Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
sin B=AH/AB
=>6/AB=sin60
=>\(AB=4\sqrt{3}\left(cm\right)\)
=>HB=2 căn 3(cm)
=>HC=8 căn 3(cm)
\(S_{AHC}=\dfrac{1}{2}\cdot8\sqrt{3}\cdot6=24\sqrt{3}\left(cm^2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Vì $AB: AC=3:7$ nên đặt $AB=3a; AC=7a$. Áp dụng hệ thức lượng trong tam giác vuông:
$\frac{1}{AH^2}=\frac{1}{AB^2}+\frac{1}{AC^2}$
$\frac{1}{42^2}=\frac{1}{(3a)^2}+\frac{1}{(7a)^2}$
$\frac{1}{42^2}=\frac{58}{441a^2}$
$\Rightarrow a=2\sqrt{58}$ (cm)
$AB=3a=6\sqrt{58}$ (cm)
$BH=\sqrt{AB^2-AH^2}=\sqrt{(6\sqrt{58})^2-42^2}=18$ (cm)
Chu vi $ABH$: $AB+BH+AH=6\sqrt{58}+18+42=60+6\sqrt{58}$ (cm)
$AC=7a=14\sqrt{58}$ (cm)
$HC=\sqrt{AC^2-AH^2}=\sqrt{(14\sqrt{58})^2-42^2}=98$ (cm)
$S_{AHC}=\frac{AH.HC}{2}=\frac{42.98}{2}=2058$ (cm vuông)
![](https://rs.olm.vn/images/avt/0.png?1311)
sin B=4/5
=>AH/AB=4/5
=>8/AB=4/5
=>AB=10cm
HB=căn 10^2-8^2=6cm
=>BC=10^2/6=50/3(cm)
S ABC=1/2*8*50/3=4*50/3=200/3cm2
![](https://rs.olm.vn/images/avt/0.png?1311)
bạn tự vẽ hình giúp mik nha
\(AH=\sqrt{AB^2-BH^2}\left(pytago\right)=\sqrt{6^2-3^2}=3\sqrt{3}\)
trong \(\Delta ABC\) vuông tại A có
\(AH^2=BH.HC\Rightarrow HC=\dfrac{AH^2}{BH}=\dfrac{\left(3\sqrt{3}\right)^2}{3}=9\)
\(AC=\sqrt{AH^2+HC^2}=\sqrt{\left(3\sqrt{3}\right)^2+9^2}=6\sqrt{3}\)
chu vi \(\Delta ABC\)
=AB+BC+AC=6+12+6\(\sqrt{3}\)=28,4
chu vi \(\Delta ABH\)
=AB+BH+AH=6+3+3\(\sqrt{3}\)=14,2
chu vi \(\Delta AHC\)
=AH+HC+AC=3\(\sqrt{3}\)+9+\(6\sqrt{3}\)=24,6
![](https://rs.olm.vn/images/avt/0.png?1311)
a, AB = 7,5cm, AC = 10cm, BC = 12,5cm, HC = 8cm
b, AH = 3 3 cm; P A B C = 18 + 6 3 c m ; P A B H = 9 + 3 3 c m ; P A C H = 9 + 9 3 c m
\(S_{ABC}=\dfrac{1}{2}\cdot AB\cdot AC\)
=>1/2*6*AC=24
=>AC*3=24
=>AC=8cm
=>BC=10cm
AH=6*8/10=4,8cm
H=8^2/10=6,4cm
S AHC=1/2*4,8*6,4=15,36cm2