How Large is the Yard??

[Mooch22]

There is a odd shaped quadrilateral lot of land. The sides are all different in length. The left side is 146 m, the top 86 m, the right side 72 m, and the bottom unknown. The angle between the 146 m and the 86 m is 113 degrees. The angle between the 86 m and the 72 m is 131 degrees.

In order to apply for a building permit, the property had to be surveyed. The surveyors took the measurements of three sides and two included angles. From the informaiton they were able to determine the area of the yard. The surveyors said that the area was 2.5 acres. However, the owners disagreed. They thought they had more land than that.

The question is to calculate the area of the yard and see if the 23.5 acres is correctly calculated. [/quote]

 

[stapel]

(Why is this a "sticky"?)

What have you tried? Does your book give you any particular formulas? How far have you gotten with this?

Eliz.

 

[soroban]

Hello, Mooch22!

It would have been considerate of you to show your progress on this problem.
. . Otherwise, you're saying, "Here's the problem. .Do it for me!" . ***

There is a odd shaped quadrilateral lot of land.
The left side is 146 m, the top 86 m, the right side 72 m, and the bottom unknown.
The angle between the 146 m and the 86 m is 113°.
The angle between the 86 m and the 72 m is 131°.

Calculate the area of the yard.

          B    86     C
          * * * * * * *
         *113°     131°*
        *               * 72
       *                 *              Draw diagonal AC.
  146 *                   *D
     *              *          
    *         *                          
   *    *   
 A*

In triangle ABC, we know two sides and the included angle.
. . A₁ .= .(1/2)(146)(86)sin 113° . . . [I got: A₁ ≈ 5779 m²]


We want A₂, the area of triangle ACD. .We need AC and angle ACD.

In triangle ABC, use the Law of Cosines to find side AC.
. . AC² .= .146² + 86² - 2(146)(86)cos 113° . . . [I got: AC ≈ 196]

. . . . . . . . . . . . . . . . . 86² + 196² - 146²
Then: . cos(<u>/</u>BCA) .= .----------------------- . . . [I got: <u>/</u>BCA ≈ 43.4°]
. . . . . . . . . . . . . . . . . . . 2(86)(196)

Hence: .<u>/</u>ACD .= .131° - 43.4° .= .87.6°


In triangle ACD, we have: AC = 196, CD = 72, <u>/</u>ACD = 87.6°

Its area is: .A₂ .= .(1/2)(96)(72)sin 87.6° . . . [I got: A₂ ≈ 7050 m²]


The total area is: .5779 + 7050 .= .12829 m² .≈ .3.17 acres . **

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** . Do a search for "conversions".

*** . Why do I bother with these comments?
. . . . . If you're like 90% of the posters at this site,
. . . . . you won't change your ways . . . ever
. . . . . and I won't even get a thank-you.