1. In the diagram below, points A, B, C, and P are situated so that PA=2, PB=3, PC=4, and BC=5. What is the maximum possible area of triangle ABC?

Diagram: https://latex.artofproblemsolving.com/f/8/c/f8c4039f30c9ec08fd75e8d55b476ded54ceb800.png (background is transparent)

2. In triangle ABC, \angle ABC = 90^\circ and AD is an angle bisector. If AB=90, BC=x, and AC=2x-6, then find the area of triangle ABC. Round your answer to the nearest integer.

3. Medians \(\overline{DP}\) and \(\overline{EQ }\) of triangle DEF are perpendicular. If \(DP=18\) and \(EQ=24\), then what is \(DF\)?

yasbib555 Jan 25, 2019

#1**+1 **

3. Medians DP and EQ of triangle DEF are perpendicular....DP = 18 and EQ = 24

Let the medians intersect at M

Then DM = (2/3)DP = (2/3)18 = 12

And EM = (1/3)EQ = (1/3)(24) = 8

So.....using the Pythagorean Theorem, ......DQ = sqrt ( DM^2 + EM^2 ) = sqrt (12^2 + 8^2) = sqrt (144 + 64) =

sqrt ( 208) = 4sqrt (13)

But DF = 2DQ = 2 * 4sqrt (13) = 8sqrt (13)

CPhill Jan 25, 2019

#3**+1 **

2. In triangle ABC, \angle ABC = 90° and AD is an angle bisector. If AB=90, BC=x, and AC=2x-6, then find the area of triangle ABC. Round your answer to the nearest integer.

Don't see where the angle bisector comes into play, here.....

We have that

BC^2 + AB^2 = AC^2

x^2 + 90^2 = (2x - 6)^2

x^2 + 90^2 = 4x^2 - 24x + 36

3x^2 - 24x - 8064 = 0

x^2 - 8x - 2688 = 0 this factors as

(x - 56) (x + 48) = 0

x = 56

So....the area is (1/2)product of leg lengths = (1/2)(AB)(BC) = (1/2)(90)(56) = 2520 units^2

CPhill Jan 26, 2019