Question

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Answer 1

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Answer 2

$\huge \boxed{ \bf{question}}$

The lengths of the diagonals of a rhombus are 9cm and 12cm.Find the distance between any two parallel sides of the rhombus.

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Let ABCD be a rhombus, where

AC = 12cm and BD = 9cm.

To find : Distance between AB and CD.

Construction : Draw DP perpendicular to AB.

Now, Area of rhombus ABCD =

$\frac{1}{2} \times d1 \times d2.$

$\implies \frac{1}{2} \times 12 \times 9$

$\implies \: {54 \:cm}^{2}$

Now, we know diagonals divides a rhombus in two equal parts

$\therefore \: AREA \: of \: triangle \: ABD \: = {27 \: cm}^{2}$

[ ---------------> 1]

Now, we know the diagonals of a rhombus bisect each other perpendicularly.

Therefore,

AO = CO = 6cm.

BO = DO = 4.5cm. and

ang.AOB = 90°.

So, by applying Pythagoras theorem in ∆ AOB, we get

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

${AB}^{2} = {4.5}^{2} + {6}^{2}$

${AB}^{2} = 20.25 \: + \: 36$

${AB}^{2} = \: 56.25$

$AB = \sqrt{56.25}$

$AB = 7.5 \: cm$

Now ,

Area ∆ ABD

$= \frac{1}{2} \times b \times h$

$\implies \: 27 = \frac{1}{2} \times 7.5 \times h \: \: \: \: \: \: \: \: (- - - > AB = b)$

$\implies \: 54 = 7.5 \times h$

$\implies \: h = 7.2 \: cm$

Therefore, the distance between both parallel sides AB and CD is 7.2 cm.