Christmas Streak 45/88: Weird Quadrilateral, I Admit

Geometry Level 5

$A$ is the midpoint of $\overline{DO}.$ $T$ is the midpoint of $\overline{CD},$ and is the point of contact of $\overline{CD}$ and quadrant circle $O.$

Find the area of $\square ABCD$ to 4 decimal places.

This problem is a part of <Christmas Streak 2017> series .

The answer is 2.92820.

2 solutions

Boi (보이)
Nov 15, 2017

Again, take a look at this beautiful diagram.

Note that $\overline{OT}=2$ and $\overline{OD}=4.$

Therefore $\angle CDH=30^{\circ}.$

Since $\overline{CD}=2\overline{DT}=4\sqrt{3},$ we figure out that $\overline{DH}=6.$

Then $\overline{OH}=2.$

$\square ABCD=\triangle ODC-\triangle ABO-\triangle BCO=4\sqrt{3}-2-2=4(\sqrt{3}-1)\approx\boxed{2.92820}.$

Do we need to find $OH$ ?

Atomsky Jahid - 3 years, 6 months ago

The length of $\overline{\mathrm{OH}}$ is needed to figure out the area of $\triangle \mathrm{BCO}.$

Boi (보이) - 3 years, 6 months ago

Oh! I used the fact $\angle BOC = \frac{\pi}{6}$ and $OC=4$ to get the area of $\triangle BCO$ .

Atomsky Jahid - 3 years, 6 months ago

@Atomsky Jahid – That also works!

Boi (보이) - 3 years, 6 months ago

Niranjan Khanderia
Jan 12, 2018

$Note~that~\Delta ~ODC~altitude~OT-(radius~at~point~of~tangentcy)~is~also~~perpendicular~ ~bisector~~of~Base~DC.\\ \therefore~Delta~ODC~is~isosceles,~~ OC=OD=4.$ .

Christmas Streak 45/88: Weird Quadrilateral, I Admit

The answer is 2.92820.

2 solutions

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