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14 Mar 2007, 12:27
Kudos
Bookmarks
for right triangle ABC, AB and AC are the legs and AB+AC=12. CB is the hypotenuse with length of 10. What is the area of the triangle?
24
11
15
2*root30
12
24
11
15
2*root30
12
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14 Mar 2007, 12:48
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First of all... Welcome to the GMATClub :D 
(B) for me
We have this 2 relationships:
o AB+AC = 12
p AB^2+AC^2 = 100 = BC^2
AB^2+AC^2 = 100
<=> AB^2+(12-AB)^2 = 100
<=> 2*AB^2 - 24*AB + 144 = 100
<=> AB^2 - 12*AB + 22 = 0
Reduiced Delta = (b/2)^2 - a*c = 36 - 22 = 14
AB=(6 - sqrt(14))
or
AB=(6 + sqrt(14))
Finally,
As the solution is symetrical to AB and BC, we can choose the values of AB and BC:
AB=(6 - sqrt(14))
BC=(6 + sqrt(14))
Area = (6 - sqrt(14))*(6 + sqrt(14))/2
= (36 - 14) / 2 = 11
(B) for me
We have this 2 relationships:
o AB+AC = 12
p AB^2+AC^2 = 100 = BC^2
AB^2+AC^2 = 100
<=> AB^2+(12-AB)^2 = 100
<=> 2*AB^2 - 24*AB + 144 = 100
<=> AB^2 - 12*AB + 22 = 0
Reduiced Delta = (b/2)^2 - a*c = 36 - 22 = 14
AB=(6 - sqrt(14))
or
AB=(6 + sqrt(14))
Finally,
As the solution is symetrical to AB and BC, we can choose the values of AB and BC:
AB=(6 - sqrt(14))
BC=(6 + sqrt(14))
Area = (6 - sqrt(14))*(6 + sqrt(14))/2
= (36 - 14) / 2 = 11
14 Mar 2007, 13:40
Kudos
Bookmarks
jarnold101
Let AB = x and AC = y
By given info
(a) x + y = 12 and
(b) x**2 + y**2 = 100
Squaring (a) we get
x**2 + y**2 + 2xy = 144
Substituting values we get
100 + 2xy = 144
or 2xy = 44
or xy = 22
or (1/2)xy = 11 = Area
