\begin{array}{l}
\text {Q1. Find the perimeters of the rectangles whose lengths and breadths are given below: } \\
\text {(i) 7 cm, 5 cm} \quad
\text {(ii) 5 cm, 4 cm} \quad
\text {(iii) 7.5 cm, 4.5 cm}
\end{array}\begin{array}{l}
\text {Sol. } \\\text {(i) We know that the perimeter of a rectangle }=2(l+b) \\
\text {Given that l=7 cm and b=5 cm} \\
\text {The perimeter of the rectangle }=2(7+5)=2 \times 12=24 \text { cm} \\ \\\text {(ii) We know that the perimeter of a rectangle }=2(l+b) \\
\text {Given that l=5 cm and b=4 cm} \\
\text {The perimeter of the rectangle }=2(5+4)=2 \times 9=18 \text { cm} \\ \\\text {(iii) We know that the perimeter of a rectangle }=2(l+b) \\
\text {Given that l=7.5 cm and b=4.5 cm} \\
\text {The perimeter of the rectangle }=2(7.5+4.5)=2 \times 12=24 \text { cm} \\ \\
\end{array}\begin{array}{l}
\text {Q2. Find the perimeters of the squares whose sides are given below: } \\
\text {(i) 10 cm} \quad
\text {(ii) 5 m} \quad
\text {(iii) 115.5 cm}
\end{array}\begin{array}{l}
\text {Sol. } \\
\text {(i) We know that the perimeter of a square }=4 \times \text {Side} \\
\text {Given that Side =10 cm} \\
\text {The perimeter of a square }=4 \times 10=40 \text { cm} \\ \\\text {(ii) We know that the perimeter of a square }=4 \times \text {Side} \\
\text {Given that Side =5 m} \\
\text {The perimeter of a square }=4 \times 5=20 \text { m} \\ \\\text {(iii) We know that the perimeter of a square }=4 \times \text {Side} \\
\text {Given that Side =115.5 cm} \\
\text {The perimeter of a square }=4 \times 115.5=462 \text { cm} \\\end{array}\begin{array}{l}
\text {Q3. Find the side of the square whose perimeter is: } \\
\text {(i) 16 m} \quad
\text {(ii) 40 cm} \quad
\text {(iii) 22 cm}
\end{array}\begin{array}{l}
\text {Sol. } \\\text {(i) We know that side of a square } =\frac {\text {Perimeter}}{4} \\
\text {Given that perimeter of squre }=16 \text { m} \\
\therefore \text {Side of the square }=\frac {16}{4}=4 \text { m} \\ \\\text {(ii) We know that side of a square } =\frac {\text {Perimeter}}{4} \\
\text {Given that perimeter of squre }=40 \text { cm} \\
\therefore \text {Side of the square }=\frac {40}{4}=10 \text { cm} \\ \\\text {(iii) We know that side of a square } =\frac {\text {Perimeter}}{4} \\
\text {Given that perimeter of squre }=22 \text { cm} \\
\therefore \text {Side of the square }=\frac {22}{4}=5.5 \text { cm} \\ \\\end{array}
\begin{array}{l}
\text {Q4. Find the breadth of the rectangle whose perimeter is 360 cm and whose length is } \\
\text {(i) 116 cm } \quad
\text {(ii) 140 cm } \quad
\text {(iii) 102 cm }
\end{array}\begin{array}{l}
\text {Sol. } \\\text {We know that the perimeter of a rectangle (P) }=2(l+ b) \\
\Rightarrow \quad \text {Breadth of rectangle } b = \frac {P}{2} – l \\\text {(i) Given perimeter } =360 \text { cm and length } = 116 \text { cm} \\
\therefore \text {Breadth of the rectangle} =\frac {360}{2} – 116 = 180 – 116 = 64 \text { cm} \\ \\\text {(ii) Given perimeter } =360 \text { cm and length } = 140 \text { cm} \\
\therefore \text {Breadth of the rectangle} =\frac {360}{2} – 140 = 180 – 140 = 40 \text { cm} \\ \\\text {(i) Given perimeter } =360 \text { cm and length } = 102 \text { cm} \\
\therefore \text {Breadth of the rectangle} =\frac {360}{2} – 102 = 180 – 102 = 78 \text { cm} \\ \\\end{array}\begin{array}{l}
\text {Q5. A rectangular piece of lawn is 55 m wide and 98 m long. Find the length of the fence around it.}
\end{array}\begin{array}{l}
\text {Sol. } \\\text {Given Breadth of the rectangular lawn }=55 \text { m} \\
\text {And lreadth of the rectangular lawn }=98 \text { m} \\\text {We know that the perimeter of a rectangle (P) }=2(l+ b) \\
\therefore \text {Perimeter of rectangular lawn } =2(98+55) =2 \times 153=306 \text { m} \\
\text {Hence, the length of the fence around it is 306 m.}\end{array}\begin{array}{l}
\text {Q6. The side of a square field is 65 m. What is the length of the fence required all around it? }
\end{array}\begin{array}{l}
\text {Sol. } \\\text {Given that side of a square field } =65 \text { m} \\
\text {Perimeter of square field } =4 \times \text {Side of the square} \\
\Rightarrow \text {Perimeter of square field }=4 \times 65=260 \text { m} \\
\text {Hence, the length of the fence required around the field is 260 m.}\end{array}\begin{array}{l}
\text {Q7. Two sides of a triangle are 15 cm and 20 cm. The perimeter of the } \\
\text { triangle is 50 cm. What is the third side? }
\end{array}\begin{array}{l}
\text {Sol. } \\\text {Given that two sides of triangle are 15 cm and 20 cm. } \\
\text {And perimeter of triangle } = 50 \text { cm} \\
\text {We know that perimeter of a triangle } = \text {Sum of all three sides of a triangle } \\
\Rightarrow 50 = 15 + 20 + \text {Length of third side} \\\text {Length of third side } = 50-(15+20)=15 \text { cm} \\
\text {Hence, the length of third side is 15 cm.}
\end{array}\begin{array}{l}
\text {Q8. A wire of length 20 m is to be folded in the form of a rectangle. How many } \\
\text { rectangles can be formed by folding the wire if the sides are positive integers in metres? }
\end{array}\begin{array}{l}
\text {Sol. } \\\text {Given that length of wire 20 m is folded in the form of rectangle.} \\
\Rightarrow \text {Perimeter of rectangle} =20 \text { m} \\\text {We know that the perimeter of a rectangle (P) }=2(l+ b) \\
\Rightarrow 20 =2(l+b) \\
\Rightarrow l+b = \frac {20}{2} \\
\Rightarrow l+b = 10 \text { m} \\\text {If the sides of rectangle are positive integers in metres then possible dimensions are } \\
\text { (1 m, 9 m),(2 m, 8 m),(3 m, 7 m), (4 m, 6 m),(5 m, 5 m) }
\text {Hence, five rectangles can be formed using the given wire.}\end{array}\begin{array}{l}
\text {Q9. A square piece of land has each side equal to 100 m. If 3 layers of } \\
\text { metal wire has to be used to fence it, what is the length of the wire needed? }
\end{array}\begin{array}{l}
\text {Sol. } \\\text {Given that side of a square land } =100 \text { m} \\
\text {Perimeter of square field } =4 \times \text {Side of the square} \\
\Rightarrow \text {Perimeter of square field }=4 \times 100=400 \text { m} \\\text {The length of wire required to fence three layers around the square field} =3 \times \text {Perimeter of square field} \\
=3 \times 400=1200 \text { m} \\\text {Hence, the length of wire required to fence 3 layers is 1200 m.}
\end{array}\begin{array}{l}
\text {Q10. Shikha runs around a square of side 75 m. Priya runs around a } \\
\text { rectangle with length 60 m and breadth 45 m. Who covers the smaller distance? }
\end{array}\begin{array}{l}
\text {Sol. } \\
\text {Given that Shikha runs around a square of side }=75 \text {m} \\
\text {Perimeter of square field } =4 \times \text {Side of the square} \\
\Rightarrow \text {Distance covered by Shikha } = \text {Perimeter of square } \\
=4 \times 75=300 \text { m} \\\text {Given that Priya runs around a rectangle having
length }=60 \text{ m} \\
\text {And Breadth }=45 \text { m} \\\text {We know that the perimeter of a rectangle (P) }=2(l+ b) \\
\Rightarrow \text {Distance covered by Priya } = \text {Perimeter of a rectangle (P) }=2(60+ 45) =2 \times 105=210 \text { m} \\\text {Hence, Priya covers the smaller distance of 210 m}
\end{array}\begin{array}{l}
\text {Q11. The dimensions of a photographs are 30 cm \times 20 cm. } \\
\text { What length of wooden frame is needed to frame the picture? }
\end{array}\begin{array}{l}
\text {Sol. } \\\text {Given that dimensions of a photographs }=30 \text { cm} \times 20 \text { cm} \\\Rightarrow \text {Required length of the wooden frame } = \text {Perimeter of the photograph } \\
=2(l+b) =2(30+20)=2 \times 50=100 \text { cm} \\
\text {Hence, the length of the wooden frame required to frame the picture is 100 cm.}\end{array}\begin{array}{l}
\text {Q12. The length of a rectangular field is 100 m. If its perimeter is 300 m, what is its breadth? }
\end{array}\begin{array}{l}
\text {Sol. } \\\text {Given that length of rectangular field} =100 \text { m} \\
\text {And Perimeter of rectangular field } =300 \text { m} \\\text {We know that the perimeter of a rectangle (P) }=2(l+ b) \\
\Rightarrow \quad \text {Breadth of rectangle } b = \frac {P}{2} – l \\
\Rightarrow \quad \text {Breadth of rectangular field} b = \frac {300}{2} – 100 =150-100=50 \text { m} \\\text {Hence, the breadth of the rectangular field is 50 m.}\end{array}\begin{array}{l}
\text {Q13. To fix fence wires in a garden, 70 m long and 50 m wide, Arvind bought } \\
\text { metal pipes for posts. he fixed a post every 5 metres apart. Each post was 2 m long. } \\
\text { What is total length of the pipes he bought for the posts? }\end{array}\begin{array}{l}
\text {Sol. } \\
\text {Given Length of garden }=70 \text { m} \\
\text {And Breadth of garden}=50 \text { m} \\\text {We know that the perimeter of a rectangle (P) }=2(l+ b) \\\therefore \text {Perimeter of garden } =2(70+50)=2 \times 120=240 \text { m} \\\text {Givne that Arvind fixes a post every 5 metres apart} \\
\therefore \text {No. of posts required } =\frac {\text {Perimter of garden}}{\text {Distance between two posts}} \\
=\frac {240}{5}=48 \text { posts} \\\text {Given that length of 1 post } =2 \text { m} \\
\Rightarrow \text {The length of 48 post } =2 \times 48 = 96 \text { m} \\\text {Hence, the total length of the pipes he bought for the posts is 96 m.}\end{array}\begin{array}{l}
\text {Q14. Find the cost of fencing a rectangular park of length 175 m } \\
\text { and breadth 125 m at the rate of ₹ 12 per meter. }
\end{array}\begin{array}{l}
\text {Sol. } \\
\text {Given Length of rectangular park}=175 \text { m} \\
\text {And Breadth of rectangular park}=125 \text { m} \\\text {We know that the perimeter of a rectangle (P) }=2(l+ b) \\\therefore \text {Perimeter of rectangular park } =2(175 + 125)=2 \times 300=600 \text { m} \\\text {Given that the cost of fencing a distance of 1 meter }=\text {Rs. } 12 \\
\Rightarrow \text {The cost of fencing a distance of 600 meters }=\text {Rs. } 12 \times 600 = \text {Rs. } 7200 \\\text {Hence, total cost of fencing the rectangular park is Rs. 7200.}\end{array}\begin{array}{l}
\text {Q15. The perimeter of a regular pentagon is 100 cm. How long is each side? }
\end{array}\begin{array}{l}
\text {Sol. } \\\text {We know that a regular pentagon is a closed polygon having 5 sides of same length. } \\\text {Given that perimeter of regular pentagon } =100 \text { cm} \\
\text {Perimeter of regular pentagon } =5 \times \text {Side of the regular pentagon} \\
\Rightarrow 100 =5 \times \text {Side of the regular pentagon} \\
\Rightarrow \text {Side of the regular pentagon} = \frac {100}{5} = 20 \text { cm} \\\text {Hence, the side of the regular pentagon is 20 cm.}\end{array}
\begin{array}{l}
\text {Q16. Find the perimeter of a regular hexagon with each side measuring 8 m.}
\end{array}\begin{array}{l}
\text {Sol. } \\\text {We know that a regular polygon is a closed polygon which has six sides of same length. } \\
\text {Given that side of the regular hexagon }=8 \text { m} \\
\text {Perimeter of regular hexagon } =6 \times \text {Side of the regular hexagon } \\
=6 \times 8=48 \text { m} \\
\text {Hence, the perimeter of a regular hexagon is 48 m.}\end{array}\begin{array}{l}
\text {Q17. A rectangular piece of land measure 0.7 km by 0.5 km. Each side } \\
\text { is to be fenced with four rows of wires. What length of the wire is needed? }
\end{array}\begin{array}{l}
\text {Sol. } \\
\text {Given Length of rectangular piece of land}=0.7 \text { km} \\
\text {And Breadth of rectangular piece of land}=0.5 \text { km} \\\text {We know that the perimeter of a rectangle (P) }=2(l+ b) \\\therefore \text {Perimeter of rectangular piece of land } =2(0.7+0.5)=2 \times 1.2=2.4 \text { km} \\\text {The length of wire needed to fence the land with 4 rows of wire }=4 \times 2.4=9.6 \text { km} \\
\text {Hence, the length of wire required to fence the land with 4 rows is 9.6 km.}\end{array}\begin{array}{l}
\text {Q18. Avneet buys 9 square paving slabs, each with a side of } \frac {1}{2} m. \\
\text {He lays them in the form of a square.} \\
\end{array}\begin{array}\text {(i) What is the perimeter of his arrangement? } \\
\text {(ii) Shari does not like his arrangement. She gets him to lay them out } \\
\text { like a cross. What is the perimeter of her arrangement? } \\
\text {(iii) Which has greater perimeter? } \\
\text {(iv) Avneet wonders, if there is a way of getting an even greater } \\
\text { perimeter. Can you find a way of doing this? } \\
\text {(The paving slabs must meet along complete edges they cannot be broken }
\end{array}\begin{array}{l}
\text {Sol. (i) Given that length of each side of the slab } =\frac {1}{2} \text { m} \\
\text {One side of the square is formed by three slabs in a square arrangement.} \\
\text {Length of side }=3 \times \frac {1}{2}=\frac {3}{2} \text { m} \\
\text {The perimeter of the square arrangement } =4 \times \frac {3}{2} \text { m} =6 \text { m} \\ \\
\end{array}\begin{array}{l}
\text {(ii) From the figure, cross arrangement has 8 sides which form } \\
\text { periphery of the arrangement and measure 1 m each.} \\\text {It also has 4 sides which measure } \frac {1}{2} \text { m} \text { each. } \\
\therefore \text {Perimeter of the cross arrangement } =1+\frac {1}{2}+1+1+\frac {1}{2}+1+1+\frac {1}{2}+1+1+\frac {1}{2}+1=8+2=10 \text { m} \\ \\\text {(iii) We know that perimeter of cross arrangement }=10 \text { m} \\
\text {Perimeter of square arrangement }=6 \text { m} \\
\text {Hence, the perimeter of cross arrangement is greater than the perimeter of square arrangement.} \\ \\
\text {(iv) No, Avneet cannot arrange the slabs having perimeter more than 1 m.}
\end{array}
