Question

find the area of a rectangular plot. one side of which is 48m and its diagnol is 50m​

Answer 1

Answer :

• Area of rectangular plot is 672m²

Given :

• One side is 48m
• Diagonal is 50m

To find :

• Area of rectangular plot

Solution :

As we know that, Pythagoras theorem,

• (AB)² = (AC)² + (BC)²

》(50)² = (48)² + (BC)²

》(BC)² = (50)² + (48)²

》(BC)² = 2500 - 2304

》(BC)² = 196

》BC = √ 196

》BC = 14cm

So,

Now we have to find the area of rectangular plot :

• Length is 48m
• Breadth is 14m

As we know that,

• Area of rectangular plot = l × b

Where, l is length and b is breadth

》 Area of rectangular plot = l × b

》48 × 14

》672m²

Hence , Area of rectangular plot is 672m²

Answer 2

${\purple{\underline{\underline{\mathsf{\mathbf{Solutíon:}}}}}}$

☯ Let ABCD be the rectangular plot.

$\:\:\:\:\:\:\:\:━━━━━━━━━━━━━━━━━━━$

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Then,

• AB ➻ 48 m
• AC ➻ 50 m

${━━━━━━━━━━━━━━━━━━━}$

$\circleddash$Need to Find : The Area of the rectangular plot.

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According to Pythagoras Theorem,

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• ${\underline{\boxed{\mathsf{\mathbf{\red{(AC)}^{\red 2}\:{\red =\:{\red \red \red (\red A \red B \red )}}^{\red 2}\:{\red +\:\red (\red B \red C\red )}^{\red 2}}}}}}$

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$\:\:\:\:\:\:\:\dashrightarrow\:\sf{(50)}^{2}\:=\:{(48)}^{2}\:+\:{(BC)}^{2}$

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$\:\:\:\:\:\:\:\dashrightarrow\:\sf{(BC)}^{2}\:=\:{(50)}^{2}\:-\:{(48)}^{2}$

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$\:\:\:\:\:\:\:\dashrightarrow\:\sf{(BC)}^{2}\:=\:{2500\:-\:2304}$

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$\:\:\:\:\:\:\:\dashrightarrow\:\sf{(BC)}^{2}\:=\:{196}$

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$\:\:\:\:\:\:\:\dashrightarrow\:\sf{BC\:=\:{\sqrt{196}}}$

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$\:\:\:\:\:\:\:\dashrightarrow\:\sf{\bold{BC\:=\:14\:m}}$

$\:\:\:\:\:\:\:\:━━━━━━━━━━━━━━━━━━━$

Hence,

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• ${\underline{\boxed{\mathsf{\mathbf{\red{The\:Area\:of\:the\:rectangle\:plot\:=\:l\:\times\:b}}}}}}$

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Where,

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• L ➻ 48 m
• B ➻ 14 m

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Then,

$\:\:\:\:\:\:\:\succcurlyeq{\sf{48\:\times\:14}}$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\succcurlyeq{\sf{672\:m}^{2}}$

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$\:\therefore\:{\underline{\sf{The\:Area\:of\:Rectangular\:Plot\:is\:{\textsf{\textbf{672\: m²}}}}}}.$

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$\:\:\:\:\:\:\:\:━━━━━━━━━━━━━━━━━━━$

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