NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

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Ans : Perimeter of square = 4 (Side of the square) = 4 (60 m) = 240 m.
Perimeter of rectangle = 2 (Length + Breadth)
= 2 (80 m + Breadth)
= 160 m + 2 x Breadth
It is given that the perimeter of the square and the rectangle are the same.
160 m + 2 x Breadth = 240 m
Breadth of the rectangle = \( \left(\frac{80}{2}\right) \mathrm{m} \) = 40 m
Area of square = \( (\text { Side })^{2}=(60 \mathrm{m})^{2}=3600 \mathrm{m}^{2} \)
Area of rectangle = Length x Breadth = (80 x 40) \( m^{2} \) = 3200 \( m^{2} \)
Thus, the area of the square field is larger than the area of the rectangular field. 

Ans : Area of the square plot = \( (25 m)^{2}=625 m^{2} \)
Area of the house = (15 m) x (20 m) = 300 \( m^{2} \)
Area of the remaining portion = Area of square plot - Area of the house
= 625 \( m^{2} \) - 300 \( m^{2} \) = 325 \( m^{2} \).
The cost of developing the garden around the house is Rs 55 per \( m^{2} \).
Total cost of developing the garden of area 325 \( m^{2} \) = Rs (55 x 325)
= Rs 17,875 

Ans : Length of the rectangle = [20 - (3.5 + 3.5)] metres = 13 m
Circumference of 1 semi-circular parts = nr \( =\left(\frac{22}{7} \times 3.5\right) \) = 11 m
Circumference of both semi-circular parts = ( 2 x 11 ) m = 22 m

Perimeter of the garden = AB + Length of both semi-circular regions BC and DA + CD
= 13m + 22m + 13m = 48 m
Area of the garden = Area of rectangle + 2 x Area of two semi-circular regions
\( =\left[(13 \times 7)+2 \times \frac{1}{2} \times \frac{22}{7} \times(3.5)^{2}\right] \mathrm{m}^{2} \)
\( =(91+38.5) \mathrm{m}^{2} \)
\( =129.5 \mathrm{m}^{2} \)