Question
$ABCD$ is a rhombus whose diagonals intersect at $O$. If $AB = 10\ cm$, diagonal $BD = 16\ cm$, find the length of diagonal $AC$.
✓
Answer
We know that the diagonals of a rhombus bisect each other at right angles.
$\therefore\text{BO}=\frac{1}{2}\text{BD}=\Big(\frac{1}{2}\times16\Big)\text{cm}$
$=8\text{cm}$
$\text{AB}=10\text{cm}$ and $\angle\text{AOB}=90^\circ$
From right $\triangle\text{AOB}$:
$\text{AB}^2=\text{AO}^2+\text{BO}^2$
$\Rightarrow\text{AO}^2=\big(\text{AB}^2-\text{BO}^2\big)$
$\Rightarrow\text{AO}^2=\big(10\big)^2-\big(8\big)^2\text{cm}^2$
$\Rightarrow\text{AO}^2=\big(100-64\big)\text{cm}^2$
$\Rightarrow\text{AO}^2=36\text{cm}^2$
$\Rightarrow\text{AO}=\sqrt{36}\text{cm}=6\text{cm}$
$\therefore\text{AC}=2\times\text{AO}$
$= \big(2\times6\big)\text{cm}$
$=12\text{cm}$
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