The Story of Geometric Shapes (Class- 6, Experience- 10)
[Exercise- page 204]
E.10.1.
Solution:
If there is a 1-meter wide road within the middle of the garden, we need to subtract the length of the road from both the width and length of the garden to find the effective dimensions of the garden itself.
Given:
Width of the garden = 30 meters
Length of the garden = 40 meters
With a 1-meter wide road, we subtract 1 meter from both the width and the length (since the road is effectively reducing the length and width of the garden by 1 meter on each side).
So, the effective dimensions of the garden become:
Width = 30 meters - 1 meter = 29 meters
Length = 40 meters - 1 meter = 39 meters
Now, to find the perimeter of the garden, we add up the lengths of all four sides:
Perimeter = 2 × (Width + Length)
Perimeter = 2 × (29 + 39)
Perimeter = 2 × 68
Perimeter = 136 meters
So, the perimeter of the garden is 136 meters.
E.10.2. The area of a rectangular field is equal to the area of a square. The length of the rectangular field is 4 times the width. The cost of rope per meter is Tk 7. The cost of the rope to make fence twice around the field is Tk 5600.
a) What will be the perimeter of the rectangular field?
b) If you sow a papaya plant in every 4 m2 area how many papaya plants will be needed?
Solution:
Given:
Length of the rectangular field is 4 times the width.
Cost of rope per meter is Tk 7.
The cost of the rope to make a fence twice around the field is Tk 5600.
Let's denote:
Width of the rectangular field = w meters
Length of the rectangular field = 4w meters
a) Perimeter of the rectangular field:
The perimeter of a rectangle is given by the formula 2 × (length + width).
So, for this rectangular field, the perimeter (P) is:
P = 2 × (4w + w)
P = 2 × (4w + w)
P = 10w
b) Number of papaya plants needed:
The total area of the field is the area of the rectangle, which is 4w × w = 4w2 square meters.
If each papaya plant needs 4 square meters, the number of plants needed is: 4w2 4 = w2
Since the length of the rectangular field is 4 times the width, the total area is 4w × w = 4w2 square meters.
So, you will need w2 papaya plants.
Now, we need to find the value of w. We can use the given information about the cost of the rope to find w.
The cost of the rope to make a fence twice around the field is Tk 5600, and the cost per meter of rope is Tk 7. Using the formula for the perimeter of the rectangular field (P=10w), we can set up the equation:
10w × 7 = 5600
w = 5600 70
w = 80
a) Perimeter of the rectangular field: P = 10w = 10 × 80 = 800 meters
b) Number of papaya plants needed: w2 = 802 = 6400 papaya plants. So, you will need 6400 papaya plants.
E.10.3.
In the diagram, the perimeter of the parallelogram field is 180 meters. The area of the parallelogram can be obtained in more than one way
a) Find the area of the parallelogram in more than one way with logical explanations.
b) Show that the area of the parallelogram field = twice the area of the triangle ABD.
Solution:
a) To find the area of the parallelogram in more than one way, let's use two different approaches:
Approach 1: Using the formula for the area of a parallelogram
Approach 2: Using the fact that the area of a parallelogram is twice the area of a triangle with the same base and height
Let's proceed with each approach:
Approach 1: Using the formula for the area of a parallelogram
The formula for the area of a parallelogram is given by base times height. Here, we can choose either AB or AD as the base, and the corresponding height would be DF.
Area = Base × Height
We'll choose AB as the base:
Area = AB × DF
Area = 2a × 5
Area = 10a square meters
Approach 2: Using the fact that the area of a parallelogram is twice the area of a triangle with the same base and height
Let's find the area of triangle ABD first:
Area of triangle ABD = 12 × Base × Height
Area of triangle ABD = 12 × 2a × 5
Area of triangle ABD = 5a square meters
Now, as per the given property, the area of the parallelogram is twice the area of triangle ABD:
Area of parallelogram = 2 × Area of triangle ABD
Area of parallelogram = 2 × 5a
Area of parallelogram = 10a square meters
So, we can see that both approaches yield the same result, which is the area of the parallelogram = 10a square meters.
b) Show that the area of the parallelogram field = twice the area of the triangle ABD.
We've already shown this in approach 2. The area of the parallelogram is twice the area of triangle ABD, as we calculated the area of the triangle ABD and then doubled it to find the area of the parallelogram.
E.10.4. The length of the floor of a room is 26 meter and the width is 20 meter. How many mats of length 4 meter and width 2.5 meter will cover the whole floor? What will be the total cost if each mat costs Tk 45?
Solution:
First, let's calculate the area of each mat:
Area of each mat = Length × Width
Area of each mat = 4 meters × 2.5 meters
Area of each mat = 10 square meters
Now, let's find the total area of the floor:
Total area of the floor = Length × Width
Total area of the floor = 26 meters × 20 meters
Total area of the floor = 520 square meters
To find out how many mats are required to cover the entire floor, we divide the total area of the floor by the area of each mat:
Number of mats = Total area of the floor Area of each mat
Number of mats = 520 10 square meters
Number of mats = 52
So, 52 mats are required to cover the entire floor.
Now, let's calculate the total cost:
Total cost = Number of mats × Cost per mat
Total cost = 52 mats × Tk 45 per mat
Total cost = Tk 2340
Therefore, the total cost to cover the entire floor with the mats will be Tk 2340.
E.10.5.
In the figure if AB = 100 cm, AC = 120 cm and BD = 80 cm, then CE =?;
Solution:
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[Exercise- page 219]
E.10.1. The length of sides of a cube is 6 cm. Find the total surface area of the object.
Solution:
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E.10.2. The length, breadth and height of a rectangular solid object are 25cm, 20cm and 15 cm respectively. Find the total surface area of this.
Solution:
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E.10.3. You want to give a gift to your friend on his/her birthday. Hence you bought a present. The present is kept in a cubic box of length 12 cm. If you want to wrap the box with wrapping paper, what is the minimum size of coloured paper needed?
Solution:
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E.10.4. The length of the gift box in the following picture is 24 cm, breadth 12 cm and height 8 cm. How much coloured/white paper will be needed to wrap the box?
Solution:
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E.10.5. Cut out some paper according to the measurement shown in the picture and then fold and attach with scotch tape to form a rectangular solid object. What will be the volume of the rectangular solid object?
Solution:
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E.10.6. The following diagram is an open rectangular box. The measurements are given centimeter unit.
a) Find values of a, x, y.
b) Find the volume of the box.
Solution:
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E.10.7. How many small cubic size pieces will be needed to make each of the shapes shown in the picture?
Solution:
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E.10.8 A truck has space of 12 feet x 6 feet x 8 feet, to fill up with cartons to carry. If the size of each carton is 2 feet x 2 feet x 1 foot, how many cartons may be possible for the truck to carry?
Solution:
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E.10.9 A pile of pages was made by putting 200 pieces of papers, like the one in the picture, with one on top of another.
a) What will be the volume of the pile of pages?
b) What is the thickness of one page?
Solution:
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E.10.10 A packet of A4 size papers is seen in the following picture:
Observe what’s written on the packet and according to that, fill up the following Table. You may take help from your teacher if necessary.
Now answer the following questions:
a) What is the weight of 1 page?
b) What is the weight of the whole packet?
c) By measuring what is the height of the packet, can you find the thickness of one page?
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