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Question Number 143003 by bramlexs22 last updated on 08/Jun/21

	Answered by EDWIN88 last updated on 08/Jun/21
	$(1) Since ΔACE and ΔACB share the same altitude,and AE=(1/3)AB ,the area of ΔACE= (1/3) the area of ΔACB . By Heron formula (1/3) the area of ΔACB = (1/3)(√(((15)/2)((7/2))((5/2))((3/2)))) =((5(√7))/4) Let CE = x . the area of ΔACE = (√((((6+x)/2))(((6−x)/2))(((x+2)/2))(((x−2)/2)))) =(1/4)(√(−(x^2 −36)(x^2 −4))) let y=x^2 ⇒((5(√7))/4)=(1/4)(√(−(y^2 −40y+144))) ⇒y^2 −49y+319=0 ⇒(y−11)(y−29)=0→ { ((y=11⇒x=(√(11)))),((y=29(rejected))) :} therefore CE = (√(11)) □$
	$(1) Since Δ ACE and Δ ACB share the same$ $altitude, and AE = \frac{1}{3} AB, the area of Δ ACE =$ $\frac{1}{3} the area of Δ ACB .$ $By Heron formula$ $\frac{1}{3} the area of Δ ACB = \frac{1}{3} \sqrt{\frac{15}{2} (\frac{7}{2}) (\frac{5}{2}) (\frac{3}{2})} = \frac{5 \sqrt{7}}{4}$ $Let CE = x . the area of Δ ACE$ $= \sqrt{(\frac{6 + x}{2}) (\frac{6 - x}{2}) (\frac{x + 2}{2}) (\frac{x - 2}{2})}$ $= \frac{1}{4} \sqrt{- (x^{2} - 36) (x^{2} - 4)}$ $let y = x^{2} \Rightarrow \frac{5 \sqrt{7}}{4} = \frac{1}{4} \sqrt{- (y^{2} - 40 y + 144)}$ $\Rightarrow y^{2} - 49 y + 319 = 0$ $\Rightarrow (y - 11) (y - 29) = 0 \to {\begin{cases} y = 11 \Rightarrow x = \sqrt{11} \\ y = 29 (rejected) \end{cases}$ $therefore CE = \sqrt{11} ◻$
	Commented by EDWIN88 last updated on 08/Jun/21

	$the other way$ $(1) by applying {Stewart}^{'} s theorem to Δ ABC$ $\Rightarrow {(AC)}^{2} (EB) + {(CB)}^{2} (AE) = AB [{(CE)}^{2} + (AE) (EB)]$ $\Rightarrow 4^{2} . 4 + 5^{2} . 2 = 6 [{(CE)}^{2} + 2.4]$ $\Rightarrow 114 = 6 {(CE)}^{2} + 48$ $\Rightarrow CE = \sqrt{\frac{114 - 48}{6}} = \sqrt{\frac{66}{6}} = \sqrt{11} ◻$
	Commented by bramlexs22 last updated on 09/Jun/21

	$n i c e$
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