Let the triangle and parallelogram have common base b,
let the Altitude of triangle is h1 and of parallelogram is h2(which is equal to 100 m), then
Area of triangle =
Area of rectangle = b*h2
As per Given,
h1=200
perimeter of window = r+2r
= [(22/7) x (63/2) + 63] = 99+63 = 162 cm
Area = (13.86 x 10000) sq.m = 138600 sq.m
Circumference =
Cost of fencing = Rs. (1320 x 4.40) = Rs. 5808.
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 of the square =
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.
We know that,
The area of a triangle with two sides given and included angle
A = 1/2 x product of sides x Sin(angle)
Here the two sides are 8 & 12
Angle = 150
Area = 1/2 x 8 x 12 x sin150
Sin(150) = sin(90+60) = cos(60) = 1/2
A = 48 x 1/2 = 24
Area of the given triangle = 24 sq units.
r =3
l=5.5m w=3.75m
area of the floor = 5.5 x 3.75 = 20.625 sq m
cost of paving = 800 x 20.625 = Rs. 16500
The length of the longest pole which can be kept in a room 12 m long, 4 m broad and 3 m high is
=> mts.
Let the diagonals of the squares be 2x and 5x respectively.
Ratio of their areas =
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