GSEB Class 7 Maths Solutions Chapter 11 Perimeter and Area Exercise 11.4

Question 1. A garden is 90 m long and 75 m broad. A path 5 m wide is to be built outside and around it. Find the area of the path. Also find the area of the garden in hectare.
Answer:
Length of the garden \( (l) = 90 \) m
Breadth of the garden \( (b) = 75 \) m
Area of the garden \( = l \times b = 90 \text{ m} \times 75 \text{ m} = 6750 \text{ m}^2 \)
To convert square meters to hectares, we use the fact that \( 1 \text{ m}^2 = \frac{1}{10000} \) ha.
So, area of the garden in hectares \( = \frac{6750}{10000} \text{ ha} = 0.675 \) ha.
Since the garden with the path forms the outer rectangle, we need its new dimensions.
Length of the outer rectangle \( = (90 + 5 + 5) \text{ m} = 100 \text{ m} \)
Breadth of the outer rectangle \( = (75 + 5 + 5) \text{ m} = 85 \text{ m} \)
Area of the outer path (rectangle) \( = 100 \text{ m} \times 85 \text{ m} = 8500 \text{ m}^2 \)
Now, the area of the path \( = \text{ [Area of outer rectangle]} - \text{ [Area of the inner rectangle]} \)
\( = 8500 \text{ m}^2 - 6750 \text{ m}^2 = 1750 \text{ m}^2 \)
The area of the path is \( 1750 \text{ m}^2 \), and the garden's area is \( 0.675 \) ha.
In simple words: First, calculate the garden's area and convert it to hectares. Then, add the path width to the garden's length and breadth to find the larger outer area. Finally, subtract the garden's area from this larger area to get the path's area.

Exam Tip: Remember to add the path width to both sides of the length and breadth when finding the outer dimensions. Also, be careful with unit conversions, especially between square meters and hectares.

Question 2. A 3 m wide path runs outside and around a rectangular park of length 125 m and breadth 65 m. Find the area of the path.
Answer:
Length of the inner region (park) \( = 125 \) m
Breadth of the inner region (park) \( = 65 \) m
Area of the inner region \( = 125 \text{ m} \times 65 \text{ m} = 8125 \text{ m}^2 \)
The path is 3 m wide and runs outside the park.
Length of the outer region \( = (125 + 3 + 3) \text{ m} = 131 \text{ m} \)
Breadth of the outer region \( = (65 + 3 + 3) \text{ m} = 71 \text{ m} \)
Area of the outer region \( = 131 \text{ m} \times 71 \text{ m} = 9301 \text{ m}^2 \)
Area of the path \( = \text{ [Area of the outer region]} - \text{ [Area of the inner region]} \)
\( = 9301 \text{ m}^2 - 8125 \text{ m}^2 = 1176 \text{ m}^2 \)
The area of the path is \( 1176 \text{ m}^2 \).
In simple words: Calculate the area of the park first. Then, add the path width to both sides of the park's length and breadth to get the dimensions of the larger area (park plus path). Find the area of this larger rectangle. The path's area is the difference between these two areas.

Exam Tip: Always make sure to add the path width twice (once for each side) to both the length and breadth when the path is outside the main figure.

Question 3. A picture is painted on a cardboard 8 cm long and 5 cm wide such that there is a margin of 1.5 cm along each of its sides. Find the total area of the margin.
Answer:
Length of the outer rectangle (cardboard) \( = 8 \) cm
Breadth of the outer rectangle (cardboard) \( = 5 \) cm
Area of the outer rectangle \( = 8 \text{ cm} \times 5 \text{ cm} = 40 \text{ cm}^2 \)
Width of the margin \( = 1.5 \) cm
For the inner rectangle (picture area, excluding margin):
Length \( = (8 - 1.5 - 1.5) \text{ cm} = (8 - 3) \text{ cm} = 5 \text{ cm} \)
Breadth \( = (5 - 1.5 - 1.5) \text{ cm} = (5 - 3) \text{ cm} = 2 \text{ cm} \)
Area of the inner rectangle \( = 5 \text{ cm} \times 2 \text{ cm} = 10 \text{ cm}^2 \)
Area of the margin \( = \text{ [Area of the outer rectangle]} - \text{ [Area of the inner rectangle]} \)
\( = 40 \text{ cm}^2 - 10 \text{ cm}^2 = 30 \text{ cm}^2 \)
The total area of the margin is \( 30 \text{ cm}^2 \).
In simple words: First, find the area of the whole cardboard. Then, subtract the margin width from both sides of the cardboard's length and breadth to find the size of the picture inside. Calculate the area of this inner picture. The margin's area is simply the outer cardboard area minus the inner picture area.

Exam Tip: When a margin is *inside* a figure, you subtract the margin width *twice* (once from each side) from both the length and breadth to get the dimensions of the inner area.

Question 4. A verandah of width 2.25 m is constructed all along outside a room which is 5.5 m long and 4 m wide. Find:
(i) The area of the verandah.
(ii) The cost of cementing the floor of the verandah at the rate of Rs. 200 per m².
Answer:
(i) Length of the room \( = 5.5 \) m
Breadth of the room \( = 4 \) m
Area of the room \( = 5.5 \text{ m} \times 4 \text{ m} = 22 \text{ m}^2 \)
Width of the verandah \( = 2.25 \) m
For the outer rectangle (room + verandah):
Length \( = [5.5 + 2.25 + 2.25] \text{ m} = [5.5 + 4.5] \text{ m} = 10 \text{ m} \)
Breadth \( = [4 + 2.25 + 2.25] \text{ m} = [4 + 4.5] \text{ m} = 8.5 \text{ m} \)
Area of the outer rectangle \( = 10 \text{ m} \times 8.5 \text{ m} = 85 \text{ m}^2 \)
Area of the verandah \( = \text{ [Area of the outer rectangle]} - \text{ [Area of the inner rectangle]} \)
\( = 85 \text{ m}^2 - 22 \text{ m}^2 = 63 \text{ m}^2 \)
The area of the verandah is \( 63 \text{ m}^2 \).
(ii) Rate of cementing \( = \text{Rs. } 200/\text{m}^2 \)
Cost of cementing the floor of the verandah \( = \text{Rs. } 200 \times 63 = \text{Rs. } 12,600 \)
Thus, the total cost for cementing the verandah floor is Rs. 12,600.
In simple words: First, calculate the area of the room. Then, add the verandah width twice to both the room's length and breadth to find the dimensions of the room plus verandah. Find the total area. The verandah's area is the difference between these two areas. For the cost, multiply the verandah's area by the given rate per square meter.

Exam Tip: Be careful with decimals in calculations and ensure you double the width when adding it to both sides of the dimensions.

Question 5. A path 1 m wide is built along the border and inside a square garden of side 30 m. Find:
(i) The area of the path.
(ii) The cost of planting grass in the remaining portion of the garden at the rate of Rs. 40 per m².
Answer:
(i) Length of the outer square (garden) \( = 30 \) m
Area of the outer square \( = 30 \text{ m} \times 30 \text{ m} = 900 \text{ m}^2 \)
Width of the path \( = 1 \) m
Side of the inner square (garden excluding path) \( = [30 - 1 - 1] \text{ m} = 28 \text{ m} \)
Area of the inner square \( = (28 \text{ m})^2 = 784 \text{ m}^2 \)
Area of the path \( = \text{ [Area of the outer square]} - \text{ [Area of the inner square]} \)
\( = 900 \text{ m}^2 - 784 \text{ m}^2 = 116 \text{ m}^2 \)
The area of the path is \( 116 \text{ m}^2 \).
(ii) Rate of planting grass \( = \text{Rs. } 40/\text{m}^2 \)
The remaining portion for grass is the area of the inner square.
Cost of planting grass \( = \text{Rs. } 40 \times 784 = \text{Rs. } 31,360 \)
The total cost for planting grass is Rs. 31,360.
In simple words: First, calculate the garden's total area. Then, subtract the path width from both sides of the garden's side to find the size of the inner area where grass will be planted. Calculate the area of this inner square; this is the area of the path. For the cost, multiply the inner square's area by the grass planting rate.

Exam Tip: For paths *inside* a figure, always subtract the path width twice from each dimension to get the inner dimensions. The "remaining portion" usually refers to the area not covered by the path.

Question 6. Two cross roads, each of width 10 m, cut at right angles through the centre of a rectangular park of length 700 m and breadth 300 m and parallel to its sides. Find the area of the roads. Also find the area of the park excluding cross roads. Give the answer in hectares.
Answer:
Length of the rectangular park \( = 700 \) m
Breadth of the rectangular park \( = 300 \) m
Width of each road \( = 10 \) m
Area of the road parallel to the length (let's call it ABCD) \( = \text{Length of park} \times \text{Width of road} \)
\( = 700 \text{ m} \times 10 \text{ m} = 7000 \text{ m}^2 \)
Area of the road parallel to the breadth (let's call it PQRS) \( = \text{Breadth of park} \times \text{Width of road} \)
\( = 300 \text{ m} \times 10 \text{ m} = 3000 \text{ m}^2 \)
The area where the two roads intersect is a square (let's call it LMNO) with side equal to the width of the road.
Area of the common square LMNO \( = 10 \text{ m} \times 10 \text{ m} = 100 \text{ m}^2 \)
Area of the roads \( = \text{ [Area of ABCD]} + \text{ [Area of PQRS]} - \text{ [Area of LMNO]} \)
\( = [7000 \text{ m}^2 + 3000 \text{ m}^2] - 100 \text{ m}^2 \)
\( = 10000 \text{ m}^2 - 100 \text{ m}^2 = 9900 \text{ m}^2 \)
To convert square meters to hectares: \( 1 \text{ m}^2 = \frac{1}{10000} \) ha
Area of roads in ha \( = \frac{9900}{10000} \text{ ha} = 0.99 \) ha
Now, the area of the park excluding cross roads:
Area of the park \( = 700 \text{ m} \times 300 \text{ m} = 210000 \text{ m}^2 \)
Area of the park excluding cross roads \( = \text{ [Area of the park]} - \text{ [Area of the cross roads]} \)
\( = 210000 \text{ m}^2 - 9900 \text{ m}^2 = 200100 \text{ m}^2 \)
Area of the park excluding cross roads in ha \( = \frac{200100}{10000} \text{ ha} = 20.01 \) ha.
In simple words: Calculate the area of each road separately. Since the roads cross, their intersection area is counted twice, so subtract it once. This gives the total road area. Then, find the total park area. Subtract the road area from the park area to get the area without roads. Convert both results to hectares by dividing by 10,000.

Exam Tip: When calculating the area of intersecting paths, always remember to subtract the area of the common overlapping region once to avoid double-counting.

Question 7. Through a rectangular field of length 90 m and breadth 60 m, two roads are constructed which are parallel to the sides and cut each other at right angles through the centre of the fields. If the width of each road is 3 m, find:
(i) the area covered by the roads.
(ii) the cost of constructing the roads at the rate of Rs. 110 per m².
Answer:
Length of the rectangular field \( = 90 \) m
Breadth of the rectangular field \( = 60 \) m
Width of each road \( = 3 \) m
(i) Area of the road parallel to the length (ABCD) \( = 90 \text{ m} \times 3 \text{ m} = 270 \text{ m}^2 \)
Area of the road parallel to the breadth (EFGH) \( = 60 \text{ m} \times 3 \text{ m} = 180 \text{ m}^2 \)
Area of the common square (PQRS) \( = 3 \text{ m} \times 3 \text{ m} = 9 \text{ m}^2 \)
Area of roads \( = \text{ [Area of ABCD]} + \text{ [Area of EFGH]} - \text{ [Area of PQRS]} \)
\( = [270 \text{ m}^2 + 180 \text{ m}^2] - 9 \text{ m}^2 \)
\( = 450 \text{ m}^2 - 9 \text{ m}^2 = 441 \text{ m}^2 \)
The area covered by the roads is \( 441 \text{ m}^2 \).
(ii) Rate of construction of roads \( = \text{Rs. } 110/\text{m}^2 \)
Cost of construction of roads \( = \text{Rs. } 110 \times 441 = \text{Rs. } 48,510 \)
The cost of constructing the roads is Rs. 48,510.
In simple words: First, calculate the area of each road. Since they cross, subtract the area of their square intersection (which is counted twice) once. This gives the total area of the roads. Then, multiply this total road area by the construction rate to find the total cost.

Exam Tip: Be consistent with units throughout your calculation and double-check the subtraction of the overlapping area for intersecting paths.

Question 8. Pragya wrapped a cord around a circular pipe of radius 4 cm and cut off the length required of the cord: Then she wrapped it around a square box of side 4 cm (also shown). Did she have any cord left? (Take \( \pi = 3.14 \))
Answer:
Radius of the circular pipe \( (r) = 4 \) cm
Circumference of the pipe \( = 2\pi r = 2 \times 3.14 \times 4 \text{ cm} = 25.12 \text{ cm} \)
Length of the cord wrapped around the circular pipe \( = 25.12 \) cm
Now, the perimeter of the square box:
Side of the square \( = 4 \) cm
Perimeter of the square \( = 4 \times \text{Side} = 4 \times 4 \text{ cm} = 16 \text{ cm} \)
Length of the cord wrapped around the square \( = 16 \) cm
Since \( 25.12 \text{ cm} > 16 \text{ cm} \), there is cord left.
The length of cord left \( = (25.12 - 16) \text{ cm} = 9.12 \text{ cm} \)
Thus, yes, Pragya has \( 9.12 \) cm of cord left.
In simple words: First, calculate the length of cord needed for the circular pipe using its radius (this is the circumference). Then, calculate the length of cord needed for the square box using its side (this is the perimeter). Compare the two lengths to see if there's any cord left, and if so, find the difference.

Exam Tip: Remember the formulas for circumference of a circle \( (2\pi r) \) and perimeter of a square \( (4 \times \text{side}) \). Pay attention to the value of \( \pi \) provided in the question.

Question 9. The figure represents a rectangular lawn with a circular flower bed in the middle. Find:
(i) the area of the whole land.
(ii) the area of the flower bed.
(iii) the area of the lawn excluding the area of the flower bed.
(iv) the circumference of the flower bed.
Answer:
Length of the rectangular plot \( = 10 \) m
Breadth of the rectangular plot \( = 5 \) m
(i) Area of the plot (whole land) \( = 10 \text{ m} \times 5 \text{ m} = 50 \text{ m}^2 \)
The area of the whole land is \( 50 \text{ m}^2 \).
(ii) Radius of the circular flower bed \( (r) = 2 \) m
Area of the flower bed \( = \pi r^2 = 3.14 \times 2 \text{ m} \times 2 \text{ m} = 12.56 \text{ m}^2 \)
The area of the flower bed is \( 12.56 \text{ m}^2 \).
(iii) Area of the lawn excluding the area of the flower bed \( = 50 \text{ m}^2 - 12.56 \text{ m}^2 = 37.44 \text{ m}^2 \)
The area of the lawn excluding the flower bed is \( 37.44 \text{ m}^2 \).
(iv) Circumference of the circular flower bed \( = 2\pi r = 2 \times 3.14 \times 2 \text{ m} = 12.56 \text{ m} \)
The circumference of the flower bed is \( 12.56 \text{ m} \).
In simple words: First, calculate the total area of the rectangular land. Then, find the area of the circular flower bed using its radius. The area of the lawn without the flower bed is the total land area minus the flower bed area. Finally, find the circumference (distance around) the flower bed.

Exam Tip: Carefully distinguish between area and circumference calculations. Ensure you use the correct formulas for rectangles and circles, and use the specified value for \( \pi \).

Question 10. In the following figures, find the area of the shaded portions:
Answer:
(i) For the first figure:
Length of the whole rectangle ABCD \( = 18 \) cm
Breadth of the whole rectangle ABCD \( = 10 \) cm
Area of the whole rectangle ABCD \( = 18 \text{ cm} \times 10 \text{ cm} = 180 \text{ cm}^2 \)
Area of right triangle AEF (base 6 cm, height 10 cm) \( = \frac{1}{2} \times 6 \times 10 \text{ cm}^2 = 30 \text{ cm}^2 \)
Area of right triangle CBE (base 8 cm, height 10 cm) \( = \frac{1}{2} \times 8 \times 10 \text{ cm}^2 = 40 \text{ cm}^2 \)
Area of the shaded portion \( = \text{ [Area of ABCD]} - \text{ [Area of } \triangle\text{AEF} + \text{ Area of } \triangle\text{CBE]} \)
\( = [180 \text{ cm}^2] - [30 \text{ cm}^2 + 40 \text{ cm}^2] \)
\( = 180 \text{ cm}^2 - 70 \text{ cm}^2 = 110 \text{ cm}^2 \)
The area of the shaded portion in figure (i) is \( 110 \text{ cm}^2 \).
(ii) For the second figure:
Side of the square PQRS \( = 20 \) cm
Area of the square PQRS \( = 20 \text{ cm} \times 20 \text{ cm} = 400 \text{ cm}^2 \)
Area of right triangle QPT (base 20 cm, height 10 cm) \( = \frac{1}{2} \times 20 \times 10 \text{ cm}^2 = 100 \text{ cm}^2 \)
Area of right triangle ATSU (base 10 cm, height 10 cm) \( = \frac{1}{2} \times 10 \times 10 \text{ cm}^2 = 50 \text{ cm}^2 \)
Area of right triangle AQRU (base 10 cm, height 20 cm) \( = \frac{1}{2} \times 10 \times 20 \text{ cm}^2 = 100 \text{ cm}^2 \)
Area of the shaded portion \( = \text{ [Area PQRS]} - \text{ [Area of } \triangle\text{QPT} + \text{ Area of } \triangle\text{ATSU} + \text{ Area of } \triangle\text{AQRU]} \)
\( = [400 \text{ cm}^2] - [100 \text{ cm}^2 + 50 \text{ cm}^2 + 100 \text{ cm}^2] \)
\( = 400 \text{ cm}^2 - 250 \text{ cm}^2 = 150 \text{ cm}^2 \)
The area of the shaded portion in figure (ii) is \( 150 \text{ cm}^2 \).
In simple words: For each figure, first find the total area of the large outer shape (rectangle or square). Then, identify the unshaded triangles within it and calculate their areas using the formula for a right triangle (half times base times height). Finally, subtract the sum of these unshaded triangle areas from the total outer area to find the area of the shaded part.

Exam Tip: When dealing with shaded regions, sometimes it's easier to calculate the area of the whole figure and subtract the areas of the unshaded parts, rather than trying to calculate the shaded area directly.

Question 11. Find the area of the quadrilateral ABCD. Here, AC = 22 cm, BM = 3 cm, DN = 3 cm, and BM \( \perp \) AC, DN \( \perp \) AC.
Answer:
For the quadrilateral ABCD, the diagonal is AC, and BM and DN are perpendiculars from B and D to AC.
The area of a quadrilateral with one diagonal and perpendiculars to it from the other two vertices is the sum of the areas of the two triangles formed.
Area of triangle ABC \( = \frac{1}{2} \times \text{Base} \times \text{Height} \)
\( = \frac{1}{2} \times \text{AC} \times \text{BM} \)
\( = \frac{1}{2} \times 22 \text{ cm} \times 3 \text{ cm}^2 = 33 \text{ cm}^2 \)
Area of triangle ACD \( = \frac{1}{2} \times \text{Base} \times \text{Height} \)
\( = \frac{1}{2} \times \text{AC} \times \text{DN} \)
\( = \frac{1}{2} \times 22 \text{ cm} \times 3 \text{ cm}^2 = 33 \text{ cm}^2 \)
Now, the area of quadrilateral ABCD \( = \text{ Area of } \triangle\text{ABC} + \text{ Area of } \triangle\text{ACD} \)
\( = 33 \text{ cm}^2 + 33 \text{ cm}^2 = 66 \text{ cm}^2 \)
The area of the quadrilateral ABCD is \( 66 \text{ cm}^2 \).
In simple words: A quadrilateral can be split into two triangles by a diagonal. If you know the diagonal's length and the heights (perpendiculars) from the other two corners to that diagonal, you can find the area of each triangle. Add these two triangle areas together to get the total area of the quadrilateral.

Exam Tip: Remember the formula for the area of a quadrilateral when a diagonal and perpendiculars to it are given: \( \text{Area} = \frac{1}{2} \times \text{diagonal} \times (\text{sum of perpendiculars}) \). This simplifies calculations when applicable.