GSEB Class 8 Maths Solutions Chapter 11 Mensuration Exercise 11.2

Question 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?
Answer: The area of a trapezium is found using a specific formula. We have the lengths of the two parallel sides and the perpendicular distance, which is the height. Let's substitute these values into the formula to get the area of the table.
Area of a trapezium \( = \frac{1}{2} \times \) [Sum of parallel sides] \( \times \) [Perpendicular distance between the parallel sides]
Here, the sum of parallel sides \( = 1.2 \, \text{m} + 1 \, \text{m} = 2.2 \, \text{m} \).
The perpendicular distance (height) \( = 0.8 \, \text{m} \).
So, Area of the table \( = \frac{1}{2} \times [1.2 + 1] \times 0.8 \, \text{m}^2 \)
\( = \frac{1}{2} \times 2.2 \times 0.8 \, \text{m}^2 \)
\( = \frac{1}{2} \times 2.2 \times 0.8 \)
\( = 1.1 \times 0.8 \)
\( = 0.88 \, \text{m}^2 \)
Thus, the area of the table is \( 0.88 \, \text{m}^2 \).
In simple words: To find the area of this table, which is shaped like a trapezium, you add the lengths of its two parallel sides, multiply that sum by its height, and then divide the whole thing by two.

Exam Tip: Always remember the formula for the area of a trapezium: \( \frac{1}{2} \times (a+b) \times h \), where 'a' and 'b' are the parallel sides and 'h' is the perpendicular height.

Question 2. The area of a trapezium is 34 cm² 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?
Answer: We are given the area of a trapezium, the length of one parallel side, and its height. We need to find the length of the other parallel side. We can use the formula for the area of a trapezium and substitute the known values to solve for the unknown side.
Let the length of the other parallel side be \( x \) cm.
Area of the trapezium \( = \frac{1}{2} \times \) [Sum of parallel sides] \( \times \) [Height]
Given, Area \( = 34 \, \text{cm}^2 \), one parallel side \( = 10 \, \text{cm} \), Height \( = 4 \, \text{cm} \).
\( 34 \, \text{cm}^2 = \frac{1}{2} \times [10 + x] \times 4 \, \text{cm}^2 \)
Multiply both sides by 2 and divide by 4:
\( \frac{2 \times 34}{4} \, \text{cm}^2 = 10 + x \)
\( \frac{68}{4} \, \text{cm}^2 = 10 + x \)
\( 17 = 10 + x \)
Now, subtract 10 from both sides to find \( x \):
\( x = 17 - 10 \)
\( x = 7 \, \text{cm} \)
Hence, the length of the other parallel side is \( 7 \, \text{cm} \).
In simple words: If you know the total area of a trapezium, one parallel side, and its height, you can use the area formula to work backward and find the missing length of the second parallel side.

Exam Tip: When finding a missing dimension, always write down the known formula first, substitute the given values, and then solve the resulting equation carefully.

Question 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?
Answer: We are given the total length of the fence, which is the perimeter of the trapezium, and the lengths of three sides (BC, CD, AD). We need to find the length of the fourth side (AB) first, as AB represents the height since it's perpendicular to the parallel sides AD and BC. Once we have the height, we can calculate the area of the trapezium.
Length of the fence \( = \) Perimeter
The perimeter of trapezium ABCD \( = AB + BC + CD + DA \).
Given Perimeter \( = 120 \, \text{m} \).
So, \( AB + 48 + 17 + 40 = 120 \)
Combine the known lengths:
\( AB + 105 = 120 \)
Subtract 105 from both sides to find AB:
\( AB = 120 - 105 \)
\( AB = 15 \, \text{m} \)
Now we have the height \( AB = 15 \, \text{m} \), and the parallel sides are \( AD = 40 \, \text{m} \) and \( BC = 48 \, \text{m} \).
Area of a trapezium \( = \frac{1}{2} \times \) [Sum of parallel sides] \( \times \) [Height]
Area \( = \frac{1}{2} \times (AD + BC) \times AB \)
\( = \frac{1}{2} \times (40 + 48) \times 15 \, \text{m}^2 \)
\( = \frac{1}{2} \times 88 \times 15 \, \text{m}^2 \)
\( = 44 \times 15 \, \text{m}^2 \)
\( = 660 \, \text{m}^2 \)
The area of the field is \( 660 \, \text{m}^2 \).
In simple words: First, use the total fence length (perimeter) and the lengths of the three given sides to find the missing side, which also acts as the height. Then, use the trapezium area formula with the parallel sides and the calculated height.

Exam Tip: In problems involving perimeter and area, always identify what each given value represents and what you need to calculate first (e.g., a missing side that acts as height) before applying area formulas.

Question 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?
Answer: To find the area of a general quadrilateral when a diagonal and the perpendiculars to it from the other two vertices are known, we use a specific formula. The field is a quadrilateral with one diagonal and two perpendiculars dropped from the opposite corners to that diagonal.
Area of a quadrilateral \( = \frac{1}{2} \times \) Diagonal \( \times \) [Sum of the perpendiculars on the diagonal from opposite vertices]
Given, Diagonal \( = 24 \, \text{m} \).
Perpendiculars \( = 8 \, \text{m} \) and \( 13 \, \text{m} \).
Sum of perpendiculars \( = 8 \, \text{m} + 13 \, \text{m} = 21 \, \text{m} \).
Area \( = \frac{1}{2} \times 24 \times (8 + 13) \, \text{m}^2 \)
\( = \frac{1}{2} \times 24 \times 21 \, \text{m}^2 \)
\( = 12 \times 21 \, \text{m}^2 \)
\( = 252 \, \text{m}^2 \)
The area of the field is \( 252 \, \text{m}^2 \).
In simple words: For a quadrilateral, you can find its area by multiplying half of a diagonal's length by the total length of the two perpendicular lines dropped from the other corners to that same diagonal.

Exam Tip: Remember that the area of a quadrilateral with one diagonal and two offsets (heights) from the other vertices is \( \frac{1}{2} d (h_1 + h_2) \). This formula simplifies finding the area of complex quadrilaterals.

Question 5. The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area?
Answer: For a rhombus, if we know the lengths of its two diagonals, we can easily calculate its area using a simple formula. The diagonals of a rhombus always intersect at right angles.
Area of the rhombus \( = \frac{1}{2} \times \) Product of diagonals
Given, diagonal \( d_1 = 7.5 \, \text{cm} \) and diagonal \( d_2 = 12 \, \text{cm} \).
Area \( = \frac{1}{2} \times 7.5 \, \text{cm} \times 12 \, \text{cm} \)
\( = \frac{1}{2} \times 90 \, \text{cm}^2 \)
\( = 45 \, \text{cm}^2 \)
The area of the rhombus is \( 45 \, \text{cm}^2 \).
In simple words: To find the area of a rhombus, just multiply the lengths of its two diagonals and then divide that result by two.

Exam Tip: Always make sure to use the product of *both* diagonals when calculating the area of a rhombus; a common mistake is to only use one diagonal.

Question 6. Find the area of a rhombus whose side is 6 cm and whose altitude is 4 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal?
Answer: We have two parts to this question. First, we need to find the area of the rhombus using its side and altitude. A rhombus is a type of parallelogram, so its area can be calculated as Base \( \times \) Height. Second, using the calculated area and the length of one diagonal, we can find the length of the other diagonal using the rhombus area formula involving diagonals.
A rhombus is a parallelogram.
Area of a parallelogram \( = \) Base \( \times \) Height
Here, Base (side of rhombus) \( = 6 \, \text{cm} \).
Height (altitude) \( = 4 \, \text{cm} \).
Therefore, Area of the rhombus \( = 6 \, \text{cm} \times 4 \, \text{cm}^2 \)
\( = 24 \, \text{cm}^2 \)
Now, we need to find the length of the other diagonal.
Let the other diagonal be \( d \).
We know that the Area of a rhombus \( = \frac{1}{2} \times d_1 \times d_2 \).
We have Area \( = 24 \, \text{cm}^2 \) and one diagonal \( d_1 = 8 \, \text{cm} \).
\( 24 \, \text{cm}^2 = \frac{1}{2} \times 8 \times d \)
\( 24 = 4d \)
Divide both sides by 4 to find \( d \):
\( d = \frac{24}{4} \)
\( d = 6 \, \text{cm} \)
Thus, the required other diagonal is \( 6 \, \text{cm} \).
In simple words: First, treat the rhombus like a parallelogram and multiply its side by its height to get the area. Then, use this area with the given diagonal in the rhombus area formula (half times product of diagonals) to find the length of the second diagonal.

Exam Tip: Remember that a rhombus is a special type of parallelogram, so both area formulas (Base × Height and \( \frac{1}{2} d_1 d_2 \)) are valid. Choose the appropriate formula based on the given information.

Question 7. 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 m² is Rs. 4?
Answer: To find the total cost of polishing the floor, we first need to determine the total area of the floor covered by the tiles. This involves calculating the area of one rhombus-shaped tile, then multiplying it by the total number of tiles. Finally, we convert this total area to square meters and multiply by the cost per square meter.
Tiles are rhombus shaped, having diagonal \( d_1 = 45 \, \text{cm} \) and diagonal \( d_2 = 30 \, \text{cm} \).
Area of a tile (rhombus) \( = \frac{1}{2} \times d_1 \times d_2 \)
\( = \frac{1}{2} \times 45 \, \text{cm} \times 30 \, \text{cm}^2 \)
\( = 45 \times 15 \, \text{cm}^2 \)
\( = 675 \, \text{cm}^2 \)
Total number of tiles \( = 3000 \).
Area of the floor \( = 675 \times 3000 \, \text{cm}^2 \)
\( = 20,25,000 \, \text{cm}^2 \)
Now, we convert the area from square centimeters to square meters. We know that \( 1 \, \text{m}^2 = 100 \times 100 \, \text{cm}^2 = 10,000 \, \text{cm}^2 \).
So, Area of the floor in \( \text{m}^2 = \frac{20,25,000}{10,000} \, \text{m}^2 \)
\( = \frac{2025}{10} \, \text{m}^2 \)
\( = 202.5 \, \text{m}^2 \)
Cost per \( \text{m}^2 \) is Rs. 4.
Cost of polishing the floor \( = 4 \times 202.5 \)
\( = \text{Rs. } 810 \)
Thus, the total cost of polishing the floor is Rs. 810.
In simple words: First, find the area of one tile. Then multiply that by the number of tiles to get the total floor area. Convert the total area from square centimeters to square meters, and finally, multiply by the cost per square meter to get the total polishing cost.

Exam Tip: Be very careful with unit conversions, especially between square centimeters and square meters. Remember that \( 1 \, \text{m}^2 \) is not \( 100 \, \text{cm}^2 \), but \( (100 \times 100) \, \text{cm}^2 \).

Question 8. Mohan wants to buy a trapezium shaped field. Its side along the river field. Its side along the river is parallel to and twice the side along road. If the area of this field is 10500 m² and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river?
Answer: We are dealing with a trapezium-shaped field where one parallel side (along the river) is twice the length of the other parallel side (along the road). We are given the total area of the field and the perpendicular distance between the parallel sides (height). We need to find the length of the side along the river. We will use variables to represent the sides and apply the area of a trapezium formula.
Let the length of the side along the road be \( x \, \text{m} \).
Then, the other side (parallel to the road, along the river) \( = 2x \, \text{m} \) (since it's twice the side along the road).
Area of the trapezium shaped field \( = \frac{1}{2} \times \) [Sum of parallel sides] \( \times \) [Height]
Given, Area \( = 10,500 \, \text{m}^2 \), Height \( = 100 \, \text{m} \).
\( 10,500 = \frac{1}{2} \times [x + 2x] \times 100 \)
\( 10,500 = \frac{1}{2} \times [3x] \times 100 \)
\( 10,500 = 3x \times 50 \)
\( 10,500 = 150x \)
Divide both sides by 150 to find \( x \):
\( x = \frac{10,500}{150} \)
\( x = 70 \, \text{m} \)
The length of the side along the road is \( 70 \, \text{m} \).
We need to find the length of the side along the river, which is \( 2x \).
Length of the side along the river \( = 2 \times 70 \, \text{m} = 140 \, \text{m} \).
Thus, the length of the side along the river is \( 140 \, \text{m} \).
In simple words: First, set up an equation using the area formula for a trapezium. If one parallel side is twice the other, call the smaller one 'x' and the larger one '2x'. Solve for 'x', then double it to find the length of the side along the river.

Exam Tip: When given relationships between sides (e.g., "twice the side"), always assign a variable to the simpler length first and then express the other length in terms of that variable. This makes setting up the equation much clearer.

Question 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?
Answer: To find the area of this regular octagon, we can break it down into simpler shapes whose areas are easy to calculate. The figure shows that the octagon can be divided into two trapeziums and one rectangle in the middle.
This regular octagon can be divided into two trapeziums (each having height 4 m and parallel sides are 11 m and 5 m) and a rectangular part with length 11 m and breadth 5 m.
First, let's find the area of one trapezium:
Area of a trapezium \( = \frac{1}{2} \times \) [Sum of parallel sides] \( \times \) Height
\( = \frac{1}{2} \times [11 \, \text{m} + 5 \, \text{m}] \times 4 \, \text{m} \)
\( = \frac{1}{2} \times 16 \, \text{m} \times 4 \, \text{m}^2 \)
\( = 8 \times 4 \, \text{m}^2 \)
\( = 32 \, \text{m}^2 \)
Since there are two such trapeziums:
Area of both the trapeziums \( = 2 \times 32 \, \text{m}^2 = 64 \, \text{m}^2 \)
Next, let's find the area of the rectangular part:
Area of a rectangle \( = \) Length \( \times \) Breadth
\( = 11 \, \text{m} \times 5 \, \text{m} = 55 \, \text{m}^2 \)
Finally, add the areas of all parts to get the total area of the regular octagon:
Area of the regular octagon \( = \) Area of two trapeziums \( + \) Area of rectangle
\( = 64 \, \text{m}^2 + 55 \, \text{m}^2 = 119 \, \text{m}^2 \)
The area of the octagonal surface is \( 119 \, \text{m}^2 \).
In simple words: Break the octagon into two trapeziums and a rectangle. Calculate the area of each shape separately. Then, add all these individual areas together to find the total area of the octagon.

Exam Tip: When dealing with complex shapes, always try to decompose them into simpler geometric figures (like triangles, rectangles, or trapeziums) whose area formulas are well-known. Be careful with dimensions for each component shape.

Question 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?
Answer: We need to calculate the area of the pentagonal park using two different division methods (Jyoti's and Kavita's) and then suggest a third way.
**Jyoti's diagram:**
Jyoti split the pentagonal shape into two identical trapeziums. The parallel sides of each trapezium are 15 m and 30 m, and the height is \( \frac{15}{2} \, \text{m} \).
Area of one trapezium \( = \frac{1}{2} \times \) [Sum of parallel sides] \( \times \) Height
\( = \frac{1}{2} \times [15 + 30] \times \frac{15}{2} \, \text{m}^2 \)
\( = \frac{1}{2} \times 45 \times \frac{15}{2} \, \text{m}^2 \)
\( = \frac{675}{4} \, \text{m}^2 \)
Area of the pentagonal shape \( = 2 \times \) Area of one trapezium
\( = 2 \times \frac{675}{4} \, \text{m}^2 \)
\( = \frac{675}{2} \, \text{m}^2 = 337.5 \, \text{m}^2 \)
**Kavita's diagram:**
Kavita split the pentagonal shape into a square and a triangle.
Area of the square \( = \) side \( \times \) side
\( = 15 \, \text{m} \times 15 \, \text{m} = 225 \, \text{m}^2 \)
The base of the triangle \( = 15 \, \text{m} \).
The height of the triangle \( = 30 \, \text{m} - 15 \, \text{m} = 15 \, \text{m} \).
Area of the triangle \( = \frac{1}{2} \times \) Base \( \times \) Height
\( = \frac{1}{2} \times 15 \, \text{m} \times 15 \, \text{m} = \frac{225}{2} \, \text{m}^2 = 112.5 \, \text{m}^2 \)
Area of the pentagonal shape \( = \) Area of square \( + \) Area of triangle
\( = 225 \, \text{m}^2 + 112.5 \, \text{m}^2 = 337.5 \, \text{m}^2 \)
Both methods give the same area, \( 337.5 \, \text{m}^2 \).
**Another way of finding the area of the pentagonal shape:**
By splitting the pentagonal shape into 3 triangles, we have:
Area of triangle I \( = \frac{1}{2} \times 15 \times 15 \, \text{m}^2 = \frac{225}{2} \, \text{m}^2 \)
Area of triangle II \( = \frac{1}{2} \times 15 \times 15 \, \text{m}^2 = \frac{225}{2} \, \text{m}^2 \)
Area of triangle III \( = \frac{1}{2} \times 15 \times 15 \, \text{m}^2 = \frac{225}{2} \, \text{m}^2 \)
Therefore, Area of the pentagonal shape \( = \frac{225}{2} \, \text{m}^2 + \frac{225}{2} \, \text{m}^2 + \frac{225}{2} \, \text{m}^2 \)
\( = \frac{3 \times 225}{2} \, \text{m}^2 = \frac{675}{2} \, \text{m}^2 = 337.5 \, \text{m}^2 \)
In simple words: You can find the park's area by splitting it in different ways, like two trapeziums or a square and a triangle. Both methods should give you the same answer. Another way is to split it into three triangles and add their areas.

Exam Tip: For irregular polygons, breaking them into standard geometric shapes is a crucial strategy. Always ensure that the sum of the areas of the smaller shapes equals the area of the original polygon.

Question 11. Diagram of the adjacent picture frame has outer dimensions = 24 cm × 28 cm and inner dimensions 16cm × 20 cm. Find the area of each section of the frame, if the width of each section is same?
Answer: A picture frame has an outer rectangle and an inner rectangle, forming a border. The border can be seen as four trapeziums. Since the width of each section is uniform, we can calculate the height of these trapeziums and then their individual areas.
The frame consists of four trapeziums (I, II, III, IV).
Let's find the width of the frame first:
Width along the top/bottom: \( \frac{28 \, \text{cm} - 20 \, \text{cm}}{2} = \frac{8 \, \text{cm}}{2} = 4 \, \text{cm} \).
Width along the sides: \( \frac{24 \, \text{cm} - 16 \, \text{cm}}{2} = \frac{8 \, \text{cm}}{2} = 4 \, \text{cm} \).
So, the uniform width of each section (height of trapeziums) is \( 4 \, \text{cm} \).
**Area of trapezium I (Top section):**
Parallel sides are the outer top length (28 cm) and the inner top length (20 cm).
Height \( = 4 \, \text{cm} \).
Area of trapezium I \( = \frac{1}{2} \times [20 + 28] \times 4 \, \text{cm}^2 \)
\( = \frac{1}{2} \times 48 \times 4 \, \text{cm}^2 \)
\( = 24 \times 4 \, \text{cm}^2 \)
\( = 96 \, \text{cm}^2 \)
**Area of trapezium II (Bottom section):**
Parallel sides are the outer bottom length (28 cm) and the inner bottom length (20 cm). This is identical to trapezium I.
Area of trapezium II \( = 96 \, \text{cm}^2 \).
**Area of trapezium III (Left section):**
Parallel sides are the outer left height (24 cm) and the inner left height (16 cm).
Height \( = 4 \, \text{cm} \).
Area of trapezium III \( = \frac{1}{2} \times [16 + 24] \times 4 \, \text{cm}^2 \)
\( = \frac{1}{2} \times 40 \times 4 \, \text{cm}^2 \)
\( = 20 \times 4 \, \text{cm}^2 \)
\( = 80 \, \text{cm}^2 \)
**Area of trapezium IV (Right section):**
Parallel sides are the outer right height (24 cm) and the inner right height (16 cm). This is identical to trapezium III.
Area of trapezium IV \( = 80 \, \text{cm}^2 \).
In simple words: A picture frame can be thought of as four trapeziums. First, figure out the frame's consistent width, which acts as the height for these trapeziums. Then, calculate the area for each of the four sections using the trapezium area formula, matching the outer and inner dimensions for each section.

Exam Tip: When dividing a complex shape like a picture frame into simpler parts, always clearly identify the parallel sides and the perpendicular height for each section to apply the trapezium formula correctly. Pay close attention to which dimensions belong to which side of the trapezium.