There are $25$ points on a plane of which $7$ are collinear . How many quadrilaterals can be formed from these points ?

Deja Bradshaw
2022-10-17
Answered

There are $25$ points on a plane of which $7$ are collinear . How many quadrilaterals can be formed from these points ?

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aitorarjolia

Answered 2022-10-18
Author has **11** answers

A quadrilateral is formed by $4$ points, where at most $2$ may be colinear. Thus, we have

$\underset{\text{choose four points}}{\underset{\u23df}{{}^{25}{C}_{4}}}-\underset{\text{subtract out ways to pick four colinear points}}{\underset{\u23df}{{}^{7}{C}_{4}}}-\underset{\text{subtract out ways to pick three colinear points}}{\underset{\u23df}{{}^{7}{C}_{3}{\cdot}^{18}{C}_{1}}}$

$\underset{\text{choose four points}}{\underset{\u23df}{{}^{25}{C}_{4}}}-\underset{\text{subtract out ways to pick four colinear points}}{\underset{\u23df}{{}^{7}{C}_{4}}}-\underset{\text{subtract out ways to pick three colinear points}}{\underset{\u23df}{{}^{7}{C}_{3}{\cdot}^{18}{C}_{1}}}$

asked 2022-08-22

Find number of Quadrilaterals that can be formed in a Decagon such that no side of Quadrilateral is common to side of Decagon.

asked 2022-08-22

As in the following figure of a quadrilateral;

If the diagonals are stated to bisect each other, I thought this should hold (considering the bottommost triangle in blue lines);

$\frac{\overrightarrow{a}+\overrightarrow{b}}{2}+\frac{\overrightarrow{b}-\overrightarrow{a}}{2}=\overrightarrow{b}$

But this shows that all quadrilaterals have their diagonals bisecting each other, since this gives $\overrightarrow{b}=\overrightarrow{b}$ , which implies it's true for all $\overrightarrow{a}$ and $\overrightarrow{b}$. Which obviously isn't true.

Where did my reasoning go wrong?

If the diagonals are stated to bisect each other, I thought this should hold (considering the bottommost triangle in blue lines);

$\frac{\overrightarrow{a}+\overrightarrow{b}}{2}+\frac{\overrightarrow{b}-\overrightarrow{a}}{2}=\overrightarrow{b}$

But this shows that all quadrilaterals have their diagonals bisecting each other, since this gives $\overrightarrow{b}=\overrightarrow{b}$ , which implies it's true for all $\overrightarrow{a}$ and $\overrightarrow{b}$. Which obviously isn't true.

Where did my reasoning go wrong?

asked 2022-07-28

The dimensions of a rectangular room are 12 feet 10 inches by 10 feet 1 inch. If square tiles of the same size are to cover the floor completely without any overlapping and if only whole tiles will be use, what is the largest possible length of a side of the tile?

asked 2022-09-04

Let ${P}_{1}$ and ${P}_{2}$ be two convex quadrilaterals such that ${P}_{1}\ne {P}_{2}$ and $Area({P}_{1})\ge Area({P}_{2})$. Is it true that it is not possible that all sides and diagonals of ${P}_{1}$ are shorter than the corresponding sides and diagonals of ${P}_{2}$?

Suppose that all four sides and one diagonal of ${P}_{1}$ are shorter than the corresponding sides and diagonal of ${P}_{2}$. Then, as it is explained in the answer to my older question, the two triangles which triangulate ${P}_{2}$ should be "flatter" (otherwise it's not possible $Area({P}_{1})\ge Area({P}_{2})$). From this it should follow that the other diagonal of ${P}_{1}$ is longer than the corresponding diagonal of ${P}_{2}$.

Do you think it's correct? Can you help me formalizing it?

Suppose that all four sides and one diagonal of ${P}_{1}$ are shorter than the corresponding sides and diagonal of ${P}_{2}$. Then, as it is explained in the answer to my older question, the two triangles which triangulate ${P}_{2}$ should be "flatter" (otherwise it's not possible $Area({P}_{1})\ge Area({P}_{2})$). From this it should follow that the other diagonal of ${P}_{1}$ is longer than the corresponding diagonal of ${P}_{2}$.

Do you think it's correct? Can you help me formalizing it?

asked 2022-08-18

This is a question from Kiselev's plane geometry book:

Four points on the plane are vertices of three quadrilaterals. Explain how this happens.

How do you explain this?

Four points on the plane are vertices of three quadrilaterals. Explain how this happens.

How do you explain this?

asked 2022-08-20

Given two parallel lines, the first one containing $N$ points and the second containing $M$ points. How many triangles and convex quadrilaterals can be formed if no line contains more than $1000$ points

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Every polygon with an even number of vertices may be partitioned by diagonals into quadrilaterals.

True or False?

True or False?