In a rhombus of side 52 cm, one of the diagonals is 40 cm long. Find the length of the second diagonal.
Answer:
96 cm
- Let ABCD be the given rhombus whose diagonals intersect at O.
Then, AB=52 cm.
Let AC=40 cm and BD=2x cm.
- We know that the diagonals of a rhombus bisect each other at right angles. ∴
- 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.