Question

Giúp mình với !!!

Answer 1

Ta có: \(\widehat{B}=\widehat{A}+\)20^o ; góc C = 3\(\widehat{A}\) ; \(\widehat{D}-\widehat{C}\) = 20^o
=> góc D = góc C + 20^o = 3 \(\widehat{A}\) + 20^o
Xét tứ giác ABCD có
góc A + góc B + góc C + góc D = 360^o
<=> góc A + (góc A +20^o) + 3\(\widehat{A}\) + (3\(\widehat{A}\) + 20^o) = 360^o
<=> 8\(\widehat{A}\) = 320^o
<=> góc A = 40^o
=> góc B = góc A + 20^o = 60^o
     góc C = 3\(\widehat{A}\) = 120^o
     góc D = 3\(\widehat{A}\) + 20^o = 140^o
b) ta có: góc A + góc D = 40⁰ + 140^o = 180^o
=> AB//CD (2 góc trong cùng phía bù nhau)
=> tứ giác ABCD là hình thang

Answer 2

a) Theo bài ta có :

\(\widehat{B}=\widehat{A}+20;\widehat{C}=3\widehat{A};\) \(\widehat{D}-\widehat{C}=20=>\widehat{D}=20+3\widehat{A}\)

Ta lại có \(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360=>\widehat{A}+20+3\widehat{A}+\widehat{A}+20+3\widehat{A}=8\widehat{A}+40=360=>8\widehat{A}=320=>\widehat{A}=40\)

\(\widehat{B}=60;\widehat{C}=120;\widehat{D}=140\)

b) Ta có \(\widehat{A}+\widehat{D}=\widehat{B}+\widehat{C}=180\)

=> ABCD là hình thang ( AB//CD)