cost per meter to find the overall fencing expense.

Exam Tip: For rectangular shapes, remember the perimeter formula \( P = 2(l+b) \). Always ensure you correctly substitute the length and breadth and then perform the multiplication and addition steps accurately.

Question 15. Sweety runs around a square park of side 75 m. Bulbul runs around a rectangular park of 60 m and a breadth of 45 m. Who covers less distance?
Answer:
Side of the square park \( = 75 \text{ m} \)
Perimeter of the square park \( = 4 \times \text{side} = 4 \times 75 \text{ m} = 300 \text{ m} \)
Distance covered by Sweety \( = 300 \text{ m} \)
Length of the rectangular park \( = 60 \text{ m} \)
Breadth of the rectangular park \( = 45 \text{ m} \)
Perimeter of the rectangular park \( = 2(\text{length} + \text{breadth}) = 2(60 \text{ m} + 45 \text{ m}) \)
\( = 2(105 \text{ m}) = 210 \text{ m} \)
Distance covered by Bulbul \( = 210 \text{ m} \)
Since \( 210 < 300 \),
Thus, Bulbul covers less distance.
In simple words: First, calculate how far Sweety runs by finding the square park's perimeter. Next, figure out Bulbul's distance by calculating the rectangular park's perimeter. Then, compare these two distances to see who covered less ground.

Exam Tip: When comparing distances covered around different shapes, always calculate the perimeter for each shape first before making any comparisons.

Question 16. What is the perimeter of each of the following figures? What do you observe about the perimeters?
(a) Figure (a)
(b) Figure (b)
(c) Figure (c)
(d) Figure (d)
Answer:
(a) Length of a side of the square \( = 25 \text{ cm} \)
Perimeter \( = 4 \times \text{side} = 4 \times 25 \text{ cm} = 100 \text{ cm} \)
(b) Length of the rectangle \( = 40 \text{ cm} \)
Breadth of the rectangle \( = 10 \text{ cm} \)
Perimeter \( = 2(\text{length} + \text{breadth}) = 2(40 \text{ cm} + 10 \text{ cm}) \)
\( = 2 \times 50 \text{ cm} = 100 \text{ cm} \)
(c) Length of the rectangle \( = 30 \text{ cm} \)
Breadth of the rectangle \( = 20 \text{ cm} \)
Perimeter \( = 2 (\text{length} + \text{breadth}) = 2(30 \text{ cm} + 20 \text{ cm}) \)
\( = 2 \times 50 \text{ cm} = 100 \text{ cm} \)
(d) Sides of the given triangle are: \( 30 \text{ cm}, 40 \text{ cm, and } 30 \text{ cm} \)
Perimeter of the triangle \( = \text{sum of the sides} = 30 \text{ cm} + 40 \text{ cm} + 30 \text{ cm} = 100 \text{ cm} \)
Here, we observe that the perimeter of each figure is \( 100 \text{ cm} \), so they all have an equal perimeter.
In simple words: We calculated the perimeter for each shape. It's interesting to see that even though the shapes look different, they all have the same perimeter of 100 cm.

Exam Tip: Sometimes different shapes can have the same perimeter. Always calculate each perimeter individually to confirm your observation rather than making assumptions based on visual appearance.

Question 17. Avneet buys 9 square paving slabs, each with a side of \( \frac{1}{2} \) m. He lays them in the form of a square.
(a) What is the perimeter of his arrangement [Fig. (i)]?
(b) Shari does not like his arrangement. She gets him to lay them out like a cross. What is the perimeter of her arrangement [(Fig.(ii)]?
(c) Which has a greater perimeter?
(d) Avneet wonders if there is a way of getting an even greater perimeter. Can you find a way of doing this? (The paving slabs must meet along complete edges i.e. they cannot be broken.)
Answer:
(a) The arrangement is in the form of a square.
The side of the square arrangement \( = \frac{1}{2} \text{ m} + \frac{1}{2} \text{ m} + \frac{1}{2} \text{ m} = \frac{3}{2} \text{ m} \) or \( 1\frac{1}{2} \text{ m} \)
The perimeter of the square arrangement \( = 4 \times \text{side} = 4 \times \frac{3}{2} \text{ m} = 6 \text{ m} \)

(b) Perimeter of the cross-arrangement (Figure (ii)):
Each square slab has a side of \( \frac{1}{2} \) m. From Figure (ii), we count 12 exposed outer edges.
Perimeter of the cross-arrangement \( = 12 \times \frac{1}{2} \text{ m} = 6 \text{ m} \)

(c) Avneet's square arrangement has a perimeter of \( 6 \text{ m} \). Shari's cross-arrangement also has a perimeter of \( 6 \text{ m} \).
Therefore, neither arrangement has a greater perimeter; they are equal.
(d) To achieve an even greater perimeter, the slabs can be arranged in a long, narrow strip.
For example, arranging all 9 tiles in a single row (a 1x9 rectangle):
Total number of tiles \( = 9 \)
This arrangement forms a rectangle with length \( = 9 \times \frac{1}{2} \text{ m} = \frac{9}{2} \text{ m} \) and breadth \( = \frac{1}{2} \text{ m} \).
Perimeter \( = 2 \times (\text{length} + \text{breadth}) = 2 \times (\frac{9}{2} \text{ m} + \frac{1}{2} \text{ m}) \)
\( = 2 \times (\frac{10}{2} \text{ m}) = 2 \times 5 \text{ m} = 10 \text{ m} \)
This perimeter of \( 10 \text{ m} \) is greater than the \( 6 \text{ m} \) perimeter of the previous arrangements.
In simple words: For the square, we added the side lengths. For the cross, we counted the outer edges and multiplied by the slab length. Both were 6 meters. To get a bigger perimeter, you can arrange the slabs in a long line, making a very thin rectangle, which exposes more outer edges.

Exam Tip: To maximize the perimeter of a fixed area made of smaller units, arrange them in a long, thin configuration to expose as many outer edges as possible. To minimize perimeter, arrange them into a compact, square-like shape.