Samacheer Kalvi Class 8 Maths Solutions Chapter 2 Measurements Exercise 2.4

Question 1. Two gates are fitted at the entrance of a library. To open the gates easily, a wheel is fixed at 6 feet distance from the wall to which the gate is fixed. If one of the gates is opened to 90°, find the distance moved by the wheel (\( \pi \) = 3.14).
Answer: Let point A be where the wall is, and AC be the gate in its starting position. AB is the gate after it has opened. The wheel moves along an arc from B to C.
The distance of the wheel from the wall is 6 feet, which is the radius of the circular path it follows.
The angle the gate opens is 90°, so the angle of the arc is \( \theta = 90^\circ \).
The radius of the arc is \( r = 6 \) feet.
We need to find the length of the arc BC.
The formula for arc length is:
\[ \text{Arc Length} = \frac{\theta}{360^\circ} \times 2\pi r \] Substitute the given values:
\[ \text{Arc Length} = \frac{90^\circ}{360^\circ} \times 2 \times 3.14 \times 6 \] Simplify the fraction:
\[ \text{Arc Length} = \frac{1}{4} \times 2 \times 3.14 \times 6 \] Multiply the numbers:
\[ \text{Arc Length} = 0.5 \times 3.14 \times 6 \] \[ \text{Arc Length} = 3.14 \times 3 \] \[ \text{Arc Length} = 9.42 \text{ feet} \] So, the wheel moves a distance of 9.42 feet.
In simple words: The wheel moves in a curved path like a part of a circle. We use a formula to find the length of this curved path, using how wide the gate opens and how far the wheel is from the wall.

🎯 Exam Tip: Remember that the distance of the wheel from the wall acts as the radius of the sector, and the angle the gate opens is the angle of the sector.

 

Question 2. With his usual speed, if a person covers a circular track of radius 150 m in 9 minutes, find the distance that he covers in 3 minutes (\( \pi \) = 3.14).
Answer: The radius of the circular track is \( r = 150 \) m.
The person covers the track in 9 minutes. This means in 9 minutes, he covers the perimeter (or circumference) of the circle.
The formula for the perimeter of a circle is:
\[ \text{Perimeter} = 2 \times \pi \times r \] Substitute the given values for \( \pi \) and \( r \):
\[ \text{Perimeter} = 2 \times 3.14 \times 150 \] \[ \text{Perimeter} = 6.28 \times 150 \] \[ \text{Perimeter} = 942 \text{ m} \] So, the distance covered in 9 minutes is 942 m.
Now, we need to find the distance covered in 1 minute. Divide the total distance by the time:
\[ \text{Distance in 1 minute} = \frac{\text{Distance in 9 minutes}}{9} \] \[ \text{Distance in 1 minute} = \frac{942}{9} \text{ m} \] To find the distance covered in 3 minutes, multiply the distance covered in 1 minute by 3:
\[ \text{Distance in 3 minutes} = \frac{942}{9} \times 3 \] Simplify the fraction (9 divided by 3 is 3):
\[ \text{Distance in 3 minutes} = \frac{942}{3} \] \[ \text{Distance in 3 minutes} = 314 \text{ m} \] Therefore, the person covers 314 m in 3 minutes.
In simple words: First, find the total distance around the circular track. This is the distance covered in 9 minutes. Then, divide this by 9 to find how far the person walks in one minute. Finally, multiply that by 3 to get the distance for 3 minutes.

🎯 Exam Tip: When dealing with circular tracks, remember that "covering the track" means completing one full perimeter or circumference. This is a common phrase in these types of problems.

 

Question 3. Find the area of the house drawing given in the figure.
Answer: To find the total area of the house drawing, we need to break it down into simpler shapes: a square, a rectangle, a triangle, and a parallelogram.
Let's find the area of each part:
1. **Square:** side = 6 cm
Area of square = \( \text{side} \times \text{side} = 6 \times 6 = 36 \text{ cm}^2 \)
2. **Rectangle:** length = 8 cm, breadth = 6 cm
Area of rectangle = \( \text{length} \times \text{breadth} = 8 \times 6 = 48 \text{ cm}^2 \)
3. **Triangle:** base = 6 cm, height = 4 cm
Area of triangle = \( \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 4 = 12 \text{ cm}^2 \)
4. **Parallelogram:** base = 8 cm, height = 4 cm
Area of parallelogram = \( \text{base} \times \text{height} = 8 \times 4 = 32 \text{ cm}^2 \)
Now, add the areas of all these shapes together to get the total area of the house drawing:
\[ \text{Total Area} = \text{Area of square} + \text{Area of rectangle} + \text{Area of triangle} + \text{Area of parallelogram} \] \[ \text{Total Area} = 36 + 48 + 12 + 32 \] \[ \text{Total Area} = 128 \text{ cm}^2 \] So, the required area of the house drawing is 128 cm\( ^2 \). Calculating areas of composite figures involves careful observation of shapes.
In simple words: Break the picture of the house into squares, rectangles, triangles, and parallelograms. Find the area of each small shape. Then, add all these areas together to get the total area of the house.

🎯 Exam Tip: Always draw lines to clearly separate the composite figure into basic geometric shapes like squares, rectangles, and triangles. Make sure to identify the correct dimensions (base, height, side) for each part.

 

Question 4. Draw the top, front and side view of the following solid shapes
(i)
Answer: For the first solid shape, we need to imagine looking at it from different directions.
(a) Top view: When you look down from the top, you see a 2x2 square arrangement, with the top-left cell empty.

(b) Front view: When you look from the front, you see a shape that looks like an 'L' turned sideways, covering 2 units in height and 2 units in width.

(c) Side view: When you look from the side (right side), you see a vertical column of two blocks.

(ii)
Answer: For the second solid shape, let's look at its views.
(a) Top view: Looking from the top, you see a 2x2 square arrangement with the top-right cell empty.

(b) Front view: Looking from the front, you see three blocks stacked in an 'L' shape.

(c) Side view: Looking from the side (right side), you see two blocks, one above the other, slightly offset.

In simple words: For each shape, draw what you would see if you looked straight down from the top, straight from the front, and straight from the side. Each view shows only the 2D outline of the object from that angle.

🎯 Exam Tip: When drawing views of 3D shapes, carefully count the number of blocks visible from each angle and represent them on a grid. Imagine shining a light from that direction and tracing the shadow.

 

Question 5. Guna fixed a single door of width 3 feet in his room where as Nathan has fixed a double door, each of width \( 1 \frac { 1 }{ 2 } \) feet in his room. From the closed position, if each of the single and double doors can open up to 120°, whose door takes a minimum area?
Answer: We need to compare the area swept by Guna's single door and Nathan's double doors.

**For Guna's single door:**
The width of Guna's door is 3 feet. This width acts as the radius of the sector when the door opens. So, \( r = 3 \) feet.
The angle the door opens is \( \theta = 120^\circ \).
The area required to open the door (Area of a sector) is given by the formula:
\[ \text{Area} = \frac{\theta}{360^\circ} \times \pi r^2 \] Substitute the values:
\[ \text{Area} = \frac{120^\circ}{360^\circ} \times \pi \times (3)^2 \] \[ \text{Area} = \frac{1}{3} \times \pi \times 9 \] \[ \text{Area} = 3\pi \text{ square feet} \]
**For Nathan's double doors:**
Nathan has double doors, and each door has a width of \( 1 \frac { 1 }{ 2 } \) feet.
\[ 1 \frac { 1 }{ 2 } = \frac{3}{2} = 1.5 \text{ feet} \] So, the radius for each door is \( r = 1.5 \) feet.
Each door can open up to \( \theta = 120^\circ \).
Since there are two doors, we need to calculate the area for one door and then multiply by 2.
Area for one door:
\[ \text{Area (one door)} = \frac{\theta}{360^\circ} \times \pi r^2 \] Substitute the values:
\[ \text{Area (one door)} = \frac{120^\circ}{360^\circ} \times \pi \times (1.5)^2 \] \[ \text{Area (one door)} = \frac{1}{3} \times \pi \times ( \frac{3}{2} )^2 \] \[ \text{Area (one door)} = \frac{1}{3} \times \pi \times \frac{9}{4} \] \[ \text{Area (one door)} = \frac{3\pi}{4} \text{ square feet} \] Now, for two doors, the total area is:
\[ \text{Total Area (double doors)} = 2 \times \frac{3\pi}{4} \] \[ \text{Total Area (double doors)} = \frac{3\pi}{2} \text{ square feet} \]
**Comparing the areas:**
Guna's door area = \( 3\pi \) square feet
Nathan's double doors area = \( \frac{3\pi}{2} \) square feet
We know that \( \frac{3\pi}{2} = 1.5\pi \).
Comparing \( 3\pi \) and \( 1.5\pi \), it is clear that \( 1.5\pi \) is smaller.
So, Nathan's double door takes the minimum area to open.
In simple words: Both Guna's single door and Nathan's double doors open like a part of a circle. We calculate how much floor space each takes up when opening. Nathan's two smaller doors take up less space together than Guna's one big door.

🎯 Exam Tip: When dealing with multiple identical objects like double doors, calculate the area or measurement for one object first, then multiply by the total number of objects to find the combined value.

 

Question 6. In a rectangular field which measures 15 m x 8m, cows are tied with a rope of length 3m at four corners of the field and also at the centre. Find the area of the field where none of the cow can graze. (\( \pi \) = 3.14).
Answer: First, let's find the total area of the rectangular field.
Length of the field = 15 m
Width of the field = 8 m
Area of the rectangle = \( \text{length} \times \text{width} = 15 \times 8 = 120 \text{ m}^2 \)

Cows are tied at the four corners with a rope of length 3m. This means each cow can graze in a quadrant (one-fourth) of a circle with a radius of 3m.
The area grazed by each cow at the corner is the area of a quadrant:
\[ \text{Area of one quadrant} = \frac{1}{4} \times \pi r^2 \] Since there are four corners, the total area grazed by the four corner cows is:
\[ \text{Area (4 quadrants)} = 4 \times \frac{1}{4} \times \pi r^2 = \pi r^2 \] Substitute \( r = 3 \) m and \( \pi = 3.14 \):
\[ \text{Area (4 quadrants)} = 3.14 \times (3)^2 = 3.14 \times 9 = 28.26 \text{ m}^2 \]
A cow is also tied at the center of the field with a rope of length 3m. This cow can graze in a full circle with a radius of 3m.
The area grazed by the cow at the center is:
\[ \text{Area (center circle)} = \pi r^2 \] Substitute \( r = 3 \) m and \( \pi = 3.14 \):
\[ \text{Area (center circle)} = 3.14 \times (3)^2 = 3.14 \times 9 = 28.26 \text{ m}^2 \]
Now, find the total area grazed by all the cows:
Total grazed area = Area (4 quadrants) + Area (center circle)
Total grazed area = \( 28.26 + 28.26 = 56.52 \text{ m}^2 \)

Finally, to find the area where none of the cows can graze, subtract the total grazed area from the total area of the field:
\[ \text{Area (ungrazed)} = \text{Area (rectangle)} - \text{Total grazed area} \] \[ \text{Area (ungrazed)} = 120 - 56.52 \] \[ \text{Area (ungrazed)} = 63.48 \text{ m}^2 \] So, the area of the field where none of the cows can graze is 63.48 m\( ^2 \). This problem shows how geometry helps calculate usable space.
In simple words: First, find the total size of the field. Then, calculate the area where the cows can eat grass. This includes four corner spots (which together make one full circle) and one spot in the middle (another full circle). Subtract the total area the cows can eat from the whole field to find the empty space.

🎯 Exam Tip: When cows are tied at corners of a rectangular field, their grazing area at each corner is a quadrant of a circle. Four quadrants combine to form one full circle, simplifying calculations.

 

Question 7. Three identical coins, each of diameter 6 cm are placed as shown. Find the area of the shaded region between the coins. (\( \pi \) = 3.14) (\( \sqrt{3} \) = 1.732)
Answer: When three identical coins are placed such that they touch each other, their centers form an equilateral triangle. The shaded region is the space within this triangle but outside the circles.
Given diameter of each coin = 6 cm.
Radius of each coin = \( \frac{\text{Diameter}}{2} = \frac{6}{2} = 3 \text{ cm} \).

The sides of the equilateral triangle formed by connecting the centers of the coins will be equal to \( \text{radius} + \text{radius} = 3 + 3 = 6 \text{ cm} \).
So, the side of the equilateral triangle is \( a = 6 \) cm.
The formula for the area of an equilateral triangle is:
\[ \text{Area of equilateral triangle} = \frac{\sqrt{3}}{4} a^2 \] Substitute \( a = 6 \) cm and \( \sqrt{3} = 1.732 \):
\[ \text{Area of equilateral triangle} = \frac{1.732}{4} \times (6)^2 \] \[ \text{Area of equilateral triangle} = \frac{1.732}{4} \times 36 \] \[ \text{Area of equilateral triangle} = 1.732 \times 9 \] \[ \text{Area of equilateral triangle} = 15.588 \text{ cm}^2 \]
Inside the equilateral triangle, there are three sectors, one from each coin. The angle of each corner of an equilateral triangle is \( 60^\circ \). So, each sector has an angle \( \theta = 60^\circ \) and radius \( r = 3 \) cm.
The area of one sector is:
\[ \text{Area of one sector} = \frac{\theta}{360^\circ} \times \pi r^2 \] Substitute \( \theta = 60^\circ \), \( r = 3 \) cm, and \( \pi = 3.14 \):
\[ \text{Area of one sector} = \frac{60^\circ}{360^\circ} \times 3.14 \times (3)^2 \] \[ \text{Area of one sector} = \frac{1}{6} \times 3.14 \times 9 \] \[ \text{Area of one sector} = \frac{3.14 \times 9}{6} \] \[ \text{Area of one sector} = \frac{28.26}{6} \] \[ \text{Area of one sector} = 4.71 \text{ cm}^2 \]
The total area of the three sectors is:
\[ \text{Total Area of 3 sectors} = 3 \times 4.71 \] \[ \text{Total Area of 3 sectors} = 14.13 \text{ cm}^2 \]
Finally, to find the area of the shaded region, subtract the total area of the three sectors from the area of the equilateral triangle:
\[ \text{Area of shaded region} = \text{Area of equilateral triangle} - \text{Total Area of 3 sectors} \] \[ \text{Area of shaded region} = 15.588 - 14.13 \] \[ \text{Area of shaded region} = 1.458 \text{ cm}^2 \] Thus, the area of the shaded region is approximately 1.458 cm\( ^2 \). This problem elegantly combines properties of circles and triangles.
In simple words: When three circles touch, their centers form a triangle with equal sides. The shaded part is the space inside this triangle that is not covered by the circles. We find the area of the triangle and subtract the areas of three small pie-slice shapes from the circles.

🎯 Exam Tip: Remember that when three identical circles touch, their centers form an equilateral triangle. The angle of each sector formed at the corners of this triangle is \( 60^\circ \).

 

Question 8. Using Euler's formula, find the unknowns.
Answer: Euler's formula for polyhedra states that:
\[ F + V - E = 2 \] where \( F \) is the number of faces, \( V \) is the number of vertices, and \( E \) is the number of edges.

(i) Given: \( V = 6 \), \( E = 14 \). We need to find \( F \).
Using Euler's formula:
\( F + V - E = 2 \)
\( F + 6 - 14 = 2 \)
\( F - 8 = 2 \)
To find \( F \), add 8 to both sides:
\( F = 2 + 8 \)
\( F = 10 \)

(ii) Given: \( F = 8 \), \( E = 10 \). We need to find \( V \).
Using Euler's formula:
\( F + V - E = 2 \)
\( 8 + V - 10 = 2 \)
\( V - 2 = 2 \)
To find \( V \), add 2 to both sides:
\( V = 2 + 2 \)
\( V = 4 \)

(iii) Given: \( F = 20 \), \( V = 10 \). We need to find \( E \).
Using Euler's formula:
\( F + V - E = 2 \)
\( 20 + 10 - E = 2 \)
\( 30 - E = 2 \)
To find \( E \), subtract 2 from 30:
\( E = 30 - 2 \)
\( E = 28 \)

Here is the tabulated summary of the results:

In simple words: Euler's formula is a rule for 3D shapes that says if you add the number of flat surfaces (faces) and the number of corners (vertices), then subtract the number of lines (edges), you will always get 2. We use this rule to find the missing numbers for different shapes.

🎯 Exam Tip: Make sure to correctly identify 'F' for faces, 'V' for vertices (corners), and 'E' for edges when applying Euler's formula. A common mistake is swapping 'V' and 'E'.