Geometry Help?

Question

Triangle ABC is congruent to triangle DEF. Find all possible values of x if...
Area of triangle ABC is 2x+9
Perimeter of triangle ABC is x+3

Are of triangle DEF is 2x-6
Perimeter of tirangle DEF is x-1

Answer 1

I suspect that you meant "similar" triangles rather than "congruent" triangles.

If so, the ratio of areas is the square of the ratio of corresponding lengths, so:

(x + 3)² / (x - 1)² = (2x + 9) / (2x - 6)

(x + 3)² (2x - 6) = (x - 1)² (2x + 9)

(x² + 6x + 9)(2x - 6) = (x² - 2x + 1)(2x + 9)

2x³ + 6x² - 18x - 54 = 2x³ + 5x² - 16x + 9

x² - 2x - 63 = 0

(x - 9)(x + 7) = 0

x = 9 or x = -7

Only the first answer gives a positive length, so x = 9 is the only solution.

Answer 2

If the triangles are congruent, then they must have the same angles AND sides.

That means the areas and perimeters must be the same as well.

This would mean that 2x+9 = 2x-6
and x+3 = x-1

Both of these are impossibilities.

NOW if the trianges are SIMILAR, but not CONGRUENT
then (x+3) = k*(x-1)
and (2x+9) = k^2 *(2x-6)

You can solve for k based on this and find that
x = (3 + k) / (k - 1)
and x = (3/2)* (3 + 2k^2) / (k^2 - 1)

You will get
k = 1/2 and 3/2
Only 3/2 makes sense, in which case x = 9.

Answer 3

If triangles are congruent, that means that all the sides and angles of one triangle are equal to all the sides and angles of the other triangle. So, the perimeter of triangle ABC has to be the same as the perimeter of triangle DEF.
x+3=x-1 So there are no possible values of x.

Answer 4

[(x+3)/(x-1)]^2 =(2x+9)/(2x-6) As the ratio of the areas is the square of the ratio of the perimeters
(x+3)^2*2(x-3)=(2x+9)(x-1)^2
2(x^2-9)(x+3)=(2x+9)(x^2-2x+1)
x^2-2x-63=0

x=((2+-sqrt(256))/2
so x= 9 and x = -7(nonsense) so x= 9( I supposed you meant similar)

Answer 5

x=9(nine) :) =)