Point D is the in center of triangle ABC. Write an expression for the length x in terms of the three side lengths AB, AC, BC.

Hint: Use areas of triangles...A=1/2bh

Answer :

[tex] \underline{ \boxed{x = \frac{2(A_T) }{(AB) + (AC) + (BC)} }}[/tex]

Explanation :

[tex]i \: figured \: it \: out \to \: notice \: that \\ there \: are \:( thre e\: trianges) \: in \: triangle \: ( ABC) \\ and \: they \: all \: have \: the \: same \: height \: = x \\ first \: triangle \: is \: \to \: triangle \: (ABD) \\ second\: triangle \: is \: \to \: triangle \: (ACD) \\ third \: triangle \: is \: \to \: triangle \: (BCD) \\ if \: the \: area \: of \: a \: triangle \: is \: = \frac{1}{2} bh : then \to \\ the \: area \: of \: triangle \: (ABD) = \frac{1}{2} (AB)x \\ the \: area \: of \: triangle \: (ACD) = \frac{1}{2} (AC)x \\ the \: area \: of \: triangle \: (BCD) = \frac{1}{2} (BC)x \\ let \: the \: total \: area \:(A_T) \: be \to \: \frac{1}{2} (AB)x + \frac{1}{2} (AC)x + \frac{1}{2} (BC)x \\ if \to\\ (A_T) = \frac{1}{2} (AB)x + \frac{1}{2} (AC)x + \frac{1}{2} (BC)x \\ then \to \\ (A_T) = \frac{1}{2} \{(AB)x + (AC)x + (BC)x \} \\ x \{(AB) + (AC) + (BC) \} = 2(A_T) \\ x = \frac{2(A_T) }{(AB) + (AC) + (BC)} [/tex]

 Expression representing the length 'x' will be, [tex]x=\frac{(AB)(AC)}{AC+BC+AB}[/tex].

Use the hint → Area of the large triangle = sum of the small triangles inside the larger triangle

Area of ΔABC = [tex]\frac{1}{2}(AB)(AC)[/tex]

Area of ΔADC = [tex]\frac{1}{2}(AC)(x)[/tex]

Area of ΔBDC = [tex]\frac{1}{2}(BC)(x)[/tex]

Area of ΔADB = [tex]\frac{1}{2}(AB)(x)[/tex]

Since, Area of ΔABC = Area of ΔADC + Area of ΔBDC + Area of ΔADB

[tex]\frac{1}{2}(AB)(AC)=\frac{1}{2}(AC)(x)+\frac{1}{2}(BC)(x)+\frac{1}{2}(AB)(x)[/tex]

[tex](AB)(AC)=[(AC)+(BC)+(AB)](x)[/tex]

[tex]x=\frac{(AB)(AC)}{AC+BC+AB}[/tex]

    Therefore, expression representing the length 'x' will be [tex]x=\frac{(AB)(AC)}{AC+BC+AB}[/tex].

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