TOAD is a quadrilateral with vertices (9, 10), (18, 10), (14, 5), and (5, 5), respectively. The diagonals intersect at Point S.

Mathematics

2 answers:

alekssr [168]2 years ago

7 0

⇒Plotting the Quadrilateral on two Dimensional Plane

And finding the diagonal of Quadrilateral bisect each other.

If you will look at the Quadrilateral, none of the interior angle is of 90°, so it can't be Square or Rectangle.

Finding the length of two Adjacent Sides

$TO=\sqrt{(18-9)^2+(10-10)^2}\\\\TO=9 \text{Unit}\\\\OA=\sqrt{(18-14)^2+(10-5)^2}\\\\OA=\sqrt{4^2+5^2}\\\\OA=\sqrt{41}$

As, the Length of Adjacent sides are not equal,so it can't be a Rhombus.

The Given Quadrilateral ,"TOAD" is a Parallelogram.

⇒The Point of Intersection of both Diagonal can be Obtained by

$=(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})\\\\\text{Point P}=(\frac{14+9}{2},\frac{5+10}{2})\\\\\text{Point P}=(\frac{23}{2},\frac{15}{2})\\\\\text{Point P}=(\frac{18+5}{2},\frac{10+5}{2})\\\\\text{Point P}=(\frac{23}{2},\frac{15}{2})$

Intersection of both diagonals is same, diagonals bisect each other.

Point of Intersection of Diagonal is Point P

$=(\frac{23}{2},\frac{15}{2})$

Part C

Distance formula

$=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2)}\\\\TP=\sqrt{(\frac{23}{2}-9)^2+(\frac{15}{2}-10)^2}\\\\TP=\sqrt{\frac{25}{4}+\frac{25}{4}}\\\\TP=\frac{5}{\sqrt{2}}\\\\PA=\sqrt{(\frac{23}{2}-14)^2+(\frac{15}{2}-5)^2}\\\\PA=\sqrt{\frac{25}{4}+\frac{25}{4}}\\\\TP=\frac{5}{\sqrt{2}}\\\\PD=\sqrt{(\frac{23}{2}-5)^2+(\frac{15}{2}-5)^2}\\\\PD=\sqrt{\frac{169}{4}+\frac{25}{4}}\\\\PD=\frac{\sqrt{169+25}}{2}\\\\PD=\frac{\sqrt{194}}{2}\\\\PO=\sqrt{(\frac{23}{2}-18)^2+(\frac{15}{2}-10)^2}\\\\PO=\sqrt{\frac{169}{4}+\frac{25}{4}}\\\\PO=\frac{\sqrt{169+25}}{2}\\\\PO=\frac{\sqrt{194}}{2}$