Question

★ELLOH




$\large\mathbf{HEY\: QUESTION❤}$




The lengths of the diagonals of a rhombus are 24 cm and 32 cm. Calculate the length of the altitude of the rhombus.

✪ REQUIRED QUALITY ANSWER
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Answer 1

Given:

• The lengths of the diagonals of a rhombus are 24 cm and 32 cm.

To find:

• Length of the altitude of the rhombus.

Solution:

• We know that diagonals of rhombus bisects each other.
• Let O be the intersecting point of both the diagonals.
• Let AC=32 and BD= 24
• ΔDOA Is a right angle triangle .So we can apply Pythagoras therom on this

$\therefore \: (AO) {}^{2} + (OD) {}^{2} = (AD) {}^{2} \\ 16 {}^{2} + 12 {}^{2} = AD {}^{2} \\ 256 + 144 = AD {}^{2} \\ \sqrt {400} = AD \\ Therefore \: Side(AD) = 20cm \\ \tt \: Hence, \: The \: Side \: of \: Rhombus \\ \tt is \: \ \underline \pink{20cm}$

Answer 2

Given : Lengths of diagonals of a rhombus are 24cm and 32cm.

To Find : Perimeter of Rhombus ?

_________________________

Solution : Since diagonals of rhombus bisect each other at 90°, so by Pythagaras Theoram we can say that :

$~$

$\underline{\frak{As~ we ~know ~that~:}}$

• Diagonals of a rhombus bisect each other at 90°.
• Perimeter of Rhombus is Equal to 4 × side.
• ${\sf{\bigg( \dfrac{D_1}{2} \bigg)^{2}~ +~ \bigg( \dfrac{D_2}{2} \bigg)^{2} ~=~\bigg (side\bigg)^{2}}}$

$~$

◗Putting D1 as 24cm and D2 as 32 cm we get :

$~$

$\qquad{\sf:\implies{\bigg(\dfrac{24}{2}\bigg)^{2}~+~\bigg(\dfrac{32}{2}\bigg)^{2}~=~\bigg(side\bigg)^{2}}}$

$\qquad{\sf:\implies{12^{2}~+~16^{2}~=~\bigg(side\bigg)^{2}}}$

$\qquad{\sf:\implies{144~+~256~=~\bigg(side\bigg)^{2}}}$

$\qquad{\sf:\implies{400~=~\bigg(side\bigg)^{2}}}$

$\qquad{\sf:\implies{side~=~\sqrt{400}}}$

$\qquad:\implies{\underline{\boxed{\frak{\pink{side~=~20}}}}}$★

$~$

• Perimeter of Rhombus = 4 × side.

$~$

Hence,

$\therefore\underline{\sf{Perimeter~=~4~×~20~=>~\bf{\underline{80~cm}}}}$