Let the bases of the two triangles be x and y and their heights be 3h and 4h respectively.
Then,  1 2 × x × 3 h 1 2 × y × 4 h = 4 3 ? x y = 4 3 × 4 3 = 16 9 

Required ratio = 16 : 9.

Area = (13.86 x 10000) sq.m = 138600 sq.m

?R 2 = 138600 ? R 2 = 138600 × 7 22 ? R = 210 m 

Circumference =  2 ?R 2 = 2 × 22 7 × 210 = 1320 m 

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

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    =  1 2 × b × h 1 

Area of rectangle  = b*h2 

As per Given,

1 2 * b * h 1 = b * h 2

1 2 * b * h 1 = b * 100

 h1=200

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 = 66 7

 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

=>   d 2 = 12 2 + 4 2 + 3 2 = 13mts.

Let the diagonals of the squares be 2x and 5x respectively.
Ratio of their areas =   1 2 * 2 x 2 : 1 2 * 5 x 2 = 4 : 25

area of a square = a² sq cm

length of the diagonal =  2 a cm

area of equilateral triangle with side   2 a= 3 4 * 2 a 2
required ratio =  3 4 * 2 a 2 : a 2 = 3 : 2