Question

Area of rhombus if Diagonals = 16 cm and 30 cm

Answer 1

Solution:

It's given That:

$\rm \longrightarrow d_{1} =16 \: cm$

$\rm \longrightarrow d_{2} =30 \: cm$

We know that:

$\rm \longrightarrow Area \: of \: Rhombus = \dfrac{1}{2} \times product \: of \: diagonals$

Therefore, the area of rhombus will be:

$\rm \longrightarrow Area = \dfrac{1}{2} \times 16 \times 30 \: cm^{2}$

$\rm \longrightarrow Area = 8 \times 30 \: cm^{2}$

$\rm \longrightarrow Area = 240 \: cm^{2}$

★ Therefore, the area of the rhombus with diagonals 16 cm and 30 cm is 240 cm²

Answer:

• Area of the rhombus is 240 cm²

Additional Information:

$\begin{gathered}\boxed{\begin{array}{cc}\\ \quad\Large\underline{\bf Area\:Formulas}:\\ \\ \star \:\sf Square=(Side)^2\\ \\ \star\sf\:Rectangle=Length\times Breadth\\\\ \star\sf\:Triangle=\dfrac{1}{2}\times Breadth\times Height\\\\ \star\sf\:Scalene\:Triangle=\sqrt{s(s-a)(s-b)(s-c)}\\ \\ \star\:\sf Rhombus=\dfrac{1}{2}\times d_1\times d_2\\\\ \star\sf\:Rhombus=\:\dfrac{d}{2}\sqrt {4a^2-d^2}\\ \\ \star\sf\:Parallelogram=Breadth\times Height\\\\ \star\sf\:Trapezium=\dfrac{1}{2}(a+b)\times Height\\ \\ \star\sf\:Equilateral\:Triangle=\dfrac{\sqrt{3}}{4}Side^2\end{array}}\end{gathered}$