percent of the area problem. help plz!!!?

A cow is bound at the center if a rectangular field measuring 12 feet * 24feet . If the rope with which the cow is tied is 10 feet long, approx. what percent of the area of the field can the cow graze upon?

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• 8 years ago

Jaher Wasim Hi!

Percent of the area problem. help plz!!!?

A cow is bound at the center if a rectangular field measuring 12 feet * 24feet . If the rope with which the cow is tied is 10 feet long, approx. what percent of the area of the field can the cow graze upon?

Your problem look simple but not as simple as it is solved than seen . It catches my interest.

I hope you are still here ti witness.

Area of the field = 12 x 24

Area of the field = 288 ft²

Refer to Figure: http://www.flickr.com/photos/27678773@N06/81909279...

Sin EOb = bE / Ob

Sin EOb = 12/2 /10

Sin EOb = 0.6 Ft

∢EOb = 36.87°

Cos EOb = OE / Ob

OE = Ob Cos EOb

OE = 10 (0.8)

OE = 8 Ft

EF = OE x 2

EF = 16 Ft

Area Sector ArcabOa = 1/2R²(EQb x 2)

Area Sector ArcabOa = ½(10)²(36.87 x 2) x π/180

Area Sector ArcabOa = 64.35 ft²

At the Opposite Side

Area Sector ArcabOa = Area Sector ArccdOc = 64.35 ft²

Area Δ aOG = 1/2aG(OG)

Note → aG = OE = 8 ← refer to figure

note → OG = Ea = bE = 12/2 ← refer to figure

Area Δ aOG = ½(8)(6)

Area Δ aOG = 24 ft²

Area Δ aOG = Area Δ dOG = 24 Ft²

Area Δ aOd = Area Δ aOG + Area Δ dOG

Area Δ aOd = 24 + 24

Area Δ aOd = 48 ft²

The Oppodite Side Area Δ bOc = Area Δ aOd = 48 ft²

Graze Land Area = Area Sector ArcabOa + Area Sector ArcccOc + Area Δ aOd + Area Δ bOc

Graze Land Area = 64.35 + 64.35 + 48 + 48

Graze Land Area = 224.7 ft²

The Percentage of Grazing Area = 224.7 / 288

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The Percentage of Grazing Area = 78.02 % ◄ Ans

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Hope this helps

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