Question

If the diagonals of a rhombus are 4.8cm and 1.4cm, then find the perimeter of the
rhombus.​

Answer 1

let abcd be rhombus and ac and and bd be diagonals.

given,ac=4.8cm and bd=1.4cm.

let the point where diagonals bisect be O.

we know that its diagonals bisect at 90 degree.

so ao=oc=4.8/2=2.4cm.

and bo=od=1.4/2=0.7cm.

ab^2=ao^2+ob^2(pythagoras theorom).

=(2.4)^2+(0.7)^2=5.76+0.49=6.25.

Therefore,ab= root 6.25m= 2.5m.

ab=bc=cd=ad=2.5m(all sides of a rhombus are equal).

perimeter of rhombus=4*side of rhombus=4*2.5=10cm.

therefore perimeter of rhombus is 10cm.

Answer 2

Question:

If the diagonals of a rhombus are 4.8cm and 1.4cm, then find the perimeter of the  rhombus.​

Answer:

The perimeter of the rhombus is 10 cm.

Given:

The diagonals of a rhombous are 4.8 cm and 1.4 cm.

To find:

The perimeter of the rhombus.

Explanation:

Let, ABCD is a rhombus.

AC and BD are the diagonals and they intersects each other at the point O.

AC = 1.4 cm and

BD = 4.8 cm.

We know, that the length of the sides of a rhombos are equal.

Hence, we get,

AB=BC=CD=AD.

The diagonals bisects each other.

Hence,

AO=OC.

BO=OD

AO = OC

∴ AO = (1.4÷2) cm

        = 0.7 cm

∴ OC = 0.7 cm [∵ AO=OC]

BO=(4.8÷2) cm

     = 2.4  cm

∴ OD = 2.4 cm [∵ OD =BO]

Now for Δ AOD_

AO= 0.7 cm

OD = 2.4 cm

∵ We know that, the angles formed by the intersection of the diagonals is 90°.

Hence,

AD = √AO²+OC² [∵ Pythagoras theorem.]

     = √(0.7)²+(2.4)²

     = $\sqrt{(\frac{7}{10})^{2}+(\frac{24}{10} )^{2} }$

    = $\sqrt{\frac{49}{100}+\frac{576}{100} }$

   = $\sqrt{\frac{49+576}{100} }$ [∵ The L.C.M. of the denominators is 100.]

   = $\sqrt{\frac{625}{100} }$

   = $\frac{25}{10}$

   = 2.5 cm.

∴ AD= 2.5 cm.

∴ The length of the each side of the rhombus is 2.5 cm.

∴ Perimeter = (sum of the sides)

                   =(side×4)

                  =(2.5×4) cm

                  = 10 cm

∴ The perimeter is 10 cm.

Refer to the attachment for the figure and explanation.