Question

Find the area of rhombus each side of whichmeasures20 cm and one of whose diagnols is 24 cm

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

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

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• Each side of rhombus is 20cm

• Diagonal is 24 cm

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$\red{\underline \bold{To \: Find:}}$

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• Area of rhombus.

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$\large{\orange{\underline{\tt{Solution :-}}}}$

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Properties of rhombus.

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⚝ All the properties of a parallelogram apply (the ones that matter here are parallel sides, opposite angles are congruent, and consecutive angles are supplementary).

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⚝ All sides are congruent by definition.

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⚝ The diagonals bisect the angles.

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⚝ The diagonals are perpendicular bisectors of each other.

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$\setlength{\unitlength}{1.6mm}\begin{picture}(5,6)\put(0,0){\line(1,1){12}}\put(20,0){\line(1,1){12}}\put(0,0){\line(1,0){20}}\end{picture}\put(6.9,12){\line(1,0){20}}\put(15,0){\line(-2,3){8}}\put(5,3){}\put(4.5,-2){}\put(11,7.5){24 cm}$

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Let ABCD be a rhombus.

BD & AC be the diagonals of the rhombus.

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$\purple{\underline \bold{According \: to \: the \ question :}}$

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AB = BC = CD = DA = 20cm

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BD = 24cm

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Let the diagonals bisect each other at 'E'

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Hence BE = 12cm

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As the diagonals of rhombus are perpendicular bisectors of each other.

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$\underline{\bold{\texttt{By pythagoras theorem}}}$

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BE² + AE² = AB²

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$\sf \longmapsto 12^2 + AE^2 = 20^2$

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$\sf \longmapsto 144 + AE^2 = 400$

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$\sf \longmapsto AE^2 = 256$

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$\sf \longmapsto AE = \sqrt {256}$

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$\sf \longmapsto AE = 16cm$ -------(1)

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$\sf \dashrightarrow AC = 2 \times AE$

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$\purple\longrightarrow$ $\sf AC = 32cm$ --------(2)

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$\underline{\bold{\texttt{Area of rhombus:}}}$

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$\red\longrightarrow$ $\bf \dfrac { 1 } { 2 } diagonal 1 \times diagonal 2$

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diagonal 1 = 24 cm [Given]

diagonal 2 = 32 cm [From (1) & (2)]

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⇛ $\sf\dfrac { 1 } { 2 } \times24 \times 32$

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$\bf \dashrightarrow Area = 384 sq cm$

$\rule{200}5$

Answer 2

Step-by-step explanation:

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