after the calculation performed and analysis we conclude that: [tex]\large \displaystyle \text { $ \mathsf{ DC = 8 } $ }[/tex].
Pythagorean theorem:
“In any right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs.”
- [tex]\large \boxed{ \displaystyle \text { $ \mathsf{ a^2 = b^2 + c^2 } $ } }[/tex]
right triangle an angle of [tex]\textstyle \sf \text {$ \sf 90^\circ \to reto $ }[/tex];
Two equal sides, i.e. congruent,
Two two angles of [tex]\textstyle \sf \text {$ \sf 45^\circ $ }[/tex].
Data provided by the statement:
- [tex]\large \displaystyle \text { $ \mathsf{ AB \cong AB ~ e ~ BC \cong BD } $ }[/tex]
- [tex]\large \displaystyle \text { $ \mathsf{ \angle \: B = D = 45^\circ } $ }[/tex]
- [tex]\large \displaystyle \text { $ \mathsf{ \angle \: C = D = 45^\circ } $ }[/tex]
First we must determine the value BC:
Applying the Pythagorean theorem, we have:
- [tex]\large \displaystyle \text { $ \mathsf{ (BC)^2 = (AB)^2 + (AD)^2 } $ }[/tex]
- [tex]\large \displaystyle \text { $ \mathsf{ (BC)^2 = 4^2 + 4^2 } $ }[/tex]
- [tex]\large \displaystyle \text { $ \mathsf{ (BC)^2 =16+16 } $ }[/tex]
- [tex]\large \displaystyle \text { $ \mathsf{ (BC)^2 = 16 \cdot 2 } $ }[/tex]
- [tex]\large \displaystyle \text { $ \mathsf{ BC = \sqrt{16 \cdot 2} } $ }[/tex]
- [tex]\large \displaystyle \text { $ \mathsf{BC =\sqrt{16} \; \cdot \sqrt{2} } $ }[/tex]
- [tex]\large \boldsymbol{ \displaystyle \sf BC = 4\sqrt{2} }[/tex]
Now we must find BC:
- [tex]\large \displaystyle \text { $ \mathsf{ (DC)^2 = (BC)^2 + (BD)^2 } $ }[/tex]
- [tex]\large \displaystyle \text { $ \mathsf{ (DC)^2 = (4\sqrt{2})^2 + (4\sqrt{2})^2} $ }[/tex]
- [tex]\large \displaystyle \text { $ \mathsf{ (DC)^2 = 16 \cdot 2 + 16 \cdot 2} $ }[/tex]
- [tex]\large \displaystyle \text { $ \mathsf{ (DC)^2 = 32 + 32} $ }[/tex]
- [tex]\large \displaystyle \text { $ \mathsf{ (DC)^2 = 64 } $ }[/tex]
- [tex]\large \displaystyle \text { $ \mathsf{ DC = \sqrt{64} } $ }[/tex]
- [tex]\large \boldsymbol{ \displaystyle \sf DC = 8 }[/tex]
#CarryOnLearning
[tex]\qquad\qquad\qquad\qquad\qquad\qquad\tt{fri \: 03-04-2022} \\ \qquad\qquad\qquad\qquad\qquad\qquad\tt{12:33 \: pm}[/tex]
