Question

The length of a rectangle is less than twice its breadth by 1cm. the length of its diagonal is 17cm . find its length and breadth

Answer 1

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

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

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• Length of rectangle is 1 less than twice its breadth.

• Diagonal is 17 cm

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

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• Length of rectangle
• Breadth of rectangle

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

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Let the length be 'x'

Let the breadth be 'y'

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

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$\purple\longrightarrow$ $\sf x + 1 = 2y$

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$\red\longrightarrow$ $\sf x = 2y - 1$ ----------(1)

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We know that diagonal of rectangle divides it into 2 right angled triangles.

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Hence,

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

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$\sf \longmapsto 17^2 = x^2 + y^2$ -----------(2)

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$\footnotesize{ \underline{\bold{\texttt{Putting x = 2y - 1 in equation (2)}}}}$

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$\sf \longmapsto 17^2 = (2y -1)^2 + y^2$

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$\sf \longmapsto 289 = 4y^2 + 1 - 4y - y^2$

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$\sf \longmapsto 289 = 5y^2 - 4y + 1$

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$\sf \longmapsto 5y^2 - 4y - 288 = 0$

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$\sf \longmapsto 5y^2 - 40y + 36y - 288 = 0$

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$\sf \longmapsto 5y(y - 8) +36(y - 8) = 0$

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$\sf \longmapsto (5y + 36)(y - 8) = 0$

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$\footnotesize{ \underline{\bold{\texttt{We got the values of 'y' as}}}}$

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• $\rm y = 8$

• $\rm y = \dfrac { -36} { 5 }$

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Since length can't be negative hence y = 8cm

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$\footnotesize{ \underline{\bold{\texttt{Putting y = 8 in equation (1)}}}}$

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$\sf \longmapsto x = 2(8) - 1$

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$\sf \longmapsto x = 15 cm$

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Hence,

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Length = 15 cm

Breadth = 8 cm

$\rule{200}5$

Answer 2

Answer:

l=15

b=8

Step-by-step explanation:

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