Question

Through a rectangular field of length 90 m and breadth 60 m, two roads are
constructed which are parallel to the sides and cut each other at right angles through
the centre of the fields. If the width of each road is 3 m, find
O the area covered by the roads.
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Answer 1

It is a nice question by the way, but have a answer,

So we begin with

Taking out the area of the whole field

By length × breadth

60x90=5400

Then take out area covered by the roads

It will be

From the road which is parallel to 60 m

Will be 3x60=180

And 90 one will be

3×90=270 add,both of these and subtract 9 as the area intersecting 2 roads is repeated

So, 180+270-9=441

And then simply subtract it from the total area

5400-441=4939

Answer 2

Answer:

4959 m square

Step by step explanation:

Let:-

• Name of rectangle be =ABCD
• Name of the first cross road = EFGH
• Name of second cross road= IJKL
• Name of the square formed by the intersection of the two cross roads=MNOP

Area of the first cross road EFGH

3x60=180 m square

Area of second cross road IJKL

90x3=270 m square

Area of the square formed =MNOP

3X3=9 m square

Total area of the two cross roads

=(sum of the area of the two cross roads)-(area of the square)

=(180+270)-9

=441 m square

Area of the two cross roads excluding the area of the rectangle

=(area of the rectangle)-(area of the two cross roads)

=(60x90)-441

=5400- 441= 4959 m square