Question

The diagonals of a rhombus are in the ratio 3:4. If its perimeter is 40 cm, find the lengths of the sides and diagonals of the rhombus

Answer 1

side = 10
digonals are 12 and 16
ihope u understand
10^2=(3x/2)^2+(4x/2)^2
100=9x^2/4 +16x^2/4
100=(25x^2)/4
x^2 =100×4/25
x=4
then d1=12
and d2=16

Answer 2

$\sf\blue{Given\:that:-}$

• Diagonals of rhombus=$\frak{3:4}$(In ratio)
• Perimeter of rhombus=$\frak{40cm}$

$\sf\red{To\:find:-}$

• The lenght of the sides.
• Diagonal of the rhombus.

$\sf\pink{Solution:-}$

Here ABCD is a rhombus.

AB=BC=CD=AD

Hence,

The Side of rhombus=$\dfrac{1}{4}×40$

$\rm\blue{(40\:is\:the\: perimeter)}$

$\frak{=10cm}$

Let BD =$\rm{3x}$ and AC =$\rm{4x}$

Therefore,OD=$\dfrac{3}{2}x$

And OC =$\dfrac{4}{2}x$

Now its clear that $\rm{∆DOC}$ is a right-angled triangle

(OD)²+(OC)²=(CD)²

$\mapsto(\dfrac{3}{2}x)²$$+(\dfrac{4}{2}x)²$$\frak{=10}^{2}$

$\mapsto\dfrac{9x²}{4}$$+\dfrac{16x²}{4}$$\frak{=100}$

$\mapsto\rm{25x}^{2}={100×4}$

$\mapsto\rm{x²=}\dfrac{400}{25}=16$

$\mapsto\rm{x=}\sqrt{16}$

$\mapsto\rm{x=}\sqrt{4×4}$

$\mapsto\rm{x=4}$

Therefore,the diagonals of BD and AC of the rhombus are 12cm and 16cm respectively.

And each of the rhombus is 10cm.

Hope it's helpful