Question

find the perimeter of the rhombus if the lengths of the diagonals 12cm,16cm respectively​

Answer 1

Given :

➡ Length of the diagonal =12cm and 16cm

To find :

➡ The perimeter of the rhombus =?

Solution :

$\pink \bigstar\large \red{ \bold{ \underline{ \underline{step \: by \: step \: explaination : }}}}$

➡ Rhombus is a two dimensional shape with four equal sides and four angles which can or cannot

be 90 degree , but opposite Angle are always same and both diagonal of a rhombus are perpendicular and bisect each other.

➡ Consider Shape ABCD and 0 is the intersecting point of both diagonals which is bisecting both diagonals ,and both diagonal are perpendicular to each other .

According to definition :

$\\ \implies\large\sf{ AO=\dfrac{16}{2}=8cm}$

$\\ \implies\large\sf{BO=\dfrac{12}{2}=6cm}$

So AOB is the right angled triangle.

By using Pythagoras theorem :

➡ We can find the length of AB which is a side of rhombus .

$\\ \pink \bigstar\large \red{ \bold{ \underline{ \underline{according \: to \: pythagoras \: theorem \: : }}}}$

$\\ \implies\large\sf{(AB)^2=(BO)^2+(AO)^2} \\ \\ \implies\large\sf{AB=\sqrt{(6)^2+(8)^2 cm}} \\ \\ \implies\large\sf{AB=\sqrt{36+64cm}} \\ \\ \implies\large\sf{ AB=\sqrt{100cm} =10cm}$

As we know that ,

➡ The perimeter of rhombus is equal to 4a where a is the length of rhombus .

So, perimeter of rhombus =4a

$\\ \implies \large \sf \: perimeter \: of \: rhombus \: = 4a$

$\\ \implies \large \sf \: perimeter \: of \: rhombus \: = 4 \times 10cm$

$\\ \implies \large \sf \: \: perimeter \: of \: rhombus \: = 40cm$

Hence the side is 10cm and the perimeter of rhombus is 40cm.