cho tứ giavs ABCD NHƯ HÌNH VẼ Bài 2. Cho tứ giác ABCD như hình vẽ.,Hãy tính tổng: $\widehat{A_1}+\widehat{B_1}+\widehat{C_1}+\widehat{D_1}.$,,Giải:,Ta có: $\widehat{A_1}+\widehat{A_2}=180^0$ (hai góc kề bù).,Tương tự: $\widehat{B_1}+\widehat{B_2}=.$,$\widehat{C_1}+\widehat{C_2}=...;\widehat{D_1}+\widehat{D_2}=...$,Do đó:,$\widehat{A_1}+\widehat{A_2}+\widehat{B_1}+\widehat{B_2}+\widehat{C_1}+\widehat{C_2}+\widehat{D_1}+\widehat{D_2}=.$,Mà: $\widehat{A_2}+\widehat{B_2}+\widehat{C_2}+\widehat{D_2}=...$ (tổng các góc trong tứ giác).,Nên: $\widehat{A_1}+\widehat{B_1}+\widehat{C_1}+\widehat{D_1}=...=...$,Vậy: $\widehat{A_1}+\widehat{B_1}+\widehat{C_1}+\widehat{D_1}=.$

Question

cho tứ giavs ABCD NHƯ HÌNH VẼ Bài 2. Cho tứ giác ABCD như hình vẽ.,Hãy tính tổng: $\widehat{A_1}+\widehat{B_1}+\widehat{C_1}+\widehat{D_1}.$,<img src=https://minio.ftech.ai/cvdata/fqa/prod/public/illustration_images/e623c944a8b04f239973c9a27920a39d.jpg />,Giải:,Ta có: $\widehat{A_1}+\widehat{A_2}=180^0$ (hai góc kề bù).,Tương tự: $\widehat{B_1}+\widehat{B_2}=.$,$\widehat{C_1}+\widehat{C_2}=...;\widehat{D_1}+\widehat{D_2}=...$,Do đó:,$\widehat{A_1}+\widehat{A_2}+\widehat{B_1}+\widehat{B_2}+\widehat{C_1}+\widehat{C_2}+\widehat{D_1}+\widehat{D_2}=.$,Mà: $\widehat{A_2}+\widehat{B_2}+\widehat{C_2}+\widehat{D_2}=...$ (tổng các góc trong tứ giác).,Nên: $\widehat{A_1}+\widehat{B_1}+\widehat{C_1}+\widehat{D_1}=...=...$,Vậy: $\widehat{A_1}+\widehat{B_1}+\widehat{C_1}+\widehat{D_1}=.$

Timi · Accepted Answer

Bài 2.
Ta có: $\widehat{A_1}+\widehat{A_2}=180^0$ (hai góc kề bù).
Tương tự: $\widehat{B_1}+\widehat{B_2}=180^0$;
$\widehat{C_1}+\widehat{C_2}=180^0;\widehat{D_1}+\widehat{D_2}=180^0$
Do đó:
$\widehat{A_1}+\widehat{A_2}+\widehat{B_1}+\widehat{B_2}+\widehat{C_1}+\widehat{C_2}+\widehat{D_1}+\widehat{D_2}=720^0$
Mà: $\widehat{A_2}+\widehat{B_2}+\widehat{C_2}+\widehat{D_2}=360^0$ (tổng các góc trong tứ giác).
Nên: $\widehat{A_1}+\widehat{B_1}+\widehat{C_1}+\widehat{D_1}=720^0-360^0=360^0$

wonbii · Answer

ta có: $\widehat{A_1}+\widehat{A_2}=180^0$ (hai góc kề bù).
tương tự: $\widehat{B_1}+\widehat{B_2}=180^0$;
$\widehat{C_1}+\widehat{C_2}=180^0;\widehat{D_1}+\widehat{D_2}=180^0$
do đó:
$\widehat{A_1}+\widehat{A_2}+\widehat{B_1}+\widehat{B_2}+\widehat{C_1}+\widehat{C_2}+\widehat{D_1}+\widehat{D_2}=720^0$
mà: $\widehat{A_2}+\widehat{B_2}+\widehat{C_2}+\widehat{D_2}=360^0$ (tổng các góc trong tứ giác).
nên $\widehat{A_1}+\widehat{B_1}+\widehat{C_1}+\widehat{D_1}=720^0-360^0=360^0$

vậy $\widehat{A_1}+\widehat{B_1}+\widehat{C_1}+\widehat{D_1}=360^0$