Question

give answer fast plz​

Answer 1

perimeter will be

$8 \sqrt{41}$

Answer 2

The figure made is as above.

So, we can conclude that if the diagonal's bisect each other at 90° then the figure should be a Rhombus.

Now, to find the Perimeter -

We know that the diagonal's are bisected. So, it divides it into 2 equal parts.

So, Diagonal1 and Diagonal2 are divided into 2 equal parts.

Which means

$AO = \frac{10}{2} \\ AO = 5cm$
And
$CO = 5cm$
Similarly,
$BO = \frac{24}{2} \\ BO = 12cm$
And
$DO = 12cm$
Now, we can apply Pythagoras Theorum and find the lenght of its Side as it is 90°.
${H}^{2} = {P}^{2} + {B}^{2} \\ {(AB)}^{2} = {(AO)}^{2} + {(BO)}^{2} \\ {(AB)}^{2} = {5}^{2} + {12}^{2} \\ AB = \sqrt{25 + 144} \\ AB = \sqrt{169} \\ AB = 13cm$
Now we got the Side as 13cm

So, Perimeter
$= 4 \times Side \\ = 4 \times AB \\ = 4 \times 13 \\ \boxed{= 52cm}$