$PR=\sqrt{18}$ and $QP=QR=\sqrt{34}$

$\angle QPR=\angle QRP\approx 69°$ and $\angle PQR\approx 43°$.

First draw rectangle $DBFH$ with $DB=HF=\sqrt{\left(72\right)}\approx \mathrm{8,5}$ and $DH=BF=6$ cm.
Next draw point $Q$ in such a way that $BQ=1$ and point $S$ in such a way that $FS=\frac{1}{4}\cdot HF$.
Now you can draw the pentagon $DBQSH$.
This pentagon has two angles of $90°$, one angle of $157°$ and an angle of $117°$.

$TS=\sqrt{18}$

$BC$ and $PQ$ are parallel and $BP=CQ$.
$BP=CQ=\sqrt{27}$, $BC=6$ and $PQ=3$ cm.

To be able to draw this figure it is smart to first calculate the height of the trapezium. That height is $\sqrt{18-{\mathrm{1,5}}^2}=\sqrt{\mathrm{15,75}}\approx \mathrm{4,0}$ cm.
The trapezium has two angles of approximately $69°$ and two angles of approximately$111°$.

$AE=DE=BF=CF=\sqrt{50}$

$\angle ABF\approx 65°$ and $\angle BCF\approx 68°$.

The width of the loft floor is $\frac{2}{5}\cdot 8=\mathrm{3,2}$ m.
The length of the loft floor is $6+2\cdot \frac{2}{5}\cdot 3=\mathrm{8,4}$ m.
Therefore the area of the floor is $\mathrm{3,2}\cdot \mathrm{8,4}=\mathrm{26,88}$ m².

Besides you see the (right) trapezium.
The edge denoted by $\mathrm{83,2}$ m has an exact length of $\sqrt{{45}^2+{70}^2}=\sqrt{6925}$.
The longest edge of the trapezium is $\sqrt{6925+{\mathrm{6,5}}^2}=\sqrt{\mathrm{6967,25}}\approx \mathrm{83,5}$ m.

Besides two right angles there is an angle of approximately $86°$ and an angle of approximately $94°$.

When $h$ is the height of the tree, then $\frac{6}{1000}\cdot h=2$ cm.
So $h=\frac{2000}{6}\approx \mathrm{333,3}$ cm. Therefore it is only a small tree of approximately $\mathrm{3,33}$ m.

When $a$ is the distance to the Statue of Liberty: $\frac{6}{a}\cdot 9300=2$. This means $a=27900$ cm, that is $279$ m.