LamboWalker wrote:
In triangle ABC, A=(1,2) B=(5,5), ∠ACB=90°. If area of △ABC is to be 6.5 square units, then the possible number of points for C is
(A) 0
(B) 1
(C) 2
(D) 4
(E) Cannot be answered based on the given information.
This means the hypotenuse is AB =\(\sqrt{(5-1)^2+(5-2)^2}=\sqrt{16+9}=5\)
When will the area be maximum for a triangle with hypotenuse?
It will be when the triangle is isosceles triangle so each side = \(\frac{5}{\sqrt{2}}\)
Max area = \(\frac{1}{2}* \frac{5}{\sqrt{2}}* \frac{5}{\sqrt{2}}=\frac{25}{4}=6.25\)
Thus, there will be no triangle with hypotenuse 5, that can have area more than 6.25 .
A
But it is slightly illogical to say that area is 6.5, when there are no such triangles possible. Could have been worded a bit better.
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