Question

A(3,0) B(-3,0) are extremites of the
base a AB of a triangle PAB. If the vertex
P varies such that internal angular bisector
of angleAPB of the triangle divides to opposite side
AB in the ratio 2:1 then maximum area of
triangle PAB
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Answer 1

Answer:

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Answer 2

Question $A=(-3,0)$ and $B=(3,0)$ are the extremities of the base $A B$ of triangle $PAB$. If the vertex $P$ varies such that the internal bisector of angle $APB$ of the triangle divides the opposite side $AB$ into two segments $\mathrm{AD}$ and $\mathrm{BD}$ such that $A D: B D=2: 1$, then the maximum value of the length of the altitude of the triangle drawn through the vertex $P$ is

Answer:

The required answer is $4$.

Step-by-step explanation:

Given: Vertex A(-3,0) and B(3,0) . The ratio is $A D: B D=2: 1$

To find: the greatest value of the triangle's altitude measured from its vertex, $p$.

Solution:

The point $E$ dividing $\overline{A B}$ externally in the ratio $2: 1$ is $(9,0)$.

Since $P$ lies on the circle described on $\overline{D E}$ as diameter, coordinates of $P$ are of the form $(5+4 \cos \theta, 4 \sin \theta)$

Therefore, maximum length of the altitude drawn from $P$ to the base $A B=|4 \sin \theta|_{\max }=4$.

Hence, the greatest value of the triangle's altitude measured from its vertex, $P$ is $4$.

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