Question

[Bf Hx]

Cho tứ giác ABCD có A=2B=3C=4D TÍNH D

Answer 1

[Trúc Giang]

Ta có: \(\left\{{}\begin{matrix}2\widehat{B}=4\widehat{D}\\3\widehat{C}=4\widehat{D}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\widehat{B}=2\widehat{D}\\\widehat{C}=\dfrac{4}{3}\widehat{D}\end{matrix}\right.\)

Tứ giác ABCD có:

\(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^0\)

\(\Rightarrow4\widehat{D}+2\widehat{D}+\dfrac{4}{3}\widehat{D}+\widehat{D}=360^0\)

\(\Rightarrow\widehat{D}\left(4+2+\dfrac{4}{3}+1\right)=360^0\)

\(\Rightarrow\widehat{D}.\dfrac{25}{3}=360^0\)

\(\Rightarrow\widehat{D}=360^0:\dfrac{25}{3}=43,2^0\)

Answer 2

[Minh Nhân]

\(TC:\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^0\)

\(\Rightarrow\widehat{A}+\dfrac{1}{2}\widehat{A}+\dfrac{1}{3}\widehat{A}+\dfrac{1}{4}\widehat{A}=360^0\)

\(\Rightarrow\widehat{A}=172.8^0\)

\(\widehat{D}=\dfrac{1}{4}\widehat{A}=\dfrac{1}{4}\cdot172.8=43.2^0\)