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Lot ABCD between two parallel street lines is 350.00 ft deep and has a 220.00-ft frontage (AB) on one street and a 260.00-ft frontage (CD) on the other. Interior angles at A and B are equal, as are those at C and D. What distances AE and BF should be laid off by a surveyor to divide the lot into two equal areas by means of a line EF parallel to AB?

Respuesta :

Answer:

240.83 ft

Explanation:

The distances AE and BF will be equal = 182.58 ft

The area of the lot will be the product of the depth and the average of the two frontages

= ( 350 * (220 + 260)/2) = 84000 ft

Half of the area becomes 42000 ft

<A = arctan (20/350) = 3. 3°

Hence 42000=h/2*(220 + 220 + 2h*tan3.3)

Solving, we obtain h= 182.28ft

EF = 220 + (2*182.28*tan 3.3)

= 240.83ft