Find the height of a rhombus whose area is $168 d m^{2}$ and the corresponding base is $21 d m$ .

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Solution

$\begin{matrix} A r e a = b a s e \times h e i g h t 168 = 21 \times h e i g h t ∴ h e i g h t = 8 d m [∴ 1 d m = 10 c m] \end{matrix}$

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Find the height of a rhombus whose area is 168 dm2 and the corresponding base is 21 dm.

Area=base×height168=21×height∴height=8dm[∴1dm=10cm]

Find the height of a rhombus whose area is $168 d m^{2}$ and the corresponding base is $21 d m$ .

$\begin{matrix} A r e a = b a s e \times h e i g h t 168 = 21 \times h e i g h t ∴ h e i g h t = 8 d m [∴ 1 d m = 10 c m] \end{matrix}$