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Có \(4AB=3BC\Rightarrow BC=\dfrac{4AB}{3}\)
+) Tam giác ABC vuông tại B có:
\(AC^2=AB^2+BC^2\left(pytago\right)\)
\(\Rightarrow AC=\sqrt{AB^2+BC^2}=\sqrt{AB^2+\left(\dfrac{4AB}{3}\right)^2}=\dfrac{5AB}{3}\)
+) Tam giác ABC vuông tại B, đường cao BH có:
\(AB.BC=BH.AC\\\Leftrightarrow AB.\dfrac{4AB}{3}=\dfrac{12}{5}.\dfrac{5AB}{3}\\ \Rightarrow AB=3\left(cm\right) \)
\(\Rightarrow BC=\dfrac{4AB}{3}=\dfrac{4.3}{3}=4\left(cm\right),AC=\dfrac{5AB}{3}=\dfrac{5.3}{3}=5\left(cm\right)\)
-) Có: \(AB^2=AH.AC\) (hệ thức lượng)
\(\Leftrightarrow3^2=AH.5\\ \Rightarrow AH=\dfrac{9}{5}\left(cm\right)\)
-) Có: \(BC^2=CH.AC\) (hệ thức lượng)
\(\Leftrightarrow4^2=CH.5\\ \Rightarrow CH=\dfrac{16}{5}\left(cm\right)\)
4AB=3BC
=>AB/3=BC/4=k
=>AB=3k; BC=4k
1/BA^2+1/BC^2=1/BH^2=1/2,4^2
=>k=1
=>AB=3cm; BC=4cm
AC=căn 3^2+4^2=5cm
AH=3^2/5=1,8cm
CH=5-1,8=3,2cm
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 \(\left\{{}\begin{matrix}AC^2=CH\cdot BC\\AH\cdot BC=AB\cdot AC\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}AC=15\left(cm\right)\\AH=12\left(cm\right)\end{matrix}\right.\)
a: Ta có: AB<AC
nên HB<HC
hay \(\left\{{}\begin{matrix}HB< 12.5\left(cm\right)\\HC>12.5\left(cm\right)\end{matrix}\right.\)
Ta có: HB+HC=BC
nên HB=25-HC
Ta có: \(AH^2=HB\cdot HC\)
\(\Leftrightarrow HC\left(25-HC\right)=12^2=144\)
\(\Leftrightarrow HC^2-25HC+144=0\)
\(\Leftrightarrow HC=16\left(cm\right)\)
\(\Leftrightarrow HB=9\left(cm\right)\)
Áp dụng hệ thức lượng trong tam giác vuông vào ΔABC vuông tại A có AH là đường cao ứng với cạnh huyền BC, ta được:
\(\left\{{}\begin{matrix}AB^2=HB\cdot BC\\AC^2=HC\cdot BC\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}AB=15\left(cm\right)\\AC=20\left(cm\right)\end{matrix}\right.\)
a) Áp dụng định lí Pytago trong \(\Delta\) AHC vuông tại H ta có :
\(AH^2+HC^2=AC^2\)
\(\Rightarrow HC^2=AC^2-AH^2\)
\(\Rightarrow HC=\sqrt{AC^2-AH^2}=\sqrt{40^2-24^2}=32cm\)
b) Áp dụng định lí Pytago trong \(\Delta\) AHC vuông tại H ta có :
\(AH^2+HC^2=AC^2\)
\(\Rightarrow AC=\sqrt{AH^2+HC^2}=\sqrt{9,6^2+12,8^2}=16cm\)
c) \(BC=CH+BH=72+12,5=84,5\left(cm\right)\)
Ta có: \(\left\{{}\begin{matrix}AB^2=BH.BC=12,5.84,5=1056,25\\AC^2=CH.BC=72.84,5=6084\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}AB=\dfrac{65}{2}\left(cm\right)\\AC=78\left(cm\right)\end{matrix}\right.\)
Ta có: \(AB.AC=AH.BC\Rightarrow AH=\dfrac{AB.AC}{BC}=\dfrac{78.\dfrac{65}{2}}{84,5}=30\left(cm\right)\)
\(4AB=3BC\Leftrightarrow AB=\dfrac{3}{4}BC\)
Áp dụng HTL: \(AB^2=BH\cdot BC\Leftrightarrow\dfrac{9}{16}BC^2=\dfrac{12}{5}BC\Leftrightarrow BC\left(\dfrac{9}{16}BC-\dfrac{12}{5}\right)=0\\ \Leftrightarrow BC=\dfrac{12}{5}:\dfrac{9}{16}=\dfrac{64}{15}\left(cm\right)\\ \Leftrightarrow AB=\dfrac{16}{5}\left(cm\right)\)
Áp dụng HTL và PTG: \(\left\{{}\begin{matrix}AC=\sqrt{BC^2-AB^2}=\dfrac{16\sqrt{7}}{15}\left(cm\right)\\CH=\dfrac{AC^2}{BC}=\dfrac{28}{15}\left(cm\right)\end{matrix}\right.\)
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