Question

The lengths of the diagonals of a rhombus are 18 cm and 24 cm. Then find t
length of the side of the rhombus.​

Answer 1

$\huge \boxed{ \underline{ \underline{ \bf{Answer}}}}$

GIVEN :

Diagonals of the rhombus : 18 cm and 24 cm.

TO FIND :

Sides of the rhombus.

SOLUTION :

Let ABCD be the rhombus and O be the point of intersection of the diagonals.

We know a rhombus has equal sides but unequal diagonals which bisect each other perpendicularly.

So, as the diagonals bisect each other perpendicularly

Therefore,

AO = CO = 12 cm and BO = DO = 9 cm.

Now,

In rt. triangle AOB, by Pythagoras theorem,

$\sf{ {AB}^{2} = {AO}^{2} + {BO}^{2}}$

$\sf{ {AB}^{2} = {12}^{2} + {9}^{2}}$

$\sf{ {AB}^{2} = 144 + 81}$

$\sf{ {AB}^{2} =225}$

$\sf{ {AB} = \sqrt{ 225} }$

$\sf{ {AB} =15}$

Hence, Each side of the rhombus is 15 cm respectively.

Answer 2

$\sf{\huge{\mathfrak{\pink{\underline{Question}}}}}$

• The lengths of the diagonals of a rhombus are 18 cm and 24 cm. Then find length of the side of the rhombus.

$\setlength{\unitlength}{1.3cm}\begin{picture}(8,2)\thicklines\put(8.6,3){\large{A}}\put(9,1.3){\sf{12 cm}}\put(9.9,1.3){\sf{9 cm}}\put(7.7,0.9){\large{B}}\put(9.2,0.7){\sf{\large{? cm}}}\put(11.1,0.9){\large{C}}\put(9.9,2.1){\large{O}}\put(8,1){\line(1,0){3}}\put(11,1){\line(1,2){1}}\put(9,3){\line(3,0){3}}\put(11,1){\line(-1,1){2}}\put(8,1){\line(2,1){4}}\put(8,1){\line(1,2){1}}\put(12.1,3){\large{D}}\end{picture}$

$\sf{\huge{\mathfrak{\pink{\underline{Solution}}}}}$

$\sf{\bold{\green{\underline{\underline{Given}}}}}$

• Given figure is a rhombus
• Diagonal 1 = 18cm
• Diagonal 2 = 24cm

$\sf{\bold{\blue{\underline{\underline{To\:find}}}}}$

• length of each side = ??

$\sf{\bold{\purple{\underline{\underline{explanation}}}}}$

• diagonals of the rhombus bisect each other .

therefore :-

OC = 9cm

OB = 12cm

• diagonals of the rhombus bisect at 90' therefore forms right angle triangle.
• Therefore BOC is a right angle triangle

Using Pythagoras theorem :-

$\sf BC^2 = BO^2 + CO^2$

$\sf BC^2 = 9^2 + 12^2$

$\sf BC^2 = 81 + 144$

$\sf BC^2 = 225$

$\sf BC = \sqrt{225}$

$\sf BC = 15cm$

• BC = side of rhombus

$\sf{\bold{\red{\boxed{\dag side\: of \: rhombus = 15cm}}}}$

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