The area of the triangle △ABC is 12 square units.
The question is incomplete. The actual question is
Triangle ABC is symmetric about the y-axis. Point A is located at (-3,0), and AB is the longest side of ABC. If the perimeter of ABC is 16, what is the area?
What is symmetry?
Symmetrical forms or figures are things that can have a line drawn through them so that the representations on both sides of the line mirror each other.
How to solve the problem?
As could be seen in the figure the triangle △ABC is symmetric along the y-axis. Hence, the side AC and BC must be of the same length let us call it 'a'. Also, the coordinate of B must be (3,0) because (3,0) is the reflection of (-3,0) corresponding to the y-axis.
And, the length of AB=6 units because the distance between (-3,0) and (3,0) is 6 units.
Now, the perimeter of the triangle is 16 units. Therefore,
2a+6=16
⇒a=(16-6)/2
⇒a=5 units.
Hence, we have an isosceles triangle △ABC we can find the height 'h' of the triangle using the Pythagoras theorem. Consider the triangle △ACE
then we have that
the height of the right triangle △ACE= h
the base of the right triangle △ACE= 3 units
the hypotenuse of the right triangle △ACE= a = 5 units.
Hence, by Pythagoras' theorem
h²+3²=5²
⇒h²=25-9=16
⇒h=4 (we choose the positive value of h because length is always positive)
Hence, the area of the triangle △ABC is
(1/2)×base×height
=(1/2)×AB×CE
=(1/2)×6×4 unit²
=12 unit²
Hence, the area of the triangle △ABC is 12 square units.
Learn more about the isosceles triangles in- brainly.com/question/2456591
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