Question

cao
8. The diagonals of a rhombus are 48 cm and 20 cm
long. Find the perimeter of the rhombus.
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Answer 1

Refer to the attachment

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Answer 2

$\huge\underline\mathrm{GIVEN:-}$

• Length of one diagonal = 48 cm
• Length of another diagonal = 20 cm

$\huge\underline\mathrm{TO\:FIND:-}$

Find the perimeter of the rhombus = ?

$\huge\underline\mathrm{SOLUTION:-}$

We have given that The diagonals of a rhombus are 48 cm and 20 cm.

Since, the diagonals of rhombus are intersect on each other at 'O'

So,

• DO = OB = 24 cm
• AO = OC = 10 cm

In ∆ AOD, using Pythagoras theorem

$\rm\bold{{Hypotenuse}^{2} = {base}^{2} + {perpendicular} ^{2}}$

$\large\bf{AD}$=$\bf\sqrt{ {(DO)}^{2} + {(AO)}^{2} }$

$\bf{AD}$=$\bf\sqrt{ {(24)}^{2} + {(10)}^{2} }$

$\bf{AD}$=$\bf\sqrt{576 + 100}$

$\bf{AD}$ =$\bf\sqrt{676}$

$\large{\boxed{\bf{AD = 26\:cm}}}$

The sides of rhombus are equal in length

$\rm\bold{Perimeter\: of\: rhombus=<strong> </strong>4\times side}$

Putting the values in the formula

$\bf{<strong> </strong>Perimeter = 4\times26}$

$\large{\boxed{\bf{Perimeter =104\:cm}}}$

Thus, the perimeter of rhombus is 104 cm