Question

If the sum of the lengths of the diagonals of a rhombus is 10 m and if its area is 9 m2. then, what is the sum of squares of the diagonals?

Answer 1

Let ABCD is the rhombus where AC and BD are the diagonals.
Area of rhombus=1/2×AC×BD
9=1/2×AC×BD
18=AC×BD
AC+BD=10
(AC+BD)=10
(AC+BD)^2=10^2
AC^2+BD^2+AC.BD=100
AC^2+BD^2+2×18=100
AC^2+BD^2+36=100
AC^2+BD^2=64

Answer 2

Answer:

Step-by-step explanation:

Let ABCD be a rhombus whose diagonals are AC and BD, we are given that  the sum of the lengths of the diagonals of a rhombus is 10 m and its area is 9 m2, therefore

$9=\frac{1}{2}AC{\times}BD$

⇒$AC{\times}BD=18$

Also, $AC+BD=10$

Now, squaring both sides of the above equation, we get

$(AC+BD)^2=(10)^2$

$(AC)^2+(BD)^2+2(AC)(BD)=100$

$(AC)^2+(BD)^2+2(18)=100$

$(AC)^2+(BD)^2=64$

Therefore, the sum of the squares of the diagonals is 64.