Here we need to find within which range of values the distance of ladder jacked out (not 25 ft, but jacked out distance) is placed. The jacked out ladder's top has height of half the distance this ladder is jacked out.
The geometric figure will be a right triangle with the sides: base 7, height x/2, ladder's length x (jacked out). We should solve for x, and there's proportion such as 1:sqrt(3):2 (30-60-90 degrees). Hence, 7=sqrt(3)*(x/2) and 49=(3x^2)/4, 3x^2=196, x^2=~65, x is closer to 8 but is slightly greater, some 8.1.
let's see which answer choice contains 8.1
A (5,8) no
B (9,10) no
C (2,7) no
D (3,7) no
E None of the above
Yes
e
[spoiler]on the second read after solving, i guess wording here is abra-cadabra (confusing), mainly relied on common sense about the object in q.
[/spoiler]
shankar.ashwin wrote:A 25 ft long ladder is placed against the wall with its base 7 ft from the wall. The base of the ladder is drawn out so that the top comes down by half the distance that the base is drawn out. This distance is in the range:
A (5,8)
B (9,10)
C (2,7)
D (3,7)
E None of the above
[spoiler]Don't have a OA and don't exactly get what the questions asks.[/spoiler]
COuld someone help..Tough one I assume
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