Question

find the area of rectangle l-1.5m b-50 cm​

Answer 1

Explanation:

Length = 1.5m

= 150 cm

Breadth = 50 cm

area = l × b

= 150 × 50

= 7500 cm²

= 0.75 m²

#answerwithquality #BAL

Answer 2

Answer:

Given

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}&\bf\:Given - \begin{cases} &\sf \star{Lenght \: of \: Rectangle = 1.5 m }\\& \sf \star{ Breath \: of \: Rectangle = 50 cm}\end{cases}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

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To Find

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}&\bf\:To \: Find - \begin{cases} &\sf \star Area \: of \: Rectangle \end{cases}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

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Using Formula

${ : \implies \bf{Area \: of \: Rectangle} = \sf{Length × Breadth} }$

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Solution

→ Here the length is in m and the breadth is in cm.

So we convert the length into cm.

As we know.

$: \implies\sf{1 \: m = 100 \: cm}$

$: \implies\sf{1.5 \: m = 1.5 \times 100 } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf {= 150 \: cm}$

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→ Now Finding Area of Rectangle by using formula

$: \implies \sf{Area = 150 \: cm × 50 \: cm}$

$: \implies \sf{Area = 7500} \: {cm}^{2}$

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→ Coverting Area cm to m.

As we know.

$: \implies\sf{ 1 \: cm = \dfrac{1}{100} } \: m$

$: \implies \sf 7500 \: {cm}^{2} = \dfrac{7500}{100} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf= 75 \: {cm}^{2}$

$\Large \underline{\boxed{\frak \pink{Area} = \sf\purple{75 \: {cm}^{2}}}}$

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Therefore

The area of rectangle is 75 cm².

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More Useful formulae

$\small\begin{gathered}\begin{gathered} \bigstar \: \bf \underline{More \: Useful \: formulae} \: \bigstar\\ \begin{gathered} \\ {\boxed{\begin{array}{cccc}{\star\small\sf{Perimeter \: of \: Rectangle = 2(Length + Breadth)}} \\ \\ {\star\small \sf{Diagonal \: of \: Rectangle = \sqrt{ {Length }^{2} + { Breadth}^{2} }}} \\ \\ \star\small\sf{Area \: of \: Square } = {a}^{2} \\ \\ \star\small\sf{Perimeter \: of \: Square = 4a } \\ \\ \star\small\sf{Diagonal \: of \: Square = a \sqrt{2} }\end{array}}}\end{gathered}\end{gathered}\end{gathered}$