Question

The area of rhombus whose dialoges are of length 10 cm and 8.2 cm is

Answer 1

41cm^2Answer:

Step-by-step explanation:

Area Of Rhombus=1/2d1.d2

Answer 2

Step-by-step explanation:

Question :-

• Find the area of rhombus whose diagonals measures 10 cm and 8.2 cm respectively.

To Find :-

• Area of rhombus.

Solution :-

Given :

• Diagonal (1) = 10 cm
• Diagonal (2) = 8.2 cm

By using the formula,

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

Where,

• d = length of the diagonals

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

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

${\longrightarrow{\sf \dfrac{1}{\cancel2} \times \cancel{10} \times 8.2}}$

${\longrightarrow{\sf 5 \times 8.2}}$

${\longrightarrow{\sf 41\ cm^2}}$

∴ Hence, area of rhombus is 41 cm².

More To Know :-

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