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Question-221025




Question Number 221025 by Tawa11 last updated on 22/May/25
Answered by mr W last updated on 23/May/25
20 convex quadrilaterals (red)  5 concave quadrilaterals (blue)  25 cross−quadrilaterals (green)  ⇒totally 50 quadrilaterals
$$\mathrm{20}\:{convex}\:{quadrilaterals}\:\left({red}\right) \\ $$$$\mathrm{5}\:{concave}\:{quadrilaterals}\:\left({blue}\right) \\ $$$$\mathrm{25}\:{cross}−{quadrilaterals}\:\left({green}\right) \\ $$$$\Rightarrow{totally}\:\mathrm{50}\:{quadrilaterals} \\ $$
Commented by mr W last updated on 23/May/25
Commented by Tawa11 last updated on 23/May/25
Wow, thanks sir.  I appreciate.
$$\mathrm{Wow},\:\mathrm{thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{appreciate}. \\ $$
Commented by mr W last updated on 23/May/25
have you checked if i have found  all?
$${have}\:{you}\:{checked}\:{if}\:{i}\:{have}\:{found} \\ $$$${all}? \\ $$
Commented by Tawa11 last updated on 25/May/25
Yes sir.
$$\mathrm{Yes}\:\mathrm{sir}. \\ $$

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