Let's assume length of one side is X, thus ractanguler has 2 sides of length X and another 2 sides of length 280-X...
Now Area of ract. = X(280-X) = 280X - X^2.....(1)
Diagonal will create 2 trinagle with hyotenuse of length 200...
Using pythagora will get following equation,
X^2 + (280-X)^2 = 200^2
--> 2X^2 + 280^2 - 560X = 200^2
--> X^2 + 280*140 - 280X = 20000
-->19200 = 280X - X^2
-->19200 = Area of ract. from (1)
PDF800 SET3 Question 31
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- gabriel
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cant answer ..lalitgmat wrote:If I twist the question for a 700+;
A small, rectangular park has a perimeter of 560 feet and One side of park is lesser than adjacent side by X feet. What's the Area of Park.
.. the eqn is 2( a + (a+x)) = 560 .. so two variables and one eqn ..
Plus... you already assume that one side is less than the other side by X feet... why would you ever need to state that in the question? This is like twisting it to 200-, not 700+lalitgmat wrote:If I twist the question for a 700+;
A small, rectangular park has a perimeter of 560 feet and One side of park is lesser than adjacent side by X feet. What's the Area of Park.
Let's assume a side has length X. And difference in length between 2 side is 200 means another side will be X+200.
simply X+X+200 = 280 which leads to X = 40.
Area = 40 * (40+200)
Don't you think this is more easier one....
May be as beny told 200-
simply X+X+200 = 280 which leads to X = 40.
Area = 40 * (40+200)
Don't you think this is more easier one....
May be as beny told 200-
Thanks,
UmanG - restless mind..
UmanG - restless mind..












