In rhombus ABCD,AC BD and overline{AC} cap | vedclass

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(D) In a rhombus,the diagonals bisect each other at right angles. Let $AM = x$ and $DM = y$. Since the diagonals intersect at $M$ at a $90^{\circ}$ an

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$10$

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(D) In a rhombus,the diagonals bisect each other at right angles. Let $AM = x$ and $DM = y$. Since the diagonals intersect at $M$ at a $90^{\circ}$ angle,$	riangle AMB$ is a right-angled triangle. By the Pythagorean theorem,$AM^2 + BM^2 = AB^2$. Given $AM + DM = 17$,we have $x + y = 17$,so $y = 17 - x$. In $	riangle AMB$,$x^2 + BM^2 = 13^2 = 169$. In $	riangle AMD$,$x^2 + y^2 = AD^2$. Since all sides of a rhombus are equal,$AD = AB = 13$. Thus,$x^2 + y^2 = 169$. Substituting $y = 17 - x$: $x^2 + (17 - x)^2 = 169$. $x^2 + 289 - 34x + x^2 = 169$. $2x^2 - 34x + 120 = 0$. $x^2 - 17x + 60 = 0$. $(x - 12)(x - 5) = 0$. So,$x = 12$ or $x = 5$. If $x = 12$,then $y = 17 - 12 = 5$. If $x = 5$,then $y = 17 - 5 = 12$. Since $AC > BD$,$AM > DM$,so $x > y$. Thus,$x = 12$ and $y = 5$. $DM = y = 5$. Since the diagonals bisect each other,$BD = 2 	imes DM = 2 	imes 5 = 10$.

$1$. $\square ABCD$ is a rhombus.	$a$. $\overline{AC}$ and $\overline{BD}$ bisect each other.
$2$. $\square ABCD$ is a parallelogram.	$b$. $\overline{AC}$ and $\overline{BD}$ bisect each other at right angles.
$3$. $\square ABCD$ is a rectangle.	$c$. $\overline{AC}$ and $\overline{BD}$ are congruent and bisect each other at right angles.
$4$. $\square ABCD$ is a square.	$d$. $\overline{AC}$ and $\overline{BD}$ are congruent and bisect each other.

In rhombus $ABCD$,$AC > BD$ and $\overline{AC} \cap \overline{BD} = \{M\}$. If $AM + DM = 17$ and $AB = 13$,find $BD$.

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