toán 5

T

thien7202

C

cry_with_me



Gọi h là đường cao trong $\Delta{AED}$

$S_{\Delta{AED}} = \dfrac{1}{2}.h.AD$

$ \leftrightarrow 45 = \dfrac{1}{2}.h.15$

$\rightarrow h = 6 (cm^2)$

$ \rightarrow S_{\Delta{AEC}} = \dfrac{1}{2}.6.20 = 60 (cm^2)$

Gọi h' là đường cao trong $\Delta{ABC}$

ta có :

$S_{\Delta{AEC}} = \dfrac{1}{2}.h'.10 $

$\leftrightarrow 60 = \dfrac{1}{2}.h'.10 $

$ \rightarrow h' = 12 (cm^2)$

$\rightarrow S_{\Delta{ABC}} = \dfrac{1}{2}.12.15 = 90 (cm^2)$
 
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