A rectangular parcel of land has an area of 6,000 ft2. A diagonal between opposite corners is measured to be 10 ft longer than one side of the parcel. What are the dimensions of the land, correct to the nearest foot? ft (smaller value) by ft (larger value)

50ft by 120ft

Step-by-step explanation:

Area of a rectangle = L × W

6000ft² = L × W

L = 6000/W

When a diagonal line divides a rectangle into 2 right angled triangles, the diagonal line = Hypotenuse of either of the triangle and it is the longest side.

The formula for a right angle triangle =

a² + b² = c²( c = hypotenuse)

We are told in the question that:

A diagonal between opposite corners is measured to be 10 ft longer than one side of the parcel

Let us assume the side that the hypotenuse is longer than = Width

Hence, the Diagonal = (W + 10)²

Therefore

L² + W² = (W + 10)²

Since L = 6000/W

W² + (6000/W)² = (W + 10)²

W² + (6000/W)² = (W + 10) (W + 10)

W² + (6000/W)² = W² + 10W + 10W + 100

W² + (6000/W)² = W² + 20W + 100

W² - W² + (6000/W)² = 20W+ 100

6000²/W² = 20W + 100

6000² = W²( 20W + 100)

6000² = 20W³ + 100W²

20W³ + 100W² - 6000² = 0

20W³ + 100W² - 36000000 = 0

20(W³ + 5W² - 1800000) = 0

Factorising the quadratic equation,

20(W − 120)(W² + 125W + 15000) = 0

W - 120 = 0

W = 120

Therefore,

W(Width) = 120feet

Since the Width = 120 feet

We can find the length

6000ft² = L × W

L = 6000/W

L = 6000/120

L = 50 feet

The dimensions of the land, correct to the nearest foot is 50ft by 120ft

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