1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# Andy spent Rs. 450 in fencing the parallelogram shaped land (leaving 1 m for the gate) that he had before. If Andy would start from a point 5 m away from A (on side AB) and take the shortest possible path to the opposite side, he would reach exactly at D. As you know Land Grabbers restructured Andy’s land to be a rectangle having the same base and same width. Find the cost of fencing the new land in the same way at Rs. 6 /m.

Open in App
Solution

## Let us redraw the figure given like this. With what is given in the question, we can say that DE is the height of the parallelogram. Let the unknown side of the parallelogram be ‘x’ m. Perimeter of the parallelogram field which is fenced = Total cost of fencingCost of fencing for 1m Rs. 4506 = 75 m So, Perimeter = (2×25+2x−1)=75 =2x–1=25 x=13 m So the other side of parallelogram is 13 m. In △ADE, AD = 13 m, AE = 5 m Using Pythagoras theorem, AD2=AE2+DE2 or, DE2=132−52=144=122 So, DE=12 m The restructured land is in the shape of a rectangle with same base and width. So, length of the rectangular land = 25 m Breadth of the rectangular land = 12 m Perimeter = 2 (l + b) = 2 (25 + 12) = 2 × 37 m = 74 m Perimeter to be fenced = (74 - 1) m = 73 m (∵ 1 m is kept open for entry to the farm) Cost of fencing = Rs. 6m × 73 m = Rs. 438

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Area of a Parallelogram
MATHEMATICS
Watch in App
Explore more
Join BYJU'S Learning Program