I am trying to get the intersection of 8-16-24-32 and16-32-48

Mathematics

1 answer:

kakasveta [241]2 years ago

4 0

{8, 16, 24, 32}int.{16, 32, 48}={16, 32}

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2. Given a quadrilateral with vertices (−1, 3), (1, 5), (5, 1), and (3,−1):

zlopas [31]

<h2>Explanation:</h2>

In every rectangle, the two diagonals have the same length. If a quadrilateral's diagonals have the same length, that doesn't mean it has to be a rectangle, but if a parallelogram's diagonals have the same length, then it's definitely a rectangle.

So first of all, let's prove this is a parallelogram. The basic definition of a parallelogram is that it is a quadrilateral where both pairs of opposite sides are parallel.

So let's name the vertices as:

$A(-1,3) \\ \\ B(1,5) \\ \\ C(5,1) \\ \\ D(3,-1)$

First pair of opposite sides:

<u>Slope:</u>

$\text{For AB}: \\ \\ m=\frac{5-3}{1-(-1)}=1 \\ \\ \\ \text{For CD}: \\ \\ m=\frac{1-(-1)}{5-3}=1 \\ \\ \\ \text{So AB and CD are parallel}$

Second pair of opposite sides:

<u>Slope:</u>

$\text{For BC}: \\ \\ m=\frac{1-5}{5-1}=-1 \\ \\ \\ \text{For AD}: \\ \\ m=\frac{-1-3}{3-(-1)}=-1 \\ \\ \\ \text{So BC and AD are parallel}$

So in fact this is a parallelogram. The other thing we need to prove is that the diagonals measure the same. Using distance formula:

$d=\sqrt{(y_{2}-y_{1})^2+(x_{2}-x_{1})^2} \\ \\ \\ Diagonal \ BD: \\ \\ d=\sqrt{(5-(-1))^2+(1-3)^2}=2\sqrt{10} \\ \\ \\ Diagonal \ AC: \\ \\ d=\sqrt{(3-1)^2+(-5-1)^2}=2\sqrt{10} \\ \\ \\$