IF AD=14CM;THEN FIND THE PERIMETER OF ∆ABC
Answers
Answer:
I was surprised to discover this has a Unique solution. I’ll let you do most of the work, but here is how to approach it.
I called the sides x, y, and z. X is the short side. Y is the long side. Z is the hypotenuse and the longest side.
X + Y + Z = 14 (1)
X+2=Y (2)
X^2 + Y^2 = Z^2 (3)
From the first two equations
Z = 12 - 2X. Square that gives
Z^2 = 144 - 48X + 4X^2 (4)
Substituting Eq 2 into Eq 3
Z^2 = X^2 + (X + 2)^2
Z^2 = 2X^2 + 4X + 4 (5)
Set Eq 4 equal to Eq 5
144 - 48X + 4X^2 = 2X^2 + 4X + 4
Simplifying
2X^2 - 52X + 140 = 0
The quadratic formula solves for X — about 3.05
Y is 2 more. Z is found by squaring X and Y and taking the square root
Answer:
AD=AE =14cm
AD =AC + BD
AE = AC+CE
BC = FB +FC
BF = BD [TANGENTS FROM A SINGLE POINT]
CE =CF [TANGENTS FROM A SINGLE POINT]
AD = AB +BF ----1
AE = AC +CF ---2
AD + AE = AB +FB+AC+CF
AD + AE = AB + AC + BC =PERIMETER OF TRIANGLE ABC
28 IS THE PERIETER OF TRIANGLE.