Question

Cho hình bình hành ABCD, có AD = 4cm, AB = 5cm, A = 60◦ . 

a) Tính S ABCD. b) Hạ AK⊥DB. Tính độ dài AK.

Answer 1

Lời giải:
a. Kẻ $BH\perp AD$. 

Có: $\frac{BH}{AB}=\sin A=\sin 60^0=\frac{\sqrt{3}}{2}$

$BH=AB.\frac{\sqrt{3}}{2}=\frac{5}{2}\sqrt{3}$

$S_{ABCD}=BH.AD=\frac{5}{2}\sqrt{3}.4=10\sqrt{3}$ (cm²)

b.

$AH=AB\cos A=5\cos 60^0=\frac{5}{2}$ (cm)

$HD=AD-AH=4-\frac{5}{2}=1,5$ (cm)

$\tan \widehat{BDA}=\tan \widehat{BDH}=\frac{BH}{HD}=\frac{5}{2}\sqrt{3}: 1,5=\frac{5\sqrt{3}}{3}$

$\Rightarrow \widehat{BDA}=70,9^0$
$\widehat{ABK}=\widehat{ABD}=180^0-\widehat{BAD}-\widehat{BDA}=180^0-60^0-70,9^0=49,1^0$

$AK=AB\sin \widehat{ABK}=5\sin 49,1^0=3,8$ (cm)

Answer 2

Hình vẽ: