Question

if x,y are lengths of two diagonals of rhombus and side of rhombus is a then x^2+y^2=​

Answer 1

$\textbf{Given:}$

$\textsf{x and y are lengths of the diagonals of  rhombus}$

$\textsf{Length of the side is 'a'}$

$\textbf{To find:}$

$\textsf{The value of}\;\mathsf{x^2+y^2}$

$\textbf{Solution:}$

$\textsf{Concept used:}$

$\textsf{1.\;Diagonals of rhombus intersect at right angles}$

$\textsf{2.\;Diagonals of rhombus bisect each other}$

$\textsf{From the figure, by pythagoras theorem}$

$\mathsf{a^2=\left(\dfrac{x}{2}\right)^2+\left(\dfrac{y}{2}\right)^2}$

$\mathsf{a^2=\dfrac{x^2}{4}+\dfrac{y^2}{4}}$

$\mathsf{a^2=\dfrac{x^2+y^2}{4}}$

$\implies\boxed{\mathsf{x^2+y^2=4\,a^2}}$

$\textbf{Answer:}$

$\mathsf{The\;value\;of\;x^2+y^2\;is\;4a^2}$