Question

One diagonal of a rhombus is 6 cm. If the area of the rhombus is 24 sq.cm then what is the length of it's side?​

Answer 1

Answer:-

Length of its side = 5 cm

$\rule{200}{1}$

✦ DIAGRAM:-

$\setlength{\unitlength}{1 cm}}\begin{picture}(12,4)\thicklines\put(5,7){ . }\put(6,6){\line(1,0){3}}\put(7,9){\line(1,0){3}}\put(6,6){\line(1,3){1}}\put(9,6){\line(1,3){1}}\put(6,6){\line(4,3){4}}\put(9,6){\line(-2,3){2}}\put(5.6,5.9){ B }\put(9.1,5.9){ C }\put(6.6,8.9){ A }\put(10.1,8.9){ D }\put(6.8,7.2){ 6\:cm }\end{picture}$

Here,given:-

• Length of one diagonal = 6 cm
• Area of rhombus = 24 cm²

To find:-

• Side of rhombus = ?

Solution:-

As we know, each side of rhombus is equal.

And one diagonal of rhombus divides it into two isosceles triangles.

We also know,

☛ Area of isosceles∆ = (b/4) √(4a² - b²)

Here, a = equal sides. b = inequal side.

Now according to question:-

$\longrightarrow\sf 2 \bigg[ \big(\dfrac{b}{4} \big ) \sqrt{4a^2 - b^2} \bigg] = 24 \\\\\longrightarrow\sf (\dfrac{6}{4})\sqrt{4a^2 - 6^2} = 24/2 \\\\\longrightarrow\sf (\dfrac{6}{4})\sqrt{4a^2 - 36} = 12 \\\\\qquad\scriptsize\sf{\big [\because Squaring \: on \: both \: sides \big ]} \\\\\longrightarrow\sf { \bigg [ (\dfrac{6}{4})\sqrt{4a^2 - 36} \bigg] }^2 = (12)^2 \\\\\longrightarrow\sf (\dfrac{36}{16}) (4a^2 - 36) = 144 \\\\\longrightarrow\sf (\dfrac{9}{4})(4a^2 - 36) = 144 \\\\\longrightarrow\sf 9(4a^2 - 36) = 144 \times 4 \\\\ \longrightarrow\sf 36a^2 - 324 = 576 \\\\ \longrightarrow\sf 36a^2 = 576 + 324 \\\\ \longrightarrow\sf 36a^2 = 900 \\\\ \longrightarrow\sf a^2 = 900/36 \\\\\longrightarrow\sf a^2 = 25$

$\longrightarrow\sf a = \sqrt{25}$

$\longrightarrow\large\boxed{\boxed{\sf\blue{a = 5 \: cm }}}$

Therefore,

Side of rhombus = 5 cm.

Answer 2

Solution :

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

One diagonal of a rhombus is 6 cm. If the area of the rhombus is 24 cm².

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

The length of it's side.

$\bf{\red{\underline{\underline{\bf{Explanation\::}}}}}$

We know that formula of the area of rhombus :

$\bf{\boxed{\bf{Area\:of\:rhombus=\frac{1}{2} \times d_{1}\times d_{2}}}}}$

So;

$\mapsto\sf{24cm^{2} =\dfrac{1}{\cancel{2}} \times \cancel{6}cm\times d_{2}}\\\\\mapsto\sf{24cm^{2} =3cm\times d_{2}}\\\\\mapsto\sf{d_{2}=\cancel{\dfrac{24cm^{2} }{3cm}}} \\\\\mapsto\sf{\pink{d_{2}=8\:cm}}$

∴ The other diagonal of rhombus is 8 cm.

Since, We know that diagonals of rhombus bisect each other at 90°.

• AO = 4 cm
• DO = 3 cm

$\bf\large{\underline{\underline{\bf{Using\:Pythagoras\:theorem\::}}}}}$

$\sf{(Hypotenuse)^{2} =(base)^{2} +(Perpendicular)^{2} }$

In Δ AOD :

$\mapsto\sf{AD^{2} =AO^{2} +DO^{2} }\\\\\mapsto\sf{AD^{2} =(4cm)^{2} +(3cm)^{2} }\\\\\mapsto\sf{AD^{2} =16cm^{2} +9cm^{2} }\\\\\mapsto\sf{AD^{2} =25cm^{2} }\\\\\mapsto\sf{AD=\sqrt{25cm^{2} } }\\\\\mapsto\sf{\pink{AD=5\:cm}}$

Thus;

The length of the side of rhombus is 5 cm .