Question

The area of a rhombus is 25cm² & one of its diagonals is 4cm.  

         Find its perimeter?​

Answer 1

QUESTION:-

The area of a rhombus is 25 cm² and one of the diagonal is 4 cm . find its perimeter?

SOLUTION:-

• (Refer to attachment)

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Answer 2

Answer:

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎The perimeter of the rhombus is 26.24 cm.

Explanation:

${\underline{\underline{\bf{Given:}}}}$

• Area of a rhombus = 25 cm²
• Length of one of its diagonal = 4 cm

${\underline{\underline{\bf{To\:find:}}}}$

• The perimeter of the rhombus

${\underline{\underline{\bf{Formulae\:to\:be\:used:}}}}$

$\underline{\boxed{\sf{{Area}_{[Rhombus]}=\dfrac{1}{2}d_1d_2\:sq.units}}}$

$\:\:\:\:\:\:\:\:\:\:\sf{\star\:d=diagonal}$

$\underline{\boxed{\sf{Pythagorean\: Theorem:a^2+b^2=c^2}}}$

$\:\:\:\:\:\:\:\:\:\:\sf{\star\:a=base}$

$\:\:\:\:\:\:\:\:\:\:\sf{\star\:b=perpendicular}$

$\:\:\:\:\:\:\:\:\:\:\sf{\star\:c=hypotenuse}$

$\underline{\boxed{\sf{{Perimeter}_{[Rhombus]}=4s\:units}}}$

$\:\:\:\:\:\:\:\:\:\:\sf{\star\:s=side}$

${\underline{\underline{\bf{Solution:}}}}$

To find the perimeter, we must know the measure of its side. Look at the attachment! The two diagonals intersects at point O and forms four right-angled triangles.

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎$\sf{Hypotenuse=Side\:of\:rhombus}$

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎$\sf{Base=\dfrac{1}{2}d_2}$

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎$\sf{Perpendicular=\dfrac{1}{2}d_1}$

So, we can find the side using Pythagorean Theorem.

We know the length of first diagonal. But the length of second diagonal is unknown. So, first let's find the length of second diagonal, by applying the formula:

$\leadsto{\sf{Area=\dfrac{1}{2}d_1d_2\:sq.units}}$

$\:\:\:\:\:\:\:\:\implies{\sf{25=\dfrac{1}{\cancel{2}}\times \cancel{4}\times d_2}}$

$\:\:\:\:\:\:\:\:\implies{\sf{25=2\times d_2}}$

$\:\:\:\:\:\:\:\:\implies{\sf{\dfrac{\cancel{25}}{\cancel{2}}=d_2}}$

$\:\:\:\:\:\:\:\:\implies\underline{\bf{12.5\:cm=d_2}}$

Now, by inserting the measures in Pythagorean Theorem:

$\leadsto{\sf{a^2+b^2=c^2}}$

$\implies{\sf{{\bigg(\dfrac{1}{\cancel{2}}\times \cancel{4}\bigg)}^{2}+{\bigg(\dfrac{1}{\cancel{2}}\times \cancel{12.5}\bigg)}^{2}=c^2}}$

$\implies{\sf{2^2+6.25^2=c^2}}$

$\implies{\sf{4+39.0625=c^2}}$

$\implies{\sf{43.0625=c^2}}$

$\implies{\sf{\sqrt{43.0625}=c}}$

$\implies\underline{\bf{6.56\:cm=c}}$

As the length of hypotenuse is 6.56 cm, the measure of side of the rhombus equals the same.

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{\boxed{\sf{Side=6.56\:cm}}}$

Perimeter of the rhombus:

By applying the formula,

$\leadsto{\sf{4s\:units}}$

$\:\:\:\:\:\:\:\implies{\sf{4\times 6.56}}$

$\:\:\:\:\:\:\:\implies{\underline{\underline{\frak{\red{26.24\:cm}}}}}$

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