[Tóan 9]Hệ thức lượng trong tam giác vuông

T

thanhcong1594

Áp dụng định lý pytago trong Δ ABD có:

$BD$ = $\sqrt[]{AD^2 + AB^2}$
$BD$ = $\sqrt[]{15^2 + 20^2}$
$BD$ = $25$
Ta có: $AB^2$ = $OB$ . $BD$
$OB$ = $\frac{AB^2}{BD}$ = $\frac{15^2}{25}$
$OD$ = $BD$ - $OB$ = $25$ - $9$ = $16$ cm
Ta có: $AO^2$ = $OD$ . $OB$
$AO^2$ = $16$ . $9$ = $144$
$AO$ = $12$ cm
Ta lại có: $OD^2$ = $OA$ .$OC$
$OC$ = $\frac{OD^2}{OA}$ = $\frac{16^2}{12}$
$OC$ = $21,3$ cm
b/ Ta có: $AC$ = $OA$ + $OC$ = $21,3$ +$12$ = $33,3$ cm
c/ $SABC$ = $\frac{AC.BD}{2}$ = $416,25$ cm
 
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