Question

the length of the diagonal of a rhombus is 24m and 10m then find perimeter of the rhombus and area of the rhombus​

Answer 1

Given:-

• Length of the diagonal of a rhombus is 24m and 10m.

To Find:-

• Perimeter and Area of the Rhombus.

Formula Used:-

• ${\boxed{\bf{\red{Pythagoras\: Theorem :- H^2=P^2+B^2}}}}$

• ${\boxed{\bf{\red{Perimeter\: of\: Rhombus =4a}}}}$

• ${\boxed{\bf{\red{Area\: of\: Rhombus =\dfrac{1}{2}\times d_1 \times d_2}}}}$

Here,

• a = Side of the Rhombus

• $d_1$ = First Diagonal

• $d_2$ = Second Diagonal

Solution:-

According to the figure,

Let AC be 24 cm and BD be 10 cm.

So, CE = 12 cm and BE = 5 cm

Applying Pythagoras theorem in Δ BEC,

$\sf :\implies BC^2=BE^2+CE^2$

$\sf :\implies BC^2=5^2+12^2$

$\sf :\implies BC^2=25+144$

$\sf :\implies BC^2=169$

$\sf :\implies BC=\sqrt{169}$

$\sf :\implies BC=13\:cm$

Now,

$\sf :\implies Perimeter\: of\: Rhombus =4a$

$\sf :\implies Perimeter\: of\: Rhombus =4\times 13$

$\sf :\implies Perimeter\: of\: Rhombus =52\:cm$

And,

$\sf :\implies Area\: of\: Rhombus =\dfrac{1}{2}\times d_1 \times d_2$

$\sf :\implies Area\: of\: Rhombus =\dfrac{1}{2}\times 24 \times 10$

$\sf :\implies Area\: of\: Rhombus =12 \times 10$

$\sf :\implies Area\: of\: Rhombus =120\:cm^2$

Hence, The Perimeter and Area of the Rhombus is 52 cm and 120 cm² respectively.

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