Question

A rectangular lawn is 60 m by 40 m and has two roads, each 5m wide running in the middle of it. One parallel to its length and other parallel to the breadth. Find the cost of constructing the roads at Rs.80/m2
2.The area of a right triangle is 40/m2. If one of its legs measures 8 cm, find the length of the other leg.
3. Subtract {2a-3b+4c} from the sum of (a+3b-4c),(4a-b+ac) and (-2b+3c-a)
4. Solve : [⅓y2 – 4/7 y +5] – [2/7y – 2/3y2+2] – 1/7 – 3 +2y2]
5. Find the value of (-8u2 v6) x (-2uv4) For u = 2.5, v = 1
6. [2x2 – 5y2] [x2 + 3y2]
7. By what number should we multiply 3-9 so that the product is equal to 3 ?

Answer 1

1. length of lawn= 60m , breadth = 40 m 
    area of lawn  = l*b = 60*40=2400 sq m 

length of road  parallel to length = 60m and breadth = 5m 
area of this road (1st ) =  l*b=60*5 = 300 sq m 

length of road  parallel to breadth  = 40m and breadth = 5m 
area of this road (2nd ) =  l*b=40*5 = 200 sq m 

Total area of road = area of ist road = area of road second -(area of their                                                                                                              intersection)
                                    = 300 + 200 - ( 5*5)
                                     = 500- 25 = 475 sq m 
cost of constructing 1 sq m of road = Rs. 80
cost of constructing  475  sq m of road = Rs. 80 * 475 = Rs.38000 

2. lengthh of one leg  (let it be its base )= 8 cm

 area of right angled triangle = 40 sq m .
                                      40   =  1/2 * base * perpendiicular
                                      perpendicular = 40 *2 / 8 = 10 m 
therefore  measure of another leg = 10 m .


3.ATQ.
  
   {(a+3b-4c)+(4a-b+ac)+(-2b+3c-a) } - (2a-3b+4c)
   =  {a+3b-4c+4a-b+ac+-2b+3c-a } - (2a-3b+4c)
   =   4a +ac  - c - 2a + 3b - 4c  = 2a +a c - 5c + 3b 

rest of the answers i wll post later ....... 
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