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\(1,\)
\(a,\) Áp dụng HTL tam giác
\(\left\{{}\begin{matrix}AH^2=CH\cdot BH\\AB^2=BH\cdot BC\\AC^2=CH\cdot BC\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}BH=\dfrac{AH^2}{CH}=\dfrac{25}{6}\left(cm\right)\\AB=\sqrt{\dfrac{25}{6}\left(\dfrac{25}{6}+6\right)}=\dfrac{5\sqrt{61}}{6}\left(cm\right)\\AC=\sqrt{6\left(\dfrac{25}{6}+6\right)}=\sqrt{61}\left(cm\right)\end{matrix}\right.\\ BC=\dfrac{25}{6}+6=\dfrac{61}{6}\left(cm\right)\)
\(b,S_{ABC}=\dfrac{1}{2}AH\cdot BC=\dfrac{1}{2}\cdot5\cdot\dfrac{61}{6}=\dfrac{305}{12}\left(cm^2\right)\)
Bài 3:
Ta có:
\(\widehat{M}+\widehat{N}+\widehat{P}=180^o\)
\(\Rightarrow\widehat{P}=180^o-90^o-37^o=53^o\)
Mà: \(sinN=\dfrac{MN}{NP}\)
\(\Rightarrow sin37^o=\dfrac{MN}{25}\)
\(\Rightarrow MN=25\cdot sin37^o\approx15\left(cm\right)\)
Áp dung định lý Py-ta-go ta có:
\(MP=\sqrt{NP^2-MN^2}=\sqrt{25^2-15^2}=20\left(cm\right)\)
3:
a: Xét ΔABC có AC^2=BA^2+BC^2
nên ΔBAC vuông tại B
b: Xét ΔBAC vuông tại B có
sin A=BC/AC=42/58=21/29
cos A=AB/AC=40/58=20/29
tan A=BC/BA=21/20
cot A=BA/BC=20/21
c: Xét ΔABC vuông tại B có BH là đường cao
nên BH*AC=BA*BC; BA^2=AH*AC; CB^2=CH*CA
=>BH*58=40*42=1680
=>BH=840/29(cm)
BA^2=AH*AC
=>AH=BA^2/AC=40^2/58=800/29cm
CB^2=CH*CA
=>CH=CB^2/CA=42^2/58=882/29(cm)
ΔBHA vuông tại H có HE là đường cao
nênBE*BA=BH^2
=>BE*40=(840/29)^2
=>BE=17640/841(cm)
ΔBHC vuông tại H có HF là đường cao
nênBF*BC=BH^2
=>BF*42=(840/29)^2
=>BF=16800/841(cm)
Xét tứ giác BEHF có
góc BEH=góc BFH=góc EBF=90 độ
=>BEHF là hình chữ nhật
=>góc BFE=góc BHE(=1/2*sđ cung BE)
=>góc BFE=góc BAC
Xét ΔBFE và ΔBAC có
góc BFE=góc BAC
góc FBE chung
Do đó: ΔBFE đồng dạng với ΔBAC
=>S BFE/S BAC=(BF/BA)^2=(16800/441:40)^2=(420/841)^2
=>S AECF=S ABC*(1-(420/841)^2)
=>\(S_{AECF}=\dfrac{1}{2}\cdot40\cdot42\cdot\left[1-\left(\dfrac{420}{841}\right)^2\right]\simeq630,5\left(cm^2\right)\)
Đặt BH = x (0 < x < 25) (cm) => CH = 25 - x (cm)
Ta có : \(AH^2=BH.CH\Rightarrow x\left(25-x\right)=144\Leftrightarrow x^2-25x+144=0\)
\(\left(x-9\right)\left(x-16\right)=0\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=9\\x=16\end{array}\right.\) (tm)
Nếu BH = 9 cm thì CH = 16 cm\(\Rightarrow AB=\sqrt{AH^2+BH^2}=\sqrt{9^2+12^2}=15\left(cm\right)\)
\(AC=\sqrt{AH^2+CH^2}=\sqrt{12^2+16^2}=20\left(cm\right)\)
Nếu BH = 16 cm thì CH = 9 cm
\(\Rightarrow AB=\sqrt{AH^2+BH^2}=\sqrt{12^2+16^2}=20\left(cm\right)\)
\(AC=\sqrt{AH^2+CH^2}=\sqrt{9^2+12^2}=15\left(cm\right)\)
Gỉa sử \(\Delta ABC\) có AB>AC
\(AB.AC=AH.BC=12.25=300\)
\(\Leftrightarrow2AB.AC=2.300=600\)
Áp dụng định lý Pytago cho \(\Delta ABC\) vuông tại A ta có:
\(AB^2+AC^2=BC^2=25^2=625\) (1)
\(\left(1\right)\Rightarrow AB^2+AC^2-2AB.AC=625-600\)
\(\Leftrightarrow\left(AB-AC\right)^2=25\Leftrightarrow AB-AC=5\) (a) (Vì AB>AC \(\Rightarrow AB-AC>0\))
\(\left(1\right)\Rightarrow AB^2+AC^2+2AB.AC=600+625=1225\)
\(\Leftrightarrow\left(AB+AC\right)^2=1225\Rightarrow AB+AC=35\) (b)
Cộng vế vs vế của (a) và (b) ta được: \(2AB=40\Rightarrow AB=20\)
\(\Rightarrow AC=AB-5=20-5=15\)
Xét \(\Delta ABC\) vuông tại A, \(AH\perp BC\)\(\Rightarrow\) theo hệ thức lượng trong tam giác vuông ta có:
\(AB^2=BH.BC\Rightarrow BH=\frac{AB^2}{BC}=\frac{20^2}{25}=16\)
\(\Rightarrow CH=BC-BH=25-16=9\)
Xét tam giác ABC vuông ta có:
\(BC=\sqrt{AB^2+AC^2}=\sqrt{24^2+10^2}=26\left(cm\right)\)
\(\left\{{}\begin{matrix}BH=\dfrac{AB^2}{BC}=\dfrac{10^2}{26}\approx4\left(cm\right)\\HC=\dfrac{AC^2}{BC}=\dfrac{24^2}{26}\approx22\left(cm\right)\end{matrix}\right.\)
Xét tam giác ABH vuông tại H áp dung Py-ta-go ta có:
\(\Rightarrow AH=\sqrt{AB^2-BH^2}=\sqrt{10^2-4^2}=2\sqrt{21}\left(cm\right)\)
\(\Rightarrow S_{ABC}=\dfrac{1}{2}\cdot AH\cdot BC=\dfrac{1}{2}\cdot2\sqrt{21}\cdot26=26\sqrt{21}\left(cm^2\right)\)
Ta có :
\(BC^2=AB^2+AC^2\left(Pitago\right)\)
\(\Leftrightarrow BC^2=100+576=676\)
\(\Leftrightarrow BC=26\left(cm\right)\)
\(AB^2=BH.BC\Leftrightarrow BH=\dfrac{AB^2}{BC}=\dfrac{100}{26}=\dfrac{50}{13}\left(cm\right)\)
\(BC=BH-HC\)
\(\Leftrightarrow HC=BC-BH=26-\dfrac{50}{13}=\dfrac{288}{13}\left(cm\right)\)
\(AH^2=BH.HC=\dfrac{50}{13}.\dfrac{288}{13}=\dfrac{14400}{13^2}\)
\(\Leftrightarrow AH=\dfrac{120}{13}\left(cm\right)\)
\(S_{ABC}=\dfrac{1}{2}.AB.AC=\dfrac{1}{2}.10.24=120\left(cm^2\right)\)
Hoặc : \(S_{ABC}=\dfrac{1}{2}.AH.BC=\dfrac{1}{2}.\dfrac{120}{13}.26=120\left(cm^2\right)\)
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
Bài 5:
Ta có: \(AB^2=BH\cdot BC\)
\(\Leftrightarrow BH\left(BH+9\right)=400\)
\(\Leftrightarrow BH^2+25HB-16HB-400=0\)
\(\Leftrightarrow BH=16\left(cm\right)\)
hay BC=25(cm)
Xét ΔABC vuông tại A có AH là đường cao ứng với cạnh huyền BC
nên \(\left\{{}\begin{matrix}AC^2=CH\cdot BC\\AH\cdot BC=AB\cdot AC\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}AC=15\left(cm\right)\\AH=12\left(cm\right)\end{matrix}\right.\)