Question

Adjacent sides of a rectangle are in the ratio 5 : 12, if the perimeter of the rectangle is 34 cm, find the length of the diagonal.

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Answer 1

Step-by-step explanation:

Given :

• Adjacent sides of a rectangle are in the ratio of 5:12.
• Perimeter of the rectangle is 34 cm.

To Find :

• Length of the diagnol

Solution :

To find the length of the diagnol we must find the sides.

Let the common ratio be x.

Let the sides be 5x and 12x.

Perimeter of a rectangle = 2(l + b)

$\displaystyle{ \sf{ \implies34 = 2(5x + 12x)}}$

$\displaystyle{ \sf{ \implies\frac{ \cancel{34}}{ \cancel{2}} = 17x }}$

$\displaystyle{ \sf{ \implies17 = 17x}}$

$\displaystyle{ \sf{ \implies \cancel{ \frac{17}{17} = x }}}$

$\displaystyle{ \sf{ \implies1 = x}}$

So, the sides are 5 cm and 12 cm.

Now,

We need to find the length of the diagnol.

Let the diagnol be x.

By using Pythagoras Theorem,

⟹ p² + b² = h²

⟹ 5² + 12² = x²

⟹ 25 + 144 = x²

⟹ 169 = x²

⟹ √169 = x

⟹ 13 = x

Hence, measure of the diagnol of the rectangle is 13cm.

Know More :

$\boxed{\begin {minipage}{9cm}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}p\sqrt {4a^2-p^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {minipage}}$

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Answer 2

Answer :

• 13 cm

Given :

• Adjacent sides of a rectangle are in the ratio =  5 : 12
• The perimeter of the rectangle = 34 cm

To find :

• the length of the diagonal

Solution :

⇒ Let's find the length and breadth of the rectangle. Let the length and breadth be 5x and 12x.

$\rightarrow \sf Perimeter \: of \: Rectangle = 2 (l + b)$

$\sf 34 = 2 (5x + 12x)$

$\sf 34 = 2 (17x)$

$\sf 34 = 34x$

$\sf x = \dfrac{34}{34}$

$\sf \therefore x = 1$

$\longrightarrow \sf Length = 5x = 5 \times 1 = 5 cm$

$\longrightarrow \sf Breadth = 12x = 12 \times 1 = 12 cm$

⇒ Let's find the length of the Diagonal now.

• Here, to find the Diagonal, we have to use the "Pythagoras theorem" formula.

$\sf Diagonal = \sqrt{a^2 + b^2}$

$\sf = \sqrt{5^2 + 12^2}$

$\sf = \sqrt{25 + 144}$

$\sf = \sqrt{169}$

$\sf = 13$

• Therefore, the diagonal of the rectangle is 13 cm.