Let length, breadth and height of the room be 7, 3, 1 unit respectively.
Area of walls = 2(l+b)xh = 2(7+3)x1 = 20 sq. unit.
Now, length, breadth and height of room will become 3.5, 6 and 2 respectively.
Area of walls = 2(l+b)xh = 2(3.5+6)x2 = 38 sq. unit.
% Increase in the area of walls = (38-20)x100/20 = 90%.
Let the length be 'l' and breadth be 'b'.
b = l × 3/4__________(a)
2(l+b) = 1050
l+b = 525___________(b)
From equations (a) and (b),
l = 300m, b = 225 m
Area = l × b
= 300 × 225
= 67500 sq.m.
We have: l = 20 ft and lb = 680 sq. ft.So, b = 34 ft.
Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.
Given length of the rectangle = 3 cm
Breadth of the rectangle = 4 cm
Then, the diagonal of the rectangle
Then, it implies side of square = 5 cm
We know that Area of square = S x S = 5 x 5 = 25 sq.cm.
The triangle with sides 26 cm, 24 cm and 10 cm is right angled, where the hypotenuse is 26 cm.
Area of the triangle =1/2 x base x height => 1/2 x 24 x 10 = 120 sq.cm
Distance covered in one revolution = = 88m.
Number of square units in 13 by 9 is given by the area it forms with length and breadth as 13 & 9
Area = 13 x 9 = 117
Hence, number of square units in 13 by 9 is 117 sq.units.
Area of the square = = 1/2 * 3.8 * 3.8 = 7.22 sq.m
Let the bases of the two triangles be x and y and their heights be 3h and 4h respectively.
Then,
Required ratio = 16 : 9.
Area = (13.86 x 10000) sq.m = 138600 sq.m
Circumference =
Cost of fencing = Rs. (1320 x 4.40) = Rs. 5808.
perimeter of window = r+2r
= [(22/7) x (63/2) + 63] = 99+63 = 162 cm
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