the cross section of a tunnel perpendicular to its length is a trapezium abcd given that AM=BN ,AB=7 M,CD=5 M the hight of tunnel is 2.4 m
Answers
Dear Student,
Please find below the solution to the asked query:
Given : Trepezium PQRS as shown in the figure; also given PM = QM, PQ = 7 m, RS= 5m, the height of the tunnel is 2.4 m.The tunnel is 40m long , So
RS = MN = 5 m
And
PM = QN = 1 m
Total internal surface area excluding floor = Surface area of back of tunnel ( Shape of PQRS ) + RS× Length of tunnel + RQ × Length of tunnel + PS × Length of tunnel
Here , Height of tunnel = SM = RN = 2.4 m
We apply Pythagoras theorem in triangle PMS and get
PS2 = SM2 + PM2
PS2 = 2.42 + 12
PS2 = 5.76 + 1
PS2 = 6.76
PS2 = 2.62
PS = 2.6
We know Area of trapezium = Sum of parallel sides2× Height , So
Area of PQRS = 7 + 52× 2.6 = 122×2.6 = 6 × 2.6 = 15.6 m2
Then ,
Total internal surface area excluding floor = 15.6 + 5 × 40 + 2.6 × 40 + 2.6× 40 = 15.6 + 200 + 104 + 104 = 423.6 m2
Given , The rate of painting the internal surface of the tunnel (excluding the floor ) at Rs .5 per m2
Therefore ,
i ) The cost of painting the internal surface of the tunnel (excluding the floor ) = 423.6 × 5 = Rs . 2118 ( Ans )
And
Area of floor = PQ × Length of tunnel = 7 × 40 = 280 m2
Given , The rate of paving the floor at Rs .18 per m2
Therefore ,
ii ) The rate of paving the floor = 280 × 18 = Rs . 5040 ( Ans )
Hope this information will clear your doubts about Surface Areas and Volumes .
If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.
Regards
Answer:
wtihdj you for all the best wish for your help ❤️