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Albert1
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A trapezoid $ABCD$ , $BC//AD ,\,\, \angle A=90^o$ , $AC\perp BD$
given :$\dfrac {BC}{AD}=k$
find :$ \dfrac {AC}{BD}$
given :$\dfrac {BC}{AD}=k$
find :$ \dfrac {AC}{BD}$
Albert said:A trapezoid $ABCD$ , $BC//AD ,\,\, \angle A=90^o$ , $AC\perp BD$
given :$\dfrac {BC}{AD}=k$
find :$ \dfrac {AC}{BD}$
very good :)mente oscura said:Hello.
[tex]If \ \angle{A}=90º \ and \ \overline{BC} // \overline{AD} \rightarrow{}\angle{B}=90º[/tex]
[tex]If \ \overline{BC} // \overline{AD} \rightarrow{}\angle{ADB}=\angle{DBC}= \alpha[/tex]
[tex]\sin{\alpha}=\dfrac{\overline{AB}}{\overline{BD}}=\dfrac{\overline{BC}}{\overline{AC}}[/tex]
[tex]\cos{\alpha}=\dfrac{\overline{AB}}{\overline{AC}}=\dfrac{\overline{AD}}{\overline{BD}}[/tex]
Therefore:
[tex]\overline{AB}=\dfrac{\overline{BC} \ \overline{BD}}{\overline{AC}}[/tex]
[tex]\overline{AB}=\dfrac{\overline{AD} \ \overline{AC}}{\overline{BD}}[/tex]
[tex]\dfrac{\overline{BC} \ \overline{BD}}{\overline{AC}}=\dfrac{\overline{AD} \ \overline{AC}}{\overline{BD}}[/tex][tex]\dfrac{\overline{BC}}{\overline{AD}}= \dfrac{(\overline{AC})^2}{(\overline{BD})^2}=k[/tex]
Therefore:
[tex]\dfrac{\overline{AC}}{\overline{BD}}=\sqrt{k}[/tex]
Regards.
The value of AC/BD in a trapezoid can be found by dividing the length of one of the diagonals (AC) by the length of the other diagonal (BD).
To calculate AC/BD in a trapezoid, you will need to know the lengths of both diagonals and use the formula AC/BD = AC/BD.
Yes, it is possible for AC/BD to be greater than 1 in a trapezoid. This means that the length of the diagonal AC is longer than the length of the diagonal BD.
No, AC/BD cannot be negative in a trapezoid. This ratio is always positive, as diagonals of a trapezoid cannot be negative.
The value of AC/BD in a trapezoid can give insight into the shape of the trapezoid. If the value is close to 1, it means that the trapezoid is close to being a square. If the value is greater than 1, the trapezoid is more elongated, and if the value is less than 1, the trapezoid is wider.