Question

The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.

Answer 1

Let ABCD be the given rhombus.
We know that the Diagonals of a rhombus are perpendicular bisector of each other. Let the point of intersection of diagonals AC and BD be M.
Given AC = 30 cm and BD = 16 cm
Now, $AM=\frac{AC}{2}=\frac{30}{2}=15cm$
and $DM=\frac{BD}{2}=\frac{16}{2}=8cm$
Now, in right triangle AMD, by Pythagoras theorem,
$A{D}^2=A{M}^2+M{D}^2$
$\Rightarrow A{D}^2={15}^2+8^2$
$\Rightarrow A{D}^2=225+64=289$
$\Rightarrow AD=\sqrt{289}=17cm$
Again, all the sides of a rhombus are equal.
Therefore, AB = BC = CD = AD = 17 cm
Now the perimeter of a rhombus = sum of all sides = AB+BC+CD+AD = 17+17+17+17=68 cm.