Question

a parcel is 6ft longer than it is wide. each diagonal from one corner to the opposite corner is 174ft long. what are the dimensions of the parcel?

Answer 1

Let W be the width, L be the length and D be the diagonal.

By Pythagoras's Theorem, assuming the parcel is rectangular, D² = L² + W².

Also you're given that the parcel is 6 foot longer than it is wide, which tells you that L = W + 6. Subbing into the diagonal equation gives

D² = (W+6)² + W².

You are given the length of the diagonal: D = 174.

So 174² = (W+6)² + W²

30276 = (W+6)² + W². Expanding out the (W+6)² gives

30276 = W² + 12W + 36 + W², or

30276 = 2W² + 12W + 36.

Dividing through by 2 gives

15138 = W² + 6W + 18, or

0 = W² + 6W - 15120.

Solve via the quadratic formula. I'll let you do the working but you should get solutions of W = 120 and W = -126. The width of the parcel cannot be negative and hence W = 120.

Also L = W + 6 = 120 + 6 = 126.

So the parcel of land is 120 feet by 126 feet