please help asap tysm 3

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We know:-
[tex] \bigstar \boxed{ \rm Area \: of \: rhombus = \frac{Diagonal_1 \times Diagonal_2}{2} }[/tex]
By using this formula we can find second Diagonal as area of first Diagonal already given.
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So:-
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[tex] \leadsto\sf 0.15 = \dfrac{0.5 \times Diagonal_2}{2}[/tex]
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[tex] \leadsto\sf 0.15 = \dfrac{5 \times Diagonal_2}{2 \times 10}[/tex]
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[tex] \leadsto\sf 0.15 = \dfrac{\cancel5 \times Diagonal_2}{2 \times \cancel{10}}[/tex]
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[tex] \leadsto\sf 0.15 = \dfrac{ Diagonal_2}{2 \times 2}[/tex]
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[tex] \leadsto\sf 0.15 = \dfrac{ Diagonal_2}{4}[/tex]
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[tex] \leadsto\sf \dfrac{ Diagonal_2}{4} = 0.15 [/tex]
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[tex] \leadsto\sf Diagonal_2 = 0.15 \times 4[/tex]
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[tex] \leadsto\sf Diagonal_2 = \dfrac{15}{100} \times 4[/tex]
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[tex] \leadsto\sf Diagonal_2 = \dfrac{15}{\cancel{100}} \times\cancel 4[/tex]
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[tex] \leadsto\sf Diagonal_2 = \dfrac{15}{25} \times1[/tex]
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[tex] \leadsto\sf Diagonal_2 = \dfrac{15}{25} [/tex]
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[tex] \leadsto\sf Diagonal_2 = \dfrac{3}{5} [/tex]
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[tex] \leadsto\bf Diagonal_2 = 0.6 \: feet[/tex]
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[tex]\therefore\underline{ \textsf{ \textbf{Diagonal$_2$ of rhombus \: = \red{0.6 feet}}}}[/tex]