câu hỏi là gì vậy bn

câu hỏi là đổi thành bị động nha

đề bài là gì 

23D

24A

25C

26C

27D

29D

có:

gọi giao điểm nửa đường tròn(H;BH) với AB là G

Ta có:

có: \(\Delta ABC\) vuông tại A\(=>AB^2=BH.BC=>BH=\dfrac{AB^2}{BC}=\dfrac{3^2}{6}=1,5cm=R\)

\(=>S\)(nửa đường tròn tâm H)\(=\dfrac{1}{2}\pi R^2=\dfrac{1}{2}.1,5.\pi=\dfrac{3}{4}.3,14=2,355cm^2\)

xét \(\Delta ABH\) vuông tại H\(=>\sin\angle\left(A\right)=\dfrac{BH}{AB}=\dfrac{1,5}{3}=\dfrac{1}{2}=>\angle\left(A\right)=30^o=>\angle\left(B\right)=60^o\)

xét \(\Delta BHG\) có\(\left\{{}\begin{matrix}HB=HG=R\\\angle\left(B\right)=60^o\end{matrix}\right.\)=>\(\Delta BHG\) đều\(=>S\left(\Delta BHG\right)=R^2.\dfrac{\sqrt{3}}{4}=1,5^2.\dfrac{\sqrt{3}}{4}=\dfrac{9\sqrt{3}}{16}cm^2\)

có: \(S\)(quạt BHG)\(=\dfrac{\pi R^2n^o}{360^o}=\dfrac{3,14.1,5^2.60}{360}=1,1775cm^2\)

có: \(S\left(\Delta ABC\right)=\dfrac{AB.AC}{2}=\dfrac{3\sqrt{6^2-3^2}}{2}=\dfrac{9\sqrt{3}}{2}cm^2\)

\(=>S\)(phần tô đậm)\(=S\left(\Delta ABC\right)-\)[S(nửa đường tròn tâm H)\(-S\)(quạt tròn BHG-\(S\left(\Delta BHG\right)\)]

(bạn tự thay vào đi)

1.Why(you/not take).............didn't you take......................part in the picnic last Sunday?

2. Where (your teacher/be ).....is your teacher..............?

He (explain)........explain........a new lesson in that class

3. We cant't go out now because it(rain)...is raining..........

4. This is the best wine i have ever (drink).....drunk............

1. Why ( you/ not take) .......don't you take...........part in the picnic last Sunday?

2. Where ( your teacher/ be).....is your teacher ...............?

- He (explain)......is explaining................a new lesson in that class.

3. We can't go out now because it (rain)....... is raining........... .

4. This is the best wine I ever ( drink).......... have ever drunk..............

1. didn't you take

2. is your teacher

    is explaining

3. is raining

4. have ever drunk