NCERT Class 8 - Exercise 9.1
1. The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m.
Given:
Parallel side \(a = 1\) mParallel side \(b = 1.2\) m
Height (perpendicular distance) \(h = 0.8\) m
Formula:
The area of a trapezium is given by:
\( \text{Area} = \frac{1}{2} \times (a + b) \times h \)
Calculation:
Substitute the values into the formula:\( \text{Area} = \frac{1}{2} \times (1 + 1.2) \times 0.8 \)
\( \text{Area} = \frac{1}{2} \times 2.2 \times 0.8 \)
\( \text{Area} = 1.1 \times 0.8 = 0.88 \) m\(^2\)
The area of the top surface of the table is 0.88 m\(^2\).
2. The area of a trapezium is 34 cm\(^2\) and the length of one of the parallel sides is 10 cm and its height is 4 cm. Find the length of the other parallel side.
Given:
Area = 34 cm\(^2\)Parallel side \(a = 10\) cm
Height \(h = 4\) cm
Let the other parallel side be \(b\).
Formula:
Area of a trapezium:
\( \text{Area} = \frac{1}{2} \times (a + b) \times h \)
Calculation:
Substitute the known values:\( 34 = \frac{1}{2} \times (10 + b) \times 4 \)
\( 34 = 2 \times (10 + b) \)
Divide both sides by 2:
\( 17 = 10 + b \)
Subtract 10 from both sides:
\( b = 17 - 10 = 7 \) cm
The length of the other parallel side is 7 cm.
3. Length of the fence of a trapezium shaped field ABCD is 120 m. If BC = 48 m, CD = 17 m and AD = 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.
Given:
Perimeter of field = 120 mSide BC = 48 m, CD = 17 m, AD = 40 m
Step 1: Find the length of side AB
The perimeter is the sum of all sides: \( AB + BC + CD + AD = 120 \).\( AB + 48 + 17 + 40 = 120 \)
\( AB + 105 = 120 \)
\( AB = 120 - 105 = 15 \) m.
Since AB is perpendicular to the parallel sides, AB is the height of the trapezium.
Step 2: Find the area
Parallel sides are \( a = 40 \) m and \( b = 48 \) m.Height \( h = AB = 15 \) m.
\( \text{Area} = \frac{1}{2} \times (a + b) \times h \)
\( \text{Area} = \frac{1}{2} \times (40 + 48) \times 15 \)\( \text{Area} = \frac{1}{2} \times 88 \times 15 \)
\( \text{Area} = 44 \times 15 = 660 \) m\(^2\)
The area of the field is 660 m\(^2\).
4. The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.
Given:
Diagonal \(d = 24\) mPerpendiculars \(h_1 = 8\) m and \(h_2 = 13\) m.
Formula:
The area of a quadrilateral is given by:
\( \text{Area} = \frac{1}{2} \times d \times (h_1 + h_2) \)
Calculation:
Substitute the values into the formula:\( \text{Area} = \frac{1}{2} \times 24 \times (8 + 13) \)
\( \text{Area} = 12 \times 21 \)
\( \text{Area} = 252 \) m\(^2\)
The area of the field is 252 m\(^2\).
5. The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.
Given:
Diagonal \(d_1 = 7.5\) cmDiagonal \(d_2 = 12\) cm
Formula:
The area of a rhombus is given by:
\( \text{Area} = \frac{1}{2} \times d_1 \times d_2 \)
Calculation:
Substitute the values into the formula:\( \text{Area} = \frac{1}{2} \times 7.5 \times 12 \)
\( \text{Area} = 7.5 \times 6 \)
\( \text{Area} = 45 \) cm\(^2\)
The area of the rhombus is 45 cm\(^2\).
6. Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal.
Given:
Side \(a = 5\) cmAltitude \(h = 4.8\) cm
Diagonal \(d_1 = 8\) cm
Step 1: Find the Area using side and altitude
A rhombus is a type of parallelogram. The area of a parallelogram is given by:
\( \text{Area} = \text{base} \times \text{height} = a \times h \)
\( \text{Area} = 5 \times 4.8 = 24 \) cm\(^2\).
Step 2: Find the other diagonal
We also know the area of a rhombus is \( \frac{1}{2} \times d_1 \times d_2 \). Let the other diagonal be \(d_2\).\( 24 = \frac{1}{2} \times 8 \times d_2 \)
\( 24 = 4 \times d_2 \)
\( d_2 = \frac{24}{4} = 6 \) cm
The area of the rhombus is 24 cm\(^2\), and the length of the other diagonal is 6 cm.
7. The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m\(^2\) is ₹ 4.
Given:
Number of tiles = 3000Diagonals of each tile: \( d_1 = 45 \) cm, \( d_2 = 30 \) cm
Cost of polishing = ₹ 4 per m\(^2\)
Step 1: Find the area of one tile in cm\(^2\)
\( \text{Area of one tile} = \frac{1}{2} \times d_1 \times d_2 \)
\( \text{Area of one tile} = \frac{1}{2} \times 45 \times 30 = 45 \times 15 = 675 \) cm\(^2\)
Step 2: Find the total area of the floor in cm\(^2\)
\( \text{Total Area} = 3000 \times 675 = 2,025,000 \) cm\(^2\)Step 3: Convert the total area to m\(^2\)
Since 1 m = 100 cm, then 1 m\(^2\) = \( (100 \times 100) \) cm\(^2\) = 10,000 cm\(^2\).
\( \text{Total Area in m}^2 = \frac{2,025,000}{10,000} = 202.5 \) m\(^2\)
Step 4: Find the total cost of polishing
\( \text{Total Cost} = \text{Total Area} \times \text{Cost per m}^2 \)\( \text{Total Cost} = 202.5 \times 4 = 810 \)
The total cost of polishing the floor is ₹ 810.
8. Mohan wants to buy a trapezium shaped field. Its side along the river is parallel to and twice the side along the road. If the area of this field is 10500 m\(^2\) and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.
Step 1: Define the parallel sides
Let the length of the side along the road be \(x\) m.The length of the side along the river is \(2x\) m.
Step 2: Use the Area formula to find \(x\)
\( \text{Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height} \)
\( 10500 = \frac{1}{2} \times (x + 2x) \times 100 \)\( 10500 = \frac{1}{2} \times (3x) \times 100 \)
\( 10500 = 3x \times 50 \)
\( 10500 = 150x \)
\( x = \frac{10500}{150} = 70 \) m.
Step 3: Find the length of the side along the river
The side along the river is \(2x\).Length = \(2 \times 70 = 140\) m.
The length of the side along the river is 140 m.
9. Top surface of a raised platform is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface.
Analysis of the figure:
A regular octagon can be divided into a central rectangle and two identical trapeziums. From the figure, we have:- Parallel sides of the trapeziums: 5 m and 11 m.
- Height of the trapeziums: 4 m.
- Dimensions of the central rectangle: 11 m by 5 m.
Step 1: Find the area of one trapezium
\( \text{Area of trapezium} = \frac{1}{2} \times (5 + 11) \times 4 \)
\( \text{Area} = \frac{1}{2} \times 16 \times 4 = 32 \) m\(^2\).
Step 2: Find the area of the rectangle
\( \text{Area of rectangle} = \text{length} \times \text{width} \)
\( \text{Area} = 11 \times 5 = 55 \) m\(^2\).
Step 3: Find the total area
\( \text{Total Area} = (\text{Area of 2 trapeziums}) + (\text{Area of rectangle}) \)\( \text{Total Area} = (2 \times 32) + 55 = 64 + 55 = 119 \) m\(^2\).
The area of the octagonal surface is 119 m\(^2\).
10. There is a pentagonal shaped park as shown in the figure. For finding its area Jyoti and Kavita divided it in two different ways. Find the area of this park using both ways. Can you suggest some other way of finding its area?
Jyoti's Way (Dividing into two trapeziums):
She divided the pentagon into two identical trapeziums.
Parallel sides: \(a = 15\) m and \(b = 30\) m.
Height of each trapezium: \( \frac{15}{2} = 7.5 \) m.
Area of one trapezium: \( \frac{1}{2} \times (15 + 30) \times 7.5 = 168.75 \) m\(^2\).
Total Area = \( 2 \times 168.75 = 337.5 \) m\(^2\).
Parallel sides: \(a = 15\) m and \(b = 30\) m.
Height of each trapezium: \( \frac{15}{2} = 7.5 \) m.
Area of one trapezium: \( \frac{1}{2} \times (15 + 30) \times 7.5 = 168.75 \) m\(^2\).
Total Area = \( 2 \times 168.75 = 337.5 \) m\(^2\).
Kavita's Way (Dividing into a triangle and a square):
She divided the pentagon into a triangle and a square.
Side of the square = 15 m. Area = \( 15 \times 15 = 225 \) m\(^2\).
Base of the triangle = 15 m. Height = \( (30 - 15) = 15 \) m.
Area of the triangle: \( \frac{1}{2} \times 15 \times 15 = 112.5 \) m\(^2\).
Total Area = \( 225 + 112.5 = 337.5 \) m\(^2\).
Side of the square = 15 m. Area = \( 15 \times 15 = 225 \) m\(^2\).
Base of the triangle = 15 m. Height = \( (30 - 15) = 15 \) m.
Area of the triangle: \( \frac{1}{2} \times 15 \times 15 = 112.5 \) m\(^2\).
Total Area = \( 225 + 112.5 = 337.5 \) m\(^2\).
Another way to find the area:
Another method is to enclose the pentagon in a large rectangle and subtract the areas of the outer triangles.1. Area of rectangle (30 m x 15 m) = \( 450 \) m\(^2\).
2. Area of two corner triangles = \( 2 \times (\frac{1}{2} \times 7.5 \times 15) = 112.5 \) m\(^2\).
3. Area of pentagon = \( 450 - 112.5 = 337.5 \) m\(^2\).
The area of the park is 337.5 m\(^2\) using all three methods.
11. Diagram of the adjacent picture frame has outer dimensions = 24 cm × 28 cm and inner dimensions 16 cm × 20 cm. Find the area of each section of the frame, if the width of each section is same.
Analysis of the figure:
The frame consists of four trapeziums. The top/bottom sections are identical, and the left/right sections are identical. The width of the frame is the same everywhere. We can calculate this width:Width = \( \frac{\text{Outer Width} - \text{Inner Width}}{2} = \frac{24 - 16}{2} = 4 \) cm.
Step 1: Area of the top and bottom sections
These are trapeziums with parallel sides \(a = 20\) cm and \(b = 28\) cm. The height \(h = 4\) cm.
\( \text{Area} = \frac{1}{2} \times (20 + 28) \times 4 = \frac{1}{2} \times 48 \times 4 = 96 \) cm\(^2\).
Step 2: Area of the left and right sections
These are trapeziums with parallel sides \(a = 16\) cm and \(b = 24\) cm. The height \(h = 4\) cm.
\( \text{Area} = \frac{1}{2} \times (16 + 24) \times 4 = \frac{1}{2} \times 40 \times 4 = 80 \) cm\(^2\).
The area of the top and bottom sections is 96 cm\(^2\) each, and the area of the left and right sections is 80 cm\(^2\) each.