3: 

\(BC=\sqrt{12^2+16^2}=20\left(cm\right)\)

HB=12^2/20=7,2cm

=>HC=20-7,2=12,8cm

\(AD=\dfrac{2\cdot12\cdot16}{12+16}\cdot cos45=\dfrac{48\sqrt{2}}{7}\)

\(HD=\sqrt{AD^2-AH^2}=\dfrac{48}{35}\left(cm\right)\)

27/12/2017 lúc 18:59

Ex1: Điền từ thích hợp vào chỗ trống

 This is Ba. He(1)......... a student.Every morning he(2).........up at 5.30.He(3).............. his teeth and takes a(4)............... then has breakfast at 6.15. He goes to school(5)........six thirty.His house is(6).............his house so he walks.The classes(7)............at 7.15 and finish at 11.15.In the afternoon he plays sports with his friend,Nam. They play badminton but now they(8).................soccer.In the evening he (9)......his homework and goes to(10).........at 9.30

Ex2:Cho dạng đúng của động từ trong ngoặc

1.My sister(have)...........classes from Monday to Friday

2.She(read)................a book in her room now

3.He(get)........................up at 6.00 every day?

4.There(not be)..............a big yard behind his classroom

27/12/2017 lúc 18:59

Ex1: Điền từ thích hợp vào chỗ trống

 This is Ba. He(1)......... a student.Every morning he(2).........up at 5.30.He(3).............. his teeth and takes a(4)............... then has breakfast at 6.15. He goes to school(5)........six thirty.His house is(6).............his house so he walks.The classes(7)............at 7.15 and finish at 11.15.In the afternoon he plays sports with his friend,Nam. They play badminton but now they(8).................soccer.In the evening he (9)......his homework and goes to(10).........at 9.30

Ex2:Cho dạng đúng của động từ trong ngoặc

1.My sister(have)...........classes from Monday to Friday

2.She(read)................a book in her room now

3.He(get)........................up at 6.00 every day?

4.There(not be)..............a big yard behind his classroom

27/12/2017 lúc 18:59

Ex1: Điền từ thích hợp vào chỗ trống

 This is Ba. He(1)......... a student.Every morning he(2).........up at 5.30.He(3).............. his teeth and takes a(4)............... then has breakfast at 6.15. He goes to school(5)........six thirty.His house is(6).............his house so he walks.The classes(7)............at 7.15 and finish at 11.15.In the afternoon he plays sports with his friend,Nam. They play badminton but now they(8).................soccer.In the evening he (9)......his homework and goes to(10).........at 9.30

Ex2:Cho dạng đúng của động từ trong ngoặc

1.My sister(have)...........classes from Monday to Friday

2.She(read)................a book in her room now

3.He(get)........................up at 6.00 every day?

4.There(not be)..............a big yard behind his classroom

Dễ quá đi

Lời giải:
a. Áp dụng hệ thức lượng trong tam giác vuông:

$AB^2=BH.BC$

$AC^2=CH.CB$

$\Rightarrow (\frac{AB}{AC})^2=\frac{BH.BC}{CH.CB}=\frac{BH}{CH}$

$\Leftrightarrow (\frac{7}{24})^2=\frac{49}{576}=\frac{BH}{CH}$

b.

$\frac{BH}{CH}=\frac{49}{576}$

$BH+CH=BC=625$ (cm)

$\Rightarrow BH=625:(49+576).49=49$ (cm)

$CH=BC-BH=625-49=576$ (cm)

Hình vẽ:

a) Ta có: \(\dfrac{BH}{CH}=\left(\dfrac{AB}{AC}\right)^2\)

nên \(\dfrac{BH}{CH}=\left(\dfrac{7}{24}\right)^2=\dfrac{49}{576}\)

b) Ta có: \(\dfrac{BH}{CH}=\dfrac{49}{576}\)

nên \(BH=\dfrac{49}{576}CH\)

Ta có: BH+CH=BC(H nằm giữa B và C)

nên \(CH+\dfrac{49}{576}CH=625\)

\(\Leftrightarrow CH\cdot\dfrac{625}{576}=625\)

\(\Leftrightarrow CH=576\left(cm\right)\)

\(\Leftrightarrow BH=BC-CH=625-576=49\left(cm\right)\)

Ta có: \(AB^2\) = BH . BC  ;  \(AC^2\) = CH . BC

    Ta có:

    ⇒ BH  = 49 . 1 = 49

    ⇒ CH = 576 . 1 = 576

a) Ta có: \(\dfrac{BH}{HC}=\left(\dfrac{AB}{AC}\right)^2\)

\(\Leftrightarrow\dfrac{BH}{HC}=\dfrac{49}{576}\)

hay \(BH=\dfrac{49}{576}HC\)

Ta có: BH+HC=BC(H nằm giữa B và C)

\(\Leftrightarrow HC\cdot\dfrac{625}{576}=625\)

hay HC=576(cm)

\(\Leftrightarrow HB=BC-BH=625-576=49\left(cm\right)\)

Áp dụng hệ thức giữa cạnh và góc, ta có:

\(HB=AH.cotgB;HC=AH.cotgC\)

\(\Rightarrow AH=\frac{HB+HC}{cotgB+cotgC}=\frac{20}{cotg40^o+cotg30^o}\approx6,84\left(cm\right)\)

Lại áp dụng hệ thức giữa cạnh và góc, ta có:

\(AB=\frac{AH}{sinB}\approx\frac{6,84}{sin40^o}\approx10,64\left(cm\right)\)

\(AC=\frac{AH}{sinC}\approx\frac{6,84}{sin30^o}=13,68\left(cm\right)\)

\(AB=\sqrt{BH\cdot BC}=\sqrt{1.8\cdot5}=3\)

\(AC=\sqrt{5^2-3^2}=4\)

CH=BC-BH=3,2

\(AH=\dfrac{AB\cdot AC}{BC}=2.4\)