# Triangle in a square

In a square ABCD with side a = 6 cm, point E is the center of side AB and point F is the center of side BC. Calculate the size of all angles of the triangle DEF and the lengths of its sides.

Correct result:

d =  2.121 cm
e =  6.708 cm
f =  6.708 cm
E =  71.562 °
F =  71.562 °
D =  36.876 °

#### Solution:

$a=6 \ \text{cm} \ \\ d^2=(a/2)^2 + (a/2)^2 \ \\ \ \\ d=a/\sqrt{ 8 }=6/\sqrt{ 8 }=2.121 \ \text{cm}$
$e^2=a^2 + (a/2)^2=6^2 + (6/2)^2=45 \ \\ \ \\ e=\sqrt{ 1+1/4 } \cdot \ a=\sqrt{ 1+1/4 } \cdot \ 6=3 \ \sqrt{ 5 }=6.708 \ \text{cm}$
$f=e=6.7082=\dfrac{ 1677 }{ 250 }=6.708 \ \text{cm}$
$\angle FEB=45 \ ^\circ \ \\ \ \\ \sin AED=a/e \ \\ \ \\ AED=\dfrac{ 180^\circ }{ \pi } \cdot \arcsin(a/e)=\dfrac{ 180^\circ }{ \pi } \cdot \arcsin(6/6.7082) \doteq 63.4384 \ \\ \ \\ E=180 - AED - \angle FEB=180 - 63.4384 - 45=71.562 ^\circ =71^\circ 33'42"$
$F=E=71.5616=71.562 ^\circ =71^\circ 33'43"$
$D=180-F-E=180-71.562-71.5616=\dfrac{ 9219 }{ 250 }=36.876 ^\circ =36^\circ 52'34"$

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