Q1. Find the area, in square metres, of a rectangle whose
(i) Length =5.5 m, breadth =2.4 m
(ii) Length =180 cm, breadth =150 cm
Sol. \begin{array}{l} \text {(i) Area of rectangle} = \text {Length} \times \text {Breadth} \\ =5.5 \times 2.4 =13.2 m^{2} \\ \\\text {(ii) Area of rectangle} = \text {Length} \times \text {Breadth} \\ =1.8 \times 1.5=2.7 m^{2} \\ \\\end{array}Q2. Find the area, in square centimetres, of a square whose side is
(i) 2.6 cm
(ii) 1.2 dm
Sol. \begin{array}{l}\text {(i) Area of square } =(\text {Side})^{2} =(2.6)^2 =6.76 cm^{2} \\ \\\text {(ii) Given Side of the square }=1.2 \text{ dm} \\ =1.2 \times 10 \text { cm} \quad [\because 1 \text { dm} =10 \text{ cm}] \\ =12 \text{ cm} \\\text {Area of square } =(\text {Side})^{2} ={12}^{2} =144 cm^{2} \\ \\ \end{array}Q3. Find in square metres, the area of a square of side 16.5 dam.Sol. \begin{array}{l} \text { Given Side of the square }=16.5 \text{ dam} \\ =16.5 \times 10 \text { m} \quad [\because \quad 1 \text { dam} =10 \text{ m}] \\ =165 \text{ m} \\\text {Area of square } =(\text {Side})^{2} =(165)^{2} =27225 m^{2} \\ \\ \end{array}Q4. Find the area of a rectangular field in acres whose sides are:
(i) 200 m and 125 m
(ii) 75 m 5 dm and 120 mSol. \begin{array}{l} \text {(i) Area of rectangle} = \text {Length} \times \text {Breadth} \\ =200 \times 125 =25000 m^{2} \\ =25000 \times \frac {1}{100} \text { acre} \quad [\because \quad 100 \text { m}^{2} =1 \text { acre}] \\ =250 \text { acre} \\ \\\text {(ii) Given Length of rectangle = 75 m 5 dm } \\ = 75 + 0.5 \text { m} \quad [\because 1 \text { dm} =10 \text { cm}=0.1 \text { m}] \\ = 75.5 \text { m} \\ \\\text {Breadth of rectangle } =120 \text { m} \\\text {Area of rectangle} = \text {Length} \times \text {Breadth} \\ =75.5 \times 120 =9060 m^{2} \\=9060 \times \frac {1}{100} \text { acre} \quad [\because 100 \text { m}^{2} =1 \text { acre}] \\ =90.6 \text { acre} \\\end{array}Q5. Find the area of a rectangular field in hectares whose sides are:
(i) 125 m and 400 m
(ii) 75 m 5 dm and 120 m
Sol. \begin{array}{l} \text {(i) Area of rectangle} = \text {Length} \times \text {Breadth} \\ =125 \times 400 =50000 m^{2} \\ =50000 \times \frac {1}{10000} \text { acre} \quad [\because 10000 \text { m}^{2} =1 \text { hectare}] \\ =5 \text { hectare} \\ \\\text {(ii) Given Length of rectangle = 75 m 5 dm } \\ = 75 + 0.5 \text { m} \quad \quad [\because \quad 1 \text { dm} =10 \text { cm} =0.1 \text { m}] \\ = 75.5 \text { m} \\ \\\text {Breadth of rectangle } =120 \text { m} \\\text {Area of rectangle} = \text {Length} \times \text {Breadth} \\ =75.5 \times 120 =9060 m^{2} \\ =9060 \times \frac {1}{10000} \text { hectare} \quad [\because 10000 \text { m}^{2} =1 \text { hectare}] \\ =0.906 \text { hectare} \\\end{array}Q6. A door of dimensions 3 m × 2 m is on the wall of dimension 10 m × 10 m. Find the cost of painting the wall if rate of painting is Rs. 2.50 per sq. m.
Sol. Length of the door = 3 m
Breadth of the door = 2 m
Side of the wall = 10 m
\begin{array}{l}\text {Total area of the wall } =10 \text { m} \times 10 \text { m} =100 \text { m}^{2} \\ \text {Area of the door } = \text {Length} \times \text {Breadth} =3 \text { m} \times 2 \text { m}=6 \text { m}^{2} \\ \text {Hence, available area of the wall for painting } \\ = \text {Total Area of the wall} – \text {Area of the door} \\ = 100 – 6 \text { m}^{2} =94 \text { m}^{2} \\ \end{array}Given rate of painting per square metre = Rs. 2.50
Therefore, the cost of painting the wall = 94 × 2.50 = Rs. 235
Q7. A wire is in the shape of a rectangle. Its length is 40 cm and breadth is 22 cm. If the same wire is bent in the shape of a square, what will be the measure of each side. Also, find which side encloses more area?
Sol.
Length of the wire = 40 cm
Breadth of the wire = 22 cm Therefore, perimeter of the rectangle = 2(Length + Breadth) = 2(40 + 22) = 124 cm.
When wire is bent into a square, the permiter will remain same.
Therefore, perimeter of the square = 124 cm.
Lets assume that side of square be x cm.
Perimeter of square = 4x\begin{array}{l} \Rightarrow \quad 4x = 124 \\ \Rightarrow \quad x = 31 \\ \text {Side }=\frac{124}{4}=31 \text { cm} \\ \text {Area of the rectangle } =\text { Length } \times \text { Breadth } =40 \times 22 =880 \text { cm}^{2} \\ \text {Area of the square }=(\text { Side })^{2}=(31)^{2}=961 \text { cm}^{2} \\ \therefore \text {Area of square is more.} \end{array}Q8. How many square metres of glass will be required for a window, which has 12 panes, each pane measuring 25 cm by 16 cm ?
Sol. Length of glass pane = 25 cm.
Breadth of glass pane = 16 cm.
\begin{array}{l}\text {Area of glass pane }=\text { Length } \times \text { Breadth } =25 \times 16 =400 \text { cm}^{2} \\ \Rightarrow \text {Area of 12 panes } =12 \times 400=4800 \text { cm}^{2} \\ =4800 \times \frac {1}{10000} \text { m}^{2} \quad \quad [\because 1 \text { cm}^{2} = \frac {1}{10000} \text { m}^{2}] \\ =0.48 \text { m}^{2} \end{array}Q9. A marble tile measures 10 cm x 12 cm. How many tiles will be required to cover a wall of size 3 m × 4 m ? Also, find the total cost of the tiles at the rate of Rs. 2 per tile.
Sol. \begin{array}{l} \text {Area of the wall }=\text { Length } \times \text { Breadth } =3 \times 4 =12 \text { m}^{2} \\\text {Area of 1 marble tile }=\text { Length } \times \text { Breadth } =10 \times 12 =120 \text { cm}^{2} \\=120 \times \frac {1}{10000} \text { m}^{2} \quad \quad [\because 1 \text { cm}^{2} = \frac {1}{10000} \text { m}^{2}] \\ =0.012 \text { m}^{2} \\\text {Total no. of tiles required to cover the wall} = \frac {\text {Area of the wall}}{\text {Area of marble tile}} \\ =\frac {12}{0.012} =\frac {12 \times 1000}{12} = 1000 \text { tiles}\\\text {Given cost of one tile }=\text {Rs. } 2 \\ \therefore \text {Total cost to tile the wall } = 1000 \times 2 = \text {Rs.} 2000\end{array}Q10. A table top is 9 dm 5 cm long, 6 dm 5 cm broad. What will be the cost to polish it at the rate of 20 paise per square centimetre?
Sol. \begin{array}{l} \text {Length of table top } =9 \text { dm } 5 \text { cm} =9 \times 10 + 5 ) \text { cm} \quad [\because 1 \text { dm} =10 \text{ cm}] \\ =95 \text { cm} \\\text {Breadth of table top } =6 \text { dm } 5 \text { cm} =6 \times 10 + 5 ) \text { cm} \quad [\because 1 \text { dm} =10 \text{ cm}] \\ =65 \text { cm} \\\text {Area of the table top }=\text { Length } \times \text { Breadth } =95 \times 65 =6175 \text { cm}^{2} \\\text {Rate of polishing per square centimetre }=20 \text { paisa} \\ \therefore \text {Total cost of polishing } = 6175 \times 20 \text { paisa} \\ = \text {Rs.} 6175 \times 0.20 \quad [\because 1 \text { Rs} =100 \text{ pasia}] \\ = \text {Rs.} 1235\end{array}
(i) Length =5.5 m, breadth =2.4 m
(ii) Length =180 cm, breadth =150 cm
Sol. \begin{array}{l} \text {(i) Area of rectangle} = \text {Length} \times \text {Breadth} \\ =5.5 \times 2.4 =13.2 m^{2} \\ \\\text {(ii) Area of rectangle} = \text {Length} \times \text {Breadth} \\ =1.8 \times 1.5=2.7 m^{2} \\ \\\end{array}Q2. Find the area, in square centimetres, of a square whose side is
(i) 2.6 cm
(ii) 1.2 dm
Sol. \begin{array}{l}\text {(i) Area of square } =(\text {Side})^{2} =(2.6)^2 =6.76 cm^{2} \\ \\\text {(ii) Given Side of the square }=1.2 \text{ dm} \\ =1.2 \times 10 \text { cm} \quad [\because 1 \text { dm} =10 \text{ cm}] \\ =12 \text{ cm} \\\text {Area of square } =(\text {Side})^{2} ={12}^{2} =144 cm^{2} \\ \\ \end{array}Q3. Find in square metres, the area of a square of side 16.5 dam.Sol. \begin{array}{l} \text { Given Side of the square }=16.5 \text{ dam} \\ =16.5 \times 10 \text { m} \quad [\because \quad 1 \text { dam} =10 \text{ m}] \\ =165 \text{ m} \\\text {Area of square } =(\text {Side})^{2} =(165)^{2} =27225 m^{2} \\ \\ \end{array}Q4. Find the area of a rectangular field in acres whose sides are:
(i) 200 m and 125 m
(ii) 75 m 5 dm and 120 mSol. \begin{array}{l} \text {(i) Area of rectangle} = \text {Length} \times \text {Breadth} \\ =200 \times 125 =25000 m^{2} \\ =25000 \times \frac {1}{100} \text { acre} \quad [\because \quad 100 \text { m}^{2} =1 \text { acre}] \\ =250 \text { acre} \\ \\\text {(ii) Given Length of rectangle = 75 m 5 dm } \\ = 75 + 0.5 \text { m} \quad [\because 1 \text { dm} =10 \text { cm}=0.1 \text { m}] \\ = 75.5 \text { m} \\ \\\text {Breadth of rectangle } =120 \text { m} \\\text {Area of rectangle} = \text {Length} \times \text {Breadth} \\ =75.5 \times 120 =9060 m^{2} \\=9060 \times \frac {1}{100} \text { acre} \quad [\because 100 \text { m}^{2} =1 \text { acre}] \\ =90.6 \text { acre} \\\end{array}Q5. Find the area of a rectangular field in hectares whose sides are:
(i) 125 m and 400 m
(ii) 75 m 5 dm and 120 m
Sol. \begin{array}{l} \text {(i) Area of rectangle} = \text {Length} \times \text {Breadth} \\ =125 \times 400 =50000 m^{2} \\ =50000 \times \frac {1}{10000} \text { acre} \quad [\because 10000 \text { m}^{2} =1 \text { hectare}] \\ =5 \text { hectare} \\ \\\text {(ii) Given Length of rectangle = 75 m 5 dm } \\ = 75 + 0.5 \text { m} \quad \quad [\because \quad 1 \text { dm} =10 \text { cm} =0.1 \text { m}] \\ = 75.5 \text { m} \\ \\\text {Breadth of rectangle } =120 \text { m} \\\text {Area of rectangle} = \text {Length} \times \text {Breadth} \\ =75.5 \times 120 =9060 m^{2} \\ =9060 \times \frac {1}{10000} \text { hectare} \quad [\because 10000 \text { m}^{2} =1 \text { hectare}] \\ =0.906 \text { hectare} \\\end{array}Q6. A door of dimensions 3 m × 2 m is on the wall of dimension 10 m × 10 m. Find the cost of painting the wall if rate of painting is Rs. 2.50 per sq. m.
Sol. Length of the door = 3 m
Breadth of the door = 2 m
Side of the wall = 10 m
\begin{array}{l}\text {Total area of the wall } =10 \text { m} \times 10 \text { m} =100 \text { m}^{2} \\ \text {Area of the door } = \text {Length} \times \text {Breadth} =3 \text { m} \times 2 \text { m}=6 \text { m}^{2} \\ \text {Hence, available area of the wall for painting } \\ = \text {Total Area of the wall} – \text {Area of the door} \\ = 100 – 6 \text { m}^{2} =94 \text { m}^{2} \\ \end{array}Given rate of painting per square metre = Rs. 2.50
Therefore, the cost of painting the wall = 94 × 2.50 = Rs. 235
Q7. A wire is in the shape of a rectangle. Its length is 40 cm and breadth is 22 cm. If the same wire is bent in the shape of a square, what will be the measure of each side. Also, find which side encloses more area?
Sol.
Length of the wire = 40 cm
Breadth of the wire = 22 cm Therefore, perimeter of the rectangle = 2(Length + Breadth) = 2(40 + 22) = 124 cm.
When wire is bent into a square, the permiter will remain same.
Therefore, perimeter of the square = 124 cm.
Lets assume that side of square be x cm.
Perimeter of square = 4x\begin{array}{l} \Rightarrow \quad 4x = 124 \\ \Rightarrow \quad x = 31 \\ \text {Side }=\frac{124}{4}=31 \text { cm} \\ \text {Area of the rectangle } =\text { Length } \times \text { Breadth } =40 \times 22 =880 \text { cm}^{2} \\ \text {Area of the square }=(\text { Side })^{2}=(31)^{2}=961 \text { cm}^{2} \\ \therefore \text {Area of square is more.} \end{array}Q8. How many square metres of glass will be required for a window, which has 12 panes, each pane measuring 25 cm by 16 cm ?
Sol. Length of glass pane = 25 cm.
Breadth of glass pane = 16 cm.
\begin{array}{l}\text {Area of glass pane }=\text { Length } \times \text { Breadth } =25 \times 16 =400 \text { cm}^{2} \\ \Rightarrow \text {Area of 12 panes } =12 \times 400=4800 \text { cm}^{2} \\ =4800 \times \frac {1}{10000} \text { m}^{2} \quad \quad [\because 1 \text { cm}^{2} = \frac {1}{10000} \text { m}^{2}] \\ =0.48 \text { m}^{2} \end{array}Q9. A marble tile measures 10 cm x 12 cm. How many tiles will be required to cover a wall of size 3 m × 4 m ? Also, find the total cost of the tiles at the rate of Rs. 2 per tile.
Sol. \begin{array}{l} \text {Area of the wall }=\text { Length } \times \text { Breadth } =3 \times 4 =12 \text { m}^{2} \\\text {Area of 1 marble tile }=\text { Length } \times \text { Breadth } =10 \times 12 =120 \text { cm}^{2} \\=120 \times \frac {1}{10000} \text { m}^{2} \quad \quad [\because 1 \text { cm}^{2} = \frac {1}{10000} \text { m}^{2}] \\ =0.012 \text { m}^{2} \\\text {Total no. of tiles required to cover the wall} = \frac {\text {Area of the wall}}{\text {Area of marble tile}} \\ =\frac {12}{0.012} =\frac {12 \times 1000}{12} = 1000 \text { tiles}\\\text {Given cost of one tile }=\text {Rs. } 2 \\ \therefore \text {Total cost to tile the wall } = 1000 \times 2 = \text {Rs.} 2000\end{array}Q10. A table top is 9 dm 5 cm long, 6 dm 5 cm broad. What will be the cost to polish it at the rate of 20 paise per square centimetre?
Sol. \begin{array}{l} \text {Length of table top } =9 \text { dm } 5 \text { cm} =9 \times 10 + 5 ) \text { cm} \quad [\because 1 \text { dm} =10 \text{ cm}] \\ =95 \text { cm} \\\text {Breadth of table top } =6 \text { dm } 5 \text { cm} =6 \times 10 + 5 ) \text { cm} \quad [\because 1 \text { dm} =10 \text{ cm}] \\ =65 \text { cm} \\\text {Area of the table top }=\text { Length } \times \text { Breadth } =95 \times 65 =6175 \text { cm}^{2} \\\text {Rate of polishing per square centimetre }=20 \text { paisa} \\ \therefore \text {Total cost of polishing } = 6175 \times 20 \text { paisa} \\ = \text {Rs.} 6175 \times 0.20 \quad [\because 1 \text { Rs} =100 \text{ pasia}] \\ = \text {Rs.} 1235\end{array}