Question

if the area of a rhombus is 48 cm and one of its diagonals is 12 cm find the side of the rhombus

Answer 1

Some solved examples : 

1) The side of a rhombus is 18 cm . Find its perimeter.
Solution : 
Perimeter of Rhombus = 4 x side 
⇒ = 4 x 18 
&rArr = 72 cm 
∴ Perimeter of Rhombus = 72 cm.
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2) Find the area of a rhombus having each side equal to 13 cm and one of whose diagonal is 24 cm.
Solution : 
Let ABCD is a rhombus with diagonals AC and BD which intersect each other at O.
AC = 24 ⇒ AO = 12 
Let BO = x and AB = 13 cm (given) 
By Pythagorean theorem 
c 2 = a 2 + b 2 
13 2 = 12 2 + x 2 169 = 144 + x 2 
x 2 = 169 – 144 
x 2 = 25
x = 5 cm
BO = 5 cm
Diagonal BD = 2 x 5 = 10 cm.
Area = ½ x [ product of diagonals]
= ½ x 24 x 10 
Area = 120 sq.cm

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3) If the area of a rhombus is 24 sq.cm and one of its diagonal is 4 cm find the length of the other diagonal.

Solution : 
Area = ½ x d1 x d2 
24 = ½ x 4 x d2 
24 = 2 x d2 
d2 = 24/2 
d2 = 12 cm

Answer 2

$\sf{Side = \frac{1}{2}\sqrt{p^2+4(\frac{A}{p})^2}$

where p is the diagonal, and A is the area of the rhombus.

$\sf{Side = \frac{1}{2}\sqrt{12^2+4(\frac{48}{12})^2}}$

$\sf{Side = \frac{1}{2}\sqrt{144+4(4)^2}}$

$\sf{Side = \frac{1}{2}\sqrt{144+4(16)}}$

$\sf{Side = \frac{1}{2}\sqrt{144+64}}$

$\sf{Side = \frac{1}{2}\sqrt{208}}$

$\sf{Side = \frac{1}{2}\cdot(14.4222051019)$

$\sf{Side \approx 7.2~cm}$