Note: Length of diagonal = Distance covered in 3 min.at 4 km/hr

Note: Not available

Note: Let side of a square = a. length of rect. = x & breadth of rect. = y. `4a = 2 (x + y) => (x + y)/2 = a` Area of square` = a^2 = ((x + y)^2/2)` So, A`= ((x + y)^2/2)` Area of rect. = xy. So, B = xy. `(x + y)/2 > x^1/2 y^1/2` [:. A.M. > G.M.] `=> ((x + y )^2/2)`> xy i. e.A > B.

Note: Let side of a square = a. Then, its diagonal` = sqrt(2) a`. Ratio of areas of squares with sides a & `sqrt(2) a` `= a^2/(sqrt(2) a)^2` `= 1/2 or 1 : 2`

Note: Let the sides of two squares be a and b respectively. Then, `a^2/b^2 = 9/1` or `(a/b)^2` `= (3/1)^2` or `a/b = 3/1` :. Ratio of their perimeters `= (4a)/(4b) = a/b = 3/1` or ` 3 : 1`

Note: perimeter of square = 80 cm. So, side of square = 20 cm. Area of the square` = (20 xx 20) cm^2` `= 400 cm^2` With same perimeters, area of the square is larger. :. Area of rectangle`= (400 - 100) cm^2` `= 300 cm^2` Let length of rect. = x & its breadth = y. Then, xy = 300 and x + y = 40 `:. (x - y) = sqrt((x + y))^2- 4xy` `= sqrt(1600 - 1200) = 20` Solving `x + y = 40 & x - y = 20,` we get` x = 30, y = 10` :. Sides are` 30 cm, 10 cm.`

Note: Area of first square` = 10000 sq.` metres. Side of this square` = sqrt(10000) m = 201 m^2` Side of the new square` = 101 m` Difference in areas` = (101)^2 - (100)^2` ` = 201 m^2`

Note: Let breadth = x. Then, length = 2x `(2x - 5) (x + 5) - 2x xx x` `= 75 or x = 20`. `:. Length = 40 cm.`

Note: Area `= (total cost)/Rate` `= (1215/135) hectares` `= (9 xx 10000) sq.m,` :. Side of the square` = sqrt(90000) = 300 m.` perimeter of the field` = (300 xx 4) m.` `= 1200 m.` cost of fencing` = Rs. (1200 xx 3/4)` `= Rs. 900`

Note: Let the width of walk be x metres. Then, `(20 - 2x) (10 - 2x) = 96` `or 4x^2 + 60x - 104` `= 0 or x^2 + 15x - 26 = 0` `or (x - 13) (x - 2) = 0` So, `x = 2 [:. x =/ 13]