[tex](8\sqrt{2})^2 - (8)^2 = (\overline{\rm BD})^2[/tex]The required lengths are [tex]\overline{\rm AB}[/tex] = [tex]8\sqrt{2}[/tex] units and [tex]\overline{\rm BD} = 8[/tex] units, respectively.
What is the Pythagoras theorem?
It is also known as the Pythagorean theorem. It states that the square of the hypotenuse length equals the sum of squares of lengths of the other two sides in a right-angled triangle.
In the given question, ΔABC, ΔABD, and ΔDBC are right-angled triangles. It is also given that [tex]\overline{\rm DC} = \overline{\rm AD} = 8[/tex] units.
⇒ [tex]\overline{\rm AC} = \overline{\rm DC} + \overline{\rm AD} = 8 + 8 =16[/tex] units.
In ΔABC, from Trigonometry,
sin(∠BAC) = [tex]\frac{Opposite}{hypotenuse}[/tex] = [tex]\frac{\overline{\rm AB} }{\overline{\rm AC}}[/tex]
⇒ sin(45°) = [tex]\frac{\overline{\rm AB} }{16}[/tex]
⇒ [tex]\frac{1 }{\sqrt{2}} = \frac{\overline{\rm AB} }{16}[/tex]
⇒ [tex]\overline{\rm AB} = 16/\sqrt{2} = 8\sqrt{2}[/tex] units.
Again in ΔABC, from Trigonometry,
sin(∠BCA) = [tex]\frac{Opposite}{hypotenuse}[/tex] = [tex]\frac{\overline{\rm BC} }{\overline{\rm AC}}[/tex]
⇒ sin(45°) = [tex]\frac{\overline{\rm BC} }{16}[/tex]
⇒ [tex]\frac{1 }{\sqrt{2}} = \frac{\overline{\rm BC} }{16}[/tex]
⇒ [tex]\overline{\rm BC} = 16/\sqrt{2} = 8\sqrt{2}[/tex] units.
In ΔBCD, from Pythagoras theorem,
[tex](\overline{\rm BC})^2 = (\overline{\rm CD})^2 + (\overline{\rm BD})^2[/tex]
⇒ [tex](8\sqrt{2})^2 - (8)^2 = (\overline{\rm BD})^2[/tex]
⇒ [tex](\overline{\rm BD})^2 = (64\times2) - 64 = 128 - 64 =64[/tex]
⇒[tex](\overline{\rm BD}) = \sqrt{64} = 8 units.[/tex]
∴ The required line segment lengths are [tex]\overline{\rm AB}[/tex] = [tex]8\sqrt{2}[/tex] units and [tex]\overline{\rm BD} = 8[/tex] units, respectively.
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