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 cm d2=(a/2)2+(a/2)2  d=a/8=6/8=2.121 cma=6 \ \text{cm} \ \\ d^2=(a/2)^2 + (a/2)^2 \ \\ \ \\ d=a/\sqrt{ 8 }=6/\sqrt{ 8 }=2.121 \ \text{cm}
e2=a2+(a/2)2=62+(6/2)2=45  e=1+1/4 a=1+1/4 6=3 5=6.708 cme^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=1677250=6.708 cmf=e=6.7082=\dfrac{ 1677 }{ 250 }=6.708 \ \text{cm}
FEB=45   sinAED=a/e  AED=180πarcsin(a/e)=180πarcsin(6/6.7082)63.4384  E=180AEDFEB=18063.438445=71.562=713342"\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=713343"F=E=71.5616=71.562 ^\circ =71^\circ 33'43"
D=180FE=18071.56271.5616=9219250=36.876=365234"D=180-F-E=180-71.562-71.5616=\dfrac{ 9219 }{ 250 }=36.876 ^\circ =36^\circ 52'34"



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