Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 168489 by infinityaction last updated on 11/Apr/22

Commented by infinityaction last updated on 11/Apr/22

$$pleasesolvethisproblem$$

Answered by mr W last updated on 12/Apr/22

Commented by mr W last updated on 12/Apr/22

$$sayareaofparallelogram=1$$$$g_3=\frac{1}{6}$$$$g_2+y_2=\frac{1}{4}$$$$g_1=\frac{3}{15}\times \frac{1}{4}=\frac{1}{20}$$$$y_2=\frac{2}{15}\times \frac{1}{4}=\frac{1}{30}$$$$\Rightarrow g_2=\frac{1}{4}-\frac{1}{30}=\frac{13}{60}$$$$green=g_1+g_2+g_3=\frac{1}{20}+\frac{13}{60}+\frac{1}{6}=\frac{13}{30}$$$$yellow=1-\frac{13}{30}=\frac{17}{30}$$$$\frac{green}{yellow}=\frac{13}{17}$$

Commented by Tawa11 last updated on 12/Apr/22

$$\mathrm{Great}\mathrm{sir}.$$

Commented by Ari last updated on 12/Apr/22

Commented by Ari last updated on 12/Apr/22

can you speak please?

Commented by mr W last updated on 12/Apr/22

Commented by mr W last updated on 12/Apr/22

$$\frac{HC}{AH}=\frac{FH}{HD}=\frac{FC}{AD}=\frac{1}{2}$$$$\Rightarrow \frac{HC}{AC}=\frac{1}{3}$$$$\Rightarrow \frac{FH}{FD}=\frac{1}{3}$$$$\frac{EF}{AC}=\frac{1}{2}$$$$\frac{FG}{GH}=\frac{EF}{HC}=\frac{AC/2}{AC/3}=\frac{3}{2}$$$$\frac{FG}{FH}=\frac{3}{2+3}=\frac{3}{5}$$$$\frac{FG}{FD}=\frac{3/5\ast FH}{FD}=\frac{3/5\ast 1/3\ast FD}{FD}=\frac{1}{5}$$$$\frac{GH}{FD}=\frac{2/5\ast FH}{FD}=\frac{2/5\ast 1/3\ast FD}{FD}=\frac{2}{15}$$$$\frac{HD}{FD}=\frac{2}{3}$$$$\left[FCD\right]=\frac{parallelogram}{4}=\frac{1}{4}$$$$g_1=\frac{1}{5}\times \left[FCD\right]=\frac{1}{5}\times \frac{1}{4}=\frac{1}{20}$$$$g_3=\frac{2}{3}\times \left[FCD\right]=\frac{2}{3}\times \frac{1}{4}=\frac{1}{6}$$$$y_2=\frac{2}{15}\times \left[FCD\right]=\frac{2}{15}\times \frac{1}{4}=\frac{1}{30}$$$$g_2+y_2=\frac{parallelogram}{4}=\frac{1}{4}$$$$\Rightarrow g_2=\frac{1}{4}-\frac{1}{30}=\frac{13}{60}$$$$g_1+g_2+g_3=\frac{1}{20}+\frac{13}{60}+\frac{1}{6}=\frac{13}{30}$$

Commented by Ari last updated on 12/Apr/22

thanks Mr.W!

Terms of Service

Privacy Policy

Contact: info@tinkutara.com