Asked by sonika
The parallel edges of a trapezium shaped field are 77m and 60m.If the remaining edges are 25M and 26m,what is the total cost of digging twice the land at the rate of Rs.10 persq m and fencing thr land three times at the rate of Rs.20 per m
Answers
Answered by
Reiny
From each of the ends of the 60 m side, draw altitudes to the 77 m side.
You now have a rectangle and two right-angled triangles.
Let the height of all three be h,
for the triangle with hypotenuse 25, let the base be x, then
x^2 + h^2 = 25^2 -----> h^2 = 625 - x^2
for the triangle with hypotenuse 26, let the base be 10-x
h^2 + (10-x)^2 = 26^2 ---> h^2 = 376 - (10-x)^2
625 - x^2 = 376 - (10-x)^2
a bit of simple algebra yields:
x = 49/20 = 2.45 m
now you can find the base of the other triangle , 10-x
and the height h
All is now known, and you can find area, perimeter or whatever else you need.
You now have a rectangle and two right-angled triangles.
Let the height of all three be h,
for the triangle with hypotenuse 25, let the base be x, then
x^2 + h^2 = 25^2 -----> h^2 = 625 - x^2
for the triangle with hypotenuse 26, let the base be 10-x
h^2 + (10-x)^2 = 26^2 ---> h^2 = 376 - (10-x)^2
625 - x^2 = 376 - (10-x)^2
a bit of simple algebra yields:
x = 49/20 = 2.45 m
now you can find the base of the other triangle , 10-x
and the height h
All is now known, and you can find area, perimeter or whatever else you need.
Answered by
Jagarnath
Ans
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.