Singapore
In a rhombus of side ^@ 52 \space cm^@, one of the diagonals is ^@40 \space cm^@ long. Find the length of the second diagonal.
Answer:
^@ 96 \space cm^@
- Let ^@ ABCD ^@ be the given rhombus whose diagonals intersect at ^@O^@.
Then, ^@AB = 52\space cm^@.
Let ^@AC = 40\space cm^@ and ^@BD = 2x \space cm^@.
- We know that the diagonals of a rhombus bisect each other at right angles. @^ \begin{aligned} \therefore \space OA = \dfrac { 1 } { 2 } AC = 20 \space cm, OB = \dfrac { 1 } { 2 } BD = x \space cm, and \space \angle AOB = 90^\circ \end{aligned} @^
- From right ^@\Delta AOB ^@, we have @^ \begin{aligned} AB^2 =& OA^2 + OB^2 \\ \implies OB^2 =& AB^2 - OA^2 \\ =& [(52)^2 - (20)^2] \space cm^2 = [ 2704 - 400 ] \space cm^2= 2304 \space cm^2 \\ \implies\space\space \space x^2 =& \space 2304 \implies x = \sqrt{ 2304 } = 48 \space cm. \\ \therefore \space\space \space OB =& \space 48 \space cm. \\ \therefore \space\space \space BD =& 2 \times OB = 2 \times 48 \space cm = 96 \space cm. \end{aligned} @^ Hence, the length of the second diagonal is ^@ 96 \space cm.^@
