Question

find the area of a rhombus whose perimeter is 80 m and one of whose diagonal is 24 m​

Answer 1

Answer:

Refer to the attachment.........

Answer 2

GIVEN:-

• $\rm{Area\:of\:Rhombus = 80m}$

• $\rm{Diagonal\:of\:Rhombus = 24m}$.

TO FIND :-

• The area of Rhombus.

CONSTRUCTION:-

• Draw another Rhombus from B to D.

CONCEPT USED

• All Sides of Rhombus are equal.

• Diagonal Bisect each other at right angled.

FORMULAE USED:-

• ${\boxed{\rm{\blue{Area\:of\:Rhombus = \dfrac{Product\:of\:Diagonal}{2}}}}}$

• ${\boxed{\rm{\red{Pythogoras\:Theorem}}}}$

Now,

$\implies\rm{Perimeter\:of\:Rhombus = 4\times{sides}}$

$\implies\rm{80 = 4\times{sides}}$

$\implies\rm{Side = \dfrac{80}{4}}$

$\implies\rm{Side = 20m}$.

Now, In right angled triangle ∆OBC,

$\implies\rm{OC = 12m}$

$\implies\rm{BC = 20m}$

$\implies\rm{OB = ?}$.

Using Pythogoras Theorem.

$\implies\rm{(OB)^2+(OC)^2 = (AB)^2}$

$\implies\rm{(OB)^2 = (20)^2-(12)^2}$

$\implies\rm{(OB)^2 = \sqrt{256}}$

$\implies\rm{OB = 16m}$

Therefore,

$\implies\rm{2OB = BD}$

$\implies\rm{2\times{16} = BD}$

$\implies\rm{32 = BD}$.

Now,

$\rm{Area\:of\:Rhombus = \dfrac{Product\:of\:Diagonal}{2}}$

$\rm{Area\:of\:Rhombus = \dfrac{32\times{24}}{2}}$

$\rm{Area\:of\:Rhombus = \dfrac{768}{2}}$

$\rm{Area\:of\:Rhombus = 384m^2}$.

Hence, The Area of Rhombus is 384m².