Question

${ \bf{ \huge{Question:}}}$

★The diagonal of a rectangle is thrice it's smaller side.

➱Find the ratio of its sides

↝ Quality Answer needed ✅​

Answer 1

Given : The diagonal of a rectangle is thrice its small side.

Exigency to Find: The ratio of its side respectively.

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❒ Now let the smaller side be x and the diagonal be 3x

${ \underline{ \bf{ \bigstar \: According \: to \: the \: question : }}}$

In the following rectangle,

• AB is the largest side and Bc is the smaller side.

We know,

• BC × 3 = AB

Here,

• BC = x
• AC = 3x

In Triangle ABC , Using pythagoras therom we get,

$\dashrightarrow \tt \: {h}^{2} = {s}^{2} + {s}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \dashrightarrow \tt( {ac)}^{2} = ( {bc)}^{2} + ( {ab)}^{2} \: \: \: \: \\ \\ \\ \dashrightarrow \tt {(3x)}^{2} = {x}^{2} + {(ab)}^{2} \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \dashrightarrow \tt \: 9 {x}^{2} = {x}^{2} + {(ab)}^{2} \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \dashrightarrow \tt {8x}^{2} = {(ab)}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \dashrightarrow \tt \: \sqrt{8 {x}^{2} } = {(ab)}^{} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \dashrightarrow \tt \: 2 \sqrt{2x} = ab \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

• Henceforth, the longer side = 2√2x

We know,

• The smaller side = x

${ \large{ \rm{ratio : }}}$

$\longmapsto \sf 2 \sqrt{2x} : x \\ \\ \longmapsto \sf 2 \sqrt{2} : 1 \: \:$

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Thx for the question!

Answer 2

$\textbf{Answer:}$

$\text2√2 : 1$

$\textbf{Let's see the solution }$

$\boxed{\bf{Refer\: the\: attached\: figure}}$

$\sf\color{red}In\: rectangle\: ABCD$

$\sf\color{pink}AB\: is\: the\: longer\: side$

$\sf\color{green}BC\: is\: the\: smaller\: side$

$\sf\color{darkviolet}AC\: is\: the\: diagonal$

ΔABC is the right angled triangle since all angles of rectangle is of 90°.

Let the smaller side BC be x

Diagonal = AC = $\sf3 \times \text{Smaller Side}$

$\mapsto\bf{So, In ΔABC}$

$Hypotenuse^{2} = Perpendicular^2+ Base^2Hypotenuse$

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

$\sf(3x)^{2} = AB^2+ x^2(3x)$

$\sf9x^{2} = AB^2+ x^29x$

$\sf8x^{2} = AB^28x$

$\sf\sqrt{8x^{2}} = AB$

$\sf2\sqrt{2}x= AB²$

Thus the longer side is$\sf 2\sqrt{2}x2$

$\sf{Smaller\: side = x}$

$\sf {Ratio\: of\: its\: Sides = Longer\: side : smaller \:side}$

= 2√2x : x

= 2√2 : 1

$\bold\pink{\fbox{\sf{Hence\: ratio\: of \:its \:sides\: \:is \:2√2 : 1 .}}}$

Thankyou :)