Question

Find the area of a rectangular plot whose one side is 8 m long and diagonal is of length 17 m.​

Answer 1

Given:

• The one side of the rectangular plot is 8m (let it be breadth)
• The length of the diagonal of the rectangular plot is 17m

To find:

• The area of the rectangular plot.

The formula to find the area of a rectangle is (length*breadth) unit sq.

Only we know one side.

So, we have to find the other side to get the area of the rectangular plot.

Here, we can find the other side by using Pythagoras theorem, as each interior side of rectangle 90°.

So,

⇒ (length)² + (breadth)² = (diagonal)²

⇒ (length)² + (8)² = (17)²

⇒ (length)² = (17)² - (8)²

⇒ (length)² = 289 - 64

⇒ (length)² = 225

⇒ length = 15 m

So,

Area of the rectangular plot = length*breadth unit sq.

Area of the rectangular plot = 15 * 8

Area of the rectangular plot = 120 m sq.

Hence,

• The area of the rectangular plot is 120m sq.

Answer 2

$\large\underline\blue{\bold{Given \:  Question :-  }}$

• Find the area of a rectangular plot whose one side is 8 m long and diagonal is of length 17 m.

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$\bf \:\huge \red{AηsωeR : a} ✍$

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$\begin{gathered}\begin{gathered}\bf Given - \begin{cases} &\sf{one \: side \: of \: rectangle \: = 8 \: m} \\ &\sf{diagonal \: of \: rectangle \: = 17 \: m} \end{cases}\end{gathered}\end{gathered}$

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$\begin{gathered}\begin{gathered}\bf To \: Find :- \begin{cases} &\sf{area \: of \: rectangle} \end{cases}\end{gathered}\end{gathered}$

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$\begin{gathered}\Large{\bold{{\underline{Formula \: Used \::}}}} \end{gathered}$

$\begin{gathered}\;{\boxed{\bf{{(Hypotenuse)^2 = (Base)^2 + (Perpendicular)^2}}}}\end{gathered}$

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

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$\large\underline{\bold{Solution :- }}$

✏ Let us consider a rectangle ABCD such that

✏ BC = 8 m

✏ AC = 17 m

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$\large\underline{\bold{❥︎Step :- 1}}$

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✏ In right triangle ABC,

✏ Using Pythagoras Theorem,

$\sf \:  ⟼ {AC}^{2} = {BC}^{2} + {AB}^{2}$

$\sf \:  ⟼ \: {17}^{2} \: = \: {8}^{2} \: + \: {AB}^{2}$

$\sf \:  ⟼289 \: = 64 \: + \: {AB}^{2}$

$\sf \:  ⟼ \: {AB}^{2} \: = 289 \: - \: 64$

$\sf \:  ⟼ \: {AB}^{2} \: = \: 225$

$\sf \:  ⟼ \: AB \: = \: 15 \: m$

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$\large\underline{\bold{❥︎Step :- 2 }}$

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☆ Now, In rectangle ABCD, we have

✏ AB = 15 m

✏ BC = 8 m

$\bf \:  So, \: Area_{(rectangle)} \: = AB \: \times \: BC$

$\sf \:  ⟼ \: Area_{(rectangle)} \: = \: 15 \: \times \: 8$

$\sf \:  ⟼ \: Area_{(rectangle)} \: = \: 120 \: {m}^{2}$

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