Question

area of rhombus whose diagonal are 20cm and 14 cm is.
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Answer 1

Answer:

$p \: digonal = 20cm \\ qdigonal = 14cm \\ solution \\ a = \frac{pq}{ 2} = \frac{20 \times 14}{2} = 140cm {}^{2}$

Answer 2

Step-by-step explanation:

Question :-

• Find the area of rhombus whose diagonals measures 20 cm and 14 cm.

To Find :-

• Area of rhombus.

Solution :-

Given :

• Diagonal (1) = 20 cm
• Diagonal (2) = 14 cm

By using the formula,

${\sf{\longrightarrow Area\ of\ rhombus\ =\ \dfrac{1}{2} \times d_1 \times d_2}}$

Where,

• d = length of the diagonals

According to the question, by using the formula, we get :

${\sf{\longrightarrow Area\ of\ rhombus\ =\ \dfrac{1}{2} \times d_1 \times d_2}}$

${\sf{\longrightarrow \dfrac{1}{\cancel2} \times \cancel{20} \times 14}}$

${\sf{\longrightarrow 10 \times 14}}$

${\sf{\longrightarrow 140\ cm^2}}$

$\therefore$ Hence, area of rhombus is 140 cm².

More To Know :-

$\begin{gathered}\boxed{\begin {minipage}{9cm}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}p\sqrt {4a^2-p^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {minipage}}\end{gathered}$