Question

question no 2 and 3 my dear

Answer 1

$\underline{ \underline{ \bold{ \red{ \: (2) \: \: QUESTION \: \: }}}}$ ➡

$\bold{The \: \: diagonals \: \: of \: \: a \: \: rhombus \: \: are \: \: }$

$\bold{in \: \: the \: \: ratio \: \: 5 : 12 . \: If \: \: the \: \: perimeter}$

$\bold{is \: \: 104 \: \: cm \:, \: Find \: \: the \: \: lengths \: \: of \: \: }$

$\bold{the \: \: sides \: \: and \: \: the \: \: diagonals.}$

$\underline{ \underline{ \bold{ \red{ \: \: SOLUTION\: \: }}}}$➡

$\bold{Given ,} \\ \\ \bold{The \: \: perimeter \: \:o f \: \:the \: \: rhombus = 104 \: \: cm } \\ \\ \bold{Let, \: \: } \\ \\ \bold{Sides \: \: of \: \: the \: \: rhombus = x \: \: cm} \\ \\ \bold{A.T.Q.,} \\ \\ \bold{4x = 104} \\ \\ = > \bold{x = \frac{104}{4} } \\ \\ \bold{ = > x = 26} \\ \\ \bold{So, \: \: The \: \: length \: \: of \: \: the \: \: sides \: \: of \: \: } \\ \bold{the \: \: rhombus = 26 \: \: cm}$

$\bold{Again \:, } \\ \bold{ \: \: \: \: \: \: \: \: \: \: It \: \: is \: \: given,} \\ \\ \bold{Ratio \: \: of \: \: t he\: \: diagonals = 5 : 12} \\ \\ \bold{Let,} \\ \\ \bold{The \: \: diagonals \: \: are \: \: : \: \: 5y \: \: cm \: \: \: and \: \: \: 12y \: \: cm}$

$\bold{We \: \: know \: \: that,} \\ \\ \bold{The \: \: diagonals \: \: of \: \: a \: \: rhombus \: \: bisect} \\ \bold{each \: \: other \: \: at \: \: right \: \: angle \: \: i.e \: \: 90 {}^{ \circ} }$

$\bold{So,} \\ \\ \bold{(5y/2) {}^{2} + (12y/2) {}^{2} = (26) {}^{2} } \\ \\ = > \bold{25y {}^{2}/4 + 144y {}^{2} /4= (26) {}^{2} } \\ \\ \bold{ = >16 9y {}^{2}/4 = (26) {}^{2} } \\ \\ = > \bold{(13y/2) {}^{2} = (26) {}^{2} } \\ \\ \bold{ = > 13y/2 = 26} \\ \\ = > \bold{y = \frac{26 \times 2}{13} } \\ \\ \bold{ = > y = 4} \\ \\ \\ \bold{Hence\: ,\: The \: \: diagonals \: \: are \: : } \\ \\ \bold{5y =( 5 \times 4) \: cm = 20 \: \: cm} \\ \\ \bold{12y =( 12 \times 4 )\: cm = 48 \: \: cm}$

$\bold{The \: \: sides \: \: of \: \: the \: \: rhombus = 26 \: \: cm} \\ \\ \bold{The \: \: lengths \: \: of \: \: the \: \: diagonals} \\ \bold{of \: \: the \: \: rhombus = 20 \: cm \: \: \: and \: \: 48\: cm}$

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$\underline{ \underline{ \bold{ \red{(3) \: \: QUESTION\: \: }}}}$➡

$\bold{A \: \: pair \: \: of \: \: adjacent \: \: sides \: \: of \: \: a \: \:} \\ \bold{rectangle \: \: are \: \: in \: \: the \: \: ratio \: \: 3:4. }$

$\bold{If \: \: its \: \: diagonals \: \: is \: \: 20 \: \: cm \: ,\: Find \: \: the} \\ \bold{lengths \: \: of \: \: the \: \: sides \: \: and \: \: hence} \\ \bold{the \: \: perimeter \: \: of \: \: the \: \: rectangle.}$

$\underline{ \underline{ \bold{ \red{ \: \:SOLUTION \: \: }}}}$➡

$\bold{Given,} \\ \\ \bold{The \: \: ratio \: \: of \: \: the \: \: adjacent \: \: sides \: \: of} \\ \bold{the \: \: rectangle = 3 : 4} \\ \\ \bold{Length \: \: of \: \: the\: \: diagonal = 20 \: cm}$

$\bold{Let,} \\ \\ \bold{The \: \: breadth \: \: of \: \: the \: \: rectangle = 3x \: cm} \\ \bold{And,} \\ \bold{The \: \: length \: \:of \: \: the \: \: rectangle = 4x \: cm }$

$\bold{We \: \: know \: \: that,} \\ \\ \bold{Each \: \: angle \: \: of \: \: a \: \: rectangle = 90 {}^{ \circ} } \\ \\ \bold{So,} \\ \\ \bold{(3x) {}^{2} + (4x) {}^{2} = (20) {}^{2} } \\ \\ \bold{ = > 9x {}^{2} + 16x {}^{2} = (20) {}^{2} } \\ \\ = > \bold{ 25x {}^{2} =(20) {}^{2} } \\ \\ \bold{ = > (5x) {}^{2} = (20) {}^{2} } \\ \\ = > \bold{5x = 20} \\ \\ = > \bold{x = \frac{20}{5} } \\ \\ \bold{ = > x = 4} \\ \\ \bold{Hence,} \\ \\ \bold{Length \: \: of \: \: the \: \: rectangle = 4x \: \: cm} \\ \\ \bold{ = (4 \times 4) \: \: cm } \\ \\ \bold{= 16 \: \: cm} \\ \\ \bold{Breadth \: \: of \: \: the \: \: rectangle =3x \: cm } \\ \\ \bold{ = (3 \times 4) \: \: cm} \\ \\ \bold{ = 12 \: \: cm}$

$\bold{Now,} \\ \\ \bold{Perimeter \: \: of \: \: the \: \: rectangle} \\ \bold{ =2( Length + Breadth)} \\ \\ \bold{ =2 (16 + 12) \: \:cm} \\ \\ \bold{ =56 \: \: cm}$

$\bold{Length \: \: of \: \: the \: \: rectangle = 16 \: \: cm} \\ \\ \bold{Breadth \: \: of \: \: the \: \: rectangle = 12 \: \: cm} \\ \\ \bold{Perimeter \: \: of \: \: the \: \: rectangle =56 \: \: cm }$

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✝$\mathfrak{\purple{THANKS..}}$✝