Given :
- Diagonal_1 of rhombus = 24 cm
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To find :
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Note :
Kindly keep in touch with picture
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Solution:
We know:
[tex] \bigstar \boxed{ \rm Area = \dfrac{Diagonal_1 \times Diagonal_2 }{2}}[/tex]
As we can clearly see we need both diagonals, but in question only one diagonal is given.
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So first let's find other diagonal.
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[tex] \red{\textsf{ \textbf{To find Diagonal}}} \red{\sf\pmb{_2}} \leadsto[/tex]
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In △ AOB :
AB - Hypotenuse
AO = Perpendicular
BO = Base
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Base² = Hypotenuse ² - Perpendicular²
∴ BO² = AB² - AO²
- BO² = 20² - 12²
- BO² = 400 - 12²
- BO² = 400 - 144
- BO² = 256
- BO = √(256)
- BO = √(16 × 16)
- BO = 16 cm²
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- Diagonal_2 = 2BO
- Diagonal_2 = 2 × 16
- Diagonal_2 = 32 cm
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[tex] \red{\textsf{ \textbf{To find Area}}} \leadsto[/tex]
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As we already know Area of rhombus so :
[tex]\twoheadrightarrow\sf Area = \dfrac{Diagonal_1 \times Diagonal_2 }{2} \\ [/tex]
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[tex]\twoheadrightarrow\sf Area = \dfrac{32 \times 24}{2} \\ [/tex]
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[tex]\twoheadrightarrow\sf Area = \dfrac{768}{2} \\ [/tex]
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[tex]\twoheadrightarrow\bf Area = \red {384 {cm}^{2} } \\ [/tex]
