GSEB Class 7 Maths Solutions Chapter 11 પરિમિતિ અને ક્ષેત્રફળ Exercise 11.4

 

Question 1. એક બાગ 90 m લાંબો અને 75 m પહોળો છે. તેની ફરતે ચારે તરફ બહારની બાજુએ 5 m પહોળો રસ્તો બનાવવાનો છે. આ રસ્તાનું ક્ષેત્રફળ શોધો. બાગનું ક્ષેત્રફળ કેટલા હેકટર છે?

Answer: The length (l) of the rectangular garden is 90 m.
The breadth (b) of the rectangular garden is 75 m.
Therefore, the area of the rectangular garden \( = l \times b \)
\( = 90 \text{ m} \times 75 \text{ m} = 6750 \text{ m}^2 \)
Now, 1 m\(^2 = \frac {1}{10,000}\) hectare.
So, \( 6750 \text{ m}^2 = \frac {6750}{10,000} \) hectare \( = 0.675 \) hectare.
A 5 m wide path is to be built around the garden on all four sides.
The length of the rectangular garden including the path \( = (90 + 5 + 5) \text{ m} = 100 \text{ m} \)
The breadth of the rectangular garden including the path \( = (75 + 5 + 5) \text{ m} = 85 \text{ m} \)
The area of the larger rectangle with the path \( = \text{length} \times \text{breadth} \)
\( = 100 \text{ m} \times 85 \text{ m} = 8500 \text{ m}^2 \)
Therefore, the area of the path \( = \text{Area of the larger rectangle} - \text{Area of the garden} \)
\( = 8500 \text{ m}^2 - 6750 \text{ m}^2 = 1750 \text{ m}^2 \)
Hence, the area of the garden is 0.675 hectare, and the area of the path is 1750 m\(^2\).
In simple words: First, calculate the garden's area using its length and breadth. Convert this area to hectares. Then, figure out the total length and breadth when the path is included. Calculate the area of this larger shape. Finally, subtract the garden's area from the larger area to find the path's area.

Exam Tip: Remember to add the path width to both sides (length and breadth) when calculating the outer dimensions. Also, pay attention to unit conversions, especially when asked for hectares.

 

Question 2. 125 m લંબાઈ અને 65 m પહોળાઈ ધરાવતા એક લંબચોરસ બાગની ફરતે ચારે તરફ બહારની બાજુએ 3 m પહોળો રસ્તો છે. આ રસ્તાનું ક્ષેત્રફળ શોધો.

Answer: The length (l) of the rectangular garden is 125 m.
The breadth (b) of the rectangular garden is 65 m.
Therefore, the area of the rectangular garden \( = l \times b \)
\( = 125 \text{ m} \times 65 \text{ m} = 8125 \text{ m}^2 \)
A 3 m wide path is to be built around this garden on all four sides.
The length of the garden including the path \( = (125 + 3 + 3) \text{ m} = 131 \text{ m} \)
The breadth of the garden including the path \( = (65 + 3 + 3) \text{ m} = 71 \text{ m} \)
Therefore, the area of the rectangular garden including the path \( = \text{length} \times \text{breadth} \)
\( = 131 \text{ m} \times 71 \text{ m} = 9301 \text{ m}^2 \)
Now, the area of the path \( = \text{Area of the garden with path} - \text{Area of the garden} \)
\( = 9301 \text{ m}^2 - 8125 \text{ m}^2 = 1176 \text{ m}^2 \)
Hence, the area of the path is 1176 m\(^2\).
In simple words: First, find the garden's area. Then, add the path width to both the length and breadth to get the total outer dimensions, and find that total area. Finally, subtract the garden's area from the total area to discover the area of the path itself.

Exam Tip: Always draw a quick sketch to visualize the problem; this helps avoid errors when adding or subtracting path widths from the dimensions.

 

Question 3. 8 cm લાંબા અને 5 cm પહોળા પૂઠા પર એક ચિત્ર દોરેલું છે. પૂંઠા પર ચિત્રની ફરતે અંદરની બાજુએ ચારે તરફ 1.5 cm હાંસિયો છોડેલો છે. આ હાંસિયાનું ક્ષેત્રફળ શોધો.

Answer: Here, the outer rectangle includes the margin.
The length of the outer rectangle (with margin) is 8 cm.
The breadth of the outer rectangle (with margin) is 5 cm.
Therefore, the area of the outer rectangle \( = 8 \text{ cm} \times 5 \text{ cm} = 40 \text{ cm}^2 \)
The length of the inner rectangle \( = (8 \text{ cm} - 1.5 \text{ cm} - 1.5 \text{ cm}) = 5 \text{ cm} \)
The breadth of the inner rectangle \( = (5 \text{ cm} - 1.5 \text{ cm} - 1.5 \text{ cm}) = 2 \text{ cm} \)
Therefore, the area of the inner rectangle \( = 5 \text{ cm} \times 2 \text{ cm} = 10 \text{ cm}^2 \)
Therefore, the area of the margin \( = (\text{Area of the outer rectangle}) - (\text{Area of the inner rectangle}) \)
\( = 40 \text{ cm}^2 - 10 \text{ cm}^2 = 30 \text{ cm}^2 \)
Hence, the area of the margin is 30 cm\(^2\).
In simple words: Calculate the area of the whole cardboard. Then, subtract the margin's width from both sides of the length and breadth to find the size of the inner picture area. Calculate the area of this inner picture. Finally, subtract the inner picture's area from the total cardboard area to find the margin's area.

Exam Tip: When dealing with inner margins or paths, ensure you subtract the margin width twice (once for each side) from both the length and breadth to get the dimensions of the inner area.

 

Question 4. 5.5 m લાંબા અને 4 m પહોળા ઓરડાની બહારની ચારે બાજુએ 2.25 m પહોળો વરંડો બનાવેલ છે.
(i) વરંડાનું ક્ષેત્રફળ શોધો.
(ii) વરંડાના ભોયતળિયા પર Rs 200 / m\(^2\) પ્રમાણે સિમેન્ટ પાથરવાનો ખર્ શોધો.

Answer:
(i) The length of the room is 5.5 m.
The breadth of the room is 4 m.
Therefore, the area of the room's floor \( = \text{length} \times \text{breadth} \)
\( = 5.5 \text{ m} \times 4 \text{ m} = 22 \text{ m}^2 \)
Now, a 2.25 m wide veranda is built around the room on all four sides.
Therefore, the length of the outer rectangle (with veranda) \( = (5.5 + 2.25 + 2.25) \text{ m} = 10 \text{ m} \)
The breadth of the outer rectangle (with veranda) \( = (4 + 2.25 + 2.25) \text{ m} = 8.5 \text{ m} \)
Therefore, the area of the outer rectangle \( = \text{length} \times \text{breadth} \)
\( = 10 \text{ m} \times 8.5 \text{ m} = 85 \text{ m}^2 \)
Therefore, the area of the veranda \( = \text{Area of the outer rectangle} - \text{Area of the inner rectangle} \)
\( = 85 \text{ m}^2 - 22 \text{ m}^2 = 63 \text{ m}^2 \)
(ii) Now, the cost of cementing is Rs 200 per m\(^2\).
The cost of cementing \( = \text{Rs } (200 \times 63) = \text{Rs } 12,600 \)
In simple words: For (i), first find the room's area. Then, add the veranda's width to both the room's length and breadth to get the total dimensions, and calculate that total area. Subtract the room's area from the total area to find the veranda's area. For (ii), multiply the veranda's area by the cost per square meter to get the total cost.

Exam Tip: Ensure that for outside additions like paths or verandas, you add the width to both ends of each dimension. For cost calculations, clearly state the unit cost and the total area before multiplying.

 

Question 5. 30 m લંબાઈની બાજુવાળા ચોરસ બાગની અંદરની બાજુએ ચારે તરફ 1 m પહોળો રસ્તો બનાવેલ છે.
(i) રસ્તાનું ક્ષેત્રફળ શોધો.
(ii) બાગના રસ્તા સિવાયના ભાગમાં Rs 40 પ્રતિ m\(^2\) પ્રમાણે ઘાસ ઉગાડવાનો ખર્ચ શોધો.

Answer:
(i) The side length of the square garden is 30 m.
The area of the square garden \( = (\text{side})^2 = (30 \text{ m})^2 = 900 \text{ m}^2 \)
A 1 m wide path is built inside this square garden on all four sides.
The side length of the inner square \( = (30 - 1 - 1) \text{ m} = 28 \text{ m} \)
Therefore, the area of the inner square \( = (\text{side})^2 = (28 \text{ m})^2 \)
\( = 784 \text{ m}^2 \)
Now, the area of the path \( = \text{Area of the outer square} - \text{Area of the inner square} \)
\( = 900 \text{ m}^2 - 784 \text{ m}^2 = 116 \text{ m}^2 \)
(ii) Grass is to be grown in the part of the garden without the path.
The cost of growing grass is Rs 40 per m\(^2\).
Therefore, the total cost \( = \text{Rs } (40 \times 784) = \text{Rs } 31,360 \)
Hence, the area of the path is 116 m\(^2\), and the cost of growing grass is Rs 31,360.
In simple words: For (i), calculate the area of the whole square garden. Then, subtract the path width from both sides of the garden's side length to find the inner area without the path. Calculate this inner area. The path's area is the garden's area minus this inner area. For (ii), multiply the inner garden area (where grass will grow) by the cost per square meter to find the total cost.

Exam Tip: When the path is *inside* the main area, you subtract the path's width twice from the original dimensions to find the inner area. When the path is *outside*, you add the width twice.

 

Question 6. 700 m લંબાઈ અને 300 m પહોળાઈ ધરાવતા બાગની મધ્યમાંથી પસાર થતાં અને તેની બાજુઓને સમાંતર એવા 10 m પહોળા બે પરસ્પર લંબ રસ્તા બનાવેલા છે. રસ્તાઓનું ક્ષેત્રફળ શોધો. રસ્તા સિવાયના બાગનું ક્ષેત્રફળ પણ શોધો. તમારા જવાબો હેક્ટરનાં માપમાં આપો.

Answer: The length of the rectangular garden is 700 m.
The breadth of the rectangular garden is 300 m.
The area of the rectangular garden \( = \text{length} \times \text{breadth} \)
\( = 700 \text{ m} \times 300 \text{ m} = 2,10,000 \text{ m}^2 \)
In this garden, as shown in the figure, there are two 10 m wide paths.
The area of these paths \( = (\text{Area of ABCD}) + (\text{Area of PQRS}) - (\text{Area of XYZW}) \)
\( = (700 \times 10) \text{ m}^2 + (300 \times 10) \text{ m}^2 - (10 \times 10) \text{ m}^2 \)
\( = 7000 \text{ m}^2 + 3000 \text{ m}^2 - 100 \text{ m}^2 \)
\( = 10,000 \text{ m}^2 - 100 \text{ m}^2 = 9900 \text{ m}^2 \)
Now, 1 m\(^2 = \frac {1}{10,000}\) hectare.
So, \( 9900 \text{ m}^2 = 0.99 \) hectare.
Now, the area of the garden excluding the paths \( = \text{Area of the garden} - \text{Area of the paths} \)
\( = 2,10,000 \text{ m}^2 - 9900 \text{ m}^2 = 2,00,100 \text{ m}^2 \)
Now, 1 m\(^2 = \frac {1}{10,000}\) hectare.
Therefore, \( 2,00,100 \text{ m}^2 = 20.01 \) hectare.
In simple words: First, calculate the total area of the garden. Then, calculate the area of the two paths separately, and subtract the area of their overlapping intersection to avoid counting it twice. Convert this path area to hectares. Finally, subtract the path area from the total garden area to find the remaining garden area, also converted to hectares.

Exam Tip: For intersecting paths, always subtract the area of the central overlapping square to prevent double-counting. Remember to convert square meters to hectares by dividing by 10,000.

 

Question 7. 90 m લંબાઈ અને 60 m પહોળાઈ ધરાવતા ખેતરના મધ્યમાંથી પસાર થતાં અને તેની બાજુઓને સમાંતર એવા 3 m પહોળા બે પરસ્પર લંબ રસ્તા બનાવેલા છે.
(i) રસ્તાઓએ આવરેલું ક્ષેત્રફળ શોધો.
(ii) Rs 110 / m\(^2\) પ્રમાણે રસ્તાઓ બનાવવાનો ખર્ચ શોધો.

Answer:
(i) The length of the rectangular field is 90 m.
The breadth of the rectangular field is 60 m.
As shown in the figure, there are two 3 m wide paths in this field.
The area of the rectangular path PQRS \( = 90 \text{ m} \times 3 \text{ m} = 270 \text{ m}^2 \)
The area of the rectangular path ABCD \( = 60 \text{ m} \times 3 \text{ m} = 180 \text{ m}^2 \)
The area of the square path XYZW \( = 3 \text{ m} \times 3 \text{ m} = 9 \text{ m}^2 \)
The area of the paths \( = \text{Area of PQRS} + \text{Area of ABCD} - \text{Area of XYZW} \)
\( = 270 \text{ m}^2 + 180 \text{ m}^2 - 9 \text{ m}^2 \)
\( = 450 \text{ m}^2 - 9 \text{ m}^2 = 441 \text{ m}^2 \)
(ii) Now, the cost of constructing paths is Rs 110 per m\(^2\).
Therefore, the cost of constructing 441 m\(^2\) of path \( = \text{Rs } (110 \times 441) \)
\( = \text{Rs } 48,510 \)
In simple words: For (i), calculate the area of the two main paths. Then, subtract the area of the small square where they overlap, to avoid counting it twice. For (ii), multiply the total area of the paths by the given cost per square meter to find the total construction cost.

Exam Tip: Remember to use the field's length for the length-wise path and the field's breadth for the breadth-wise path. The overlapping square's side is equal to the path's width.

 

Question 8. પ્રજ્ઞાએ 4 cm ત્રિજ્યાવાળી એક વર્તુળાકાર નળીની ફરતે દોરી વીંટાળી (નીચેની આકૃતિ) અને જરૂરી લંબાઈની દોરી કાપી લીધી. હવે તેણે એ જ દોરીને 4 cmની બાજુ ધરાવતા ચોરસ ડબાની આસપાસ વીંટાળી (આકૃતિ જુઓ). શું તેની પાસે દોરી વધી હશે? (\( \pi = 3.14 \))

Answer: The radius (r) of the circular pipe is 4 cm.
Therefore, the circumference of this pipe \( = 2 \pi r \)
\( = 2 \times 3.14 \times 4 \text{ cm} = 25.12 \text{ cm} \)
Hence, the length of the string wound around the pipe is 25.12 cm.
Now, this string is wound around a square box.
The side length of the square box is 4 cm.
Therefore, the perimeter of the square box \( = 4 \times 4 \text{ cm} = 16 \text{ cm} \)
Thus, the length of the string wound around the square is 16 cm.
Now, 25.12 cm \( > \) 16 cm and \( (25.12 - 16) \text{ cm} = 9.12 \text{ cm} \)
Yes, Pragna will have 9.12 cm of string left after winding it around the square box.
In simple words: First, calculate how much string is needed to go around the circular pipe (its circumference). Then, calculate how much string is needed to go around the square box (its perimeter). Compare the two lengths. If the string from the circle is longer than what's needed for the square, then there will be some string left over.

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

 

Question 9. બાજુની આકૃતિમાં એક લંબચોરસ જમીન પરની લોનની મધ્યમાં ફૂલોનો એક વર્તુળાકાર બાગ દર્શાવેલો છે. (\( \pi = 3.14 \))
(i) બધી જમીનનું ક્ષેત્રફળ શોધો.
(ii) બાગનું ક્ષેત્રફળ શોધો.
(iii) બાગ સિવાયની જગ્યાનું ક્ષેત્રફળ શોધો.
(iv) બાગનો પરિઘ શોધો.

Answer: The length of the rectangular land is 10 m, and the breadth is 5 m.
(i) The area of the entire land \( = \text{length} \times \text{breadth} \)
\( = 10 \text{ m} \times 5 \text{ m} = 50 \text{ m}^2 \)
(ii) The radius (r) of the circular flower garden is 2 m.
Therefore, the area of the circular flower garden \( = \pi r^2 \)
\( = 3.14 \times 2 \times 2 = 12.56 \text{ m}^2 \)
(iii) The area of the space excluding the flower garden in the land \( = 50 \text{ m}^2 - 12.56 \text{ m}^2 \)
\( = 37.44 \text{ m}^2 \)
(iv) The circumference of the circular flower garden \( = 2 \pi r \)
\( = 2 \times 3.14 \times 2 \text{ m} = 12.56 \text{ m} \)
In simple words: For (i), multiply the land's length by its breadth to get its total area. For (ii), use the formula \( \pi r^2 \) to find the area of the circular garden. For (iii), subtract the garden's area from the total land area to find the remaining space. For (iv), use the formula \( 2 \pi r \) to find the distance around the circular garden.

Exam Tip: Differentiate between area (\( \pi r^2 \)) and circumference (\( 2 \pi r \)) formulas for circles. Remember the units for area (square meters) and length (meters).

 

Question 10. નીચેની આકૃતિઓમાં છાયાંકિત ભાગનું ક્ષેત્રફળ શોધો.

Answer:
(i)

The length of rectangle ABCD \( = 10 \text{ cm} + 8 \text{ cm} = 18 \text{ cm} \)
And the breadth \( = 6 \text{ cm} + 4 \text{ cm} = 10 \text{ cm} \)
The area of rectangle ABCD \( = \text{length} \times \text{breadth} \)
\( = 18 \text{ cm} \times 10 \text{ cm} = 180 \text{ cm}^2 \)
The area of triangle AAEF \( = \frac {1}{2} \times \text{base} \times \text{perpendicular height} \)
\( = \frac {1}{2} \times 6 \times 10 = 30 \text{ cm}^2 \)
The area of triangle ACBE \( = \frac {1}{2} \times \text{base} \times \text{perpendicular height} \)
\( = \frac {1}{2} \times 8 \times 10 = 40 \text{ cm}^2 \)
Now, the area of the shaded portion \( = (\text{Area of rectangle ABCD}) - (\text{Area of AAEF} + \text{Area of ACBE}) \)
\( = (180 \text{ cm}^2) - (30 \text{ cm}^2 + 40 \text{ cm}^2) \)
\( = 180 \text{ cm}^2 - 70 \text{ cm}^2 = 110 \text{ cm}^2 \)
Hence, the area of the shaded portion is 110 cm\(^2\).
(ii)Here, the given figure PQRS is a square.
The side length of the square is 20 cm.
The area of square PQRS \( = (\text{side})^2 \)
\( = 20 \text{ cm} \times 20 \text{ cm} = 400 \text{ cm}^2 \)
The area of right-angled triangle AQPT \( = \frac {1}{2} \times \text{base} \times \text{perpendicular height} \)
\( = \frac {1}{2} \times 20 \times 10 = 100 \text{ cm}^2 \)
The area of right-angled triangle ATSU \( = \frac {1}{2} \times \text{base} \times \text{perpendicular height} \)
\( = \frac {1}{2} \times 10 \times 10 = 50 \text{ cm}^2 \)
The area of right-angled triangle AQRU \( = \frac {1}{2} \times \text{base} \times \text{perpendicular height} \)
\( = \frac {1}{2} \times 10 \times 20 = 100 \text{ cm}^2 \)
Now, the area of the shaded portion \( = (\text{Area of square PQRS}) - (\text{Area of AQPT} + \text{Area of ATSU} + \text{Area of AQRU}) \)
\( = (400 \text{ cm}^2) - (100 \text{ cm}^2 + 50 \text{ cm}^2 + 100 \text{ cm}^2) \)
\( = 400 \text{ cm}^2 - 250 \text{ cm}^2 = 150 \text{ cm}^2 \)
Hence, the area of the shaded portion is 150 cm\(^2\).
In simple words: For (i), find the area of the main rectangle. Then, identify the unshaded triangles and calculate their areas using the given base and height. Subtract the sum of these triangle areas from the rectangle's area to get the shaded area. For (ii), find the area of the main square. Identify the three unshaded triangles and calculate their areas. Subtract the sum of these three triangle areas from the square's area to find the shaded area.

Exam Tip: For composite shapes, break them down into simpler geometric figures (rectangles, squares, triangles). Clearly identify the shaded and unshaded parts to ensure correct addition or subtraction of areas.

 

Question 11. ચતુષ્કોણ ABCDનું ક્ષેત્રફળ શોધો. અહીં, AC = 22 cm, BM = 3 cm, DN = 3 cm અને \( \overline{\mathrm{BM}} \perp \overline{\mathrm{AC}} \) અને \( \overline{\mathrm{DN}} \perp \overline{\mathrm{AC}} \)

Answer: Here, \( AC = 22 \text{ cm} \), \( BM = 3 \text{ cm} \), \( DN = 3 \text{ cm} \).
The area of triangle ABC \( = \frac {1}{2} \times \text{base} \times \text{perpendicular height} \)
\( = \frac {1}{2} \times AC \times BM \)
\( = \frac {1}{2} \times 22 \times 3 = 33 \text{ cm}^2 \)
The area of triangle ADC \( = \frac {1}{2} \times \text{base} \times \text{perpendicular height} \)
\( = \frac {1}{2} \times AC \times DN \)
\( = \frac {1}{2} \times 22 \times 3 = 33 \text{ cm}^2 \)
The area of quadrilateral ABCD \( = \text{Area of triangle ABC} + \text{Area of triangle ADC} \)
\( = 33 \text{ cm}^2 + 33 \text{ cm}^2 \)
\( = 66 \text{ cm}^2 \)
Hence, the area of quadrilateral ABCD is 66 cm\(^2\).
In simple words: A quadrilateral can be split into two triangles using a diagonal. Calculate the area of each triangle using the diagonal as the base and the perpendiculars from the other vertices as heights. Add the areas of these two triangles together to get the total area of the quadrilateral.

Exam Tip: Remember that the area of a quadrilateral can be found by dividing it into two triangles along one of its diagonals. The area of each triangle is then \( \frac{1}{2} \times \text{diagonal length} \times \text{perpendicular height} \).