Question

If the area of a rhombus be 24 cm² and one of its diagonal be 4 cm, find the perimeter of the rhombus.​

Answer 1

Perimeter = 8√10 cm

Step-by-step explanation:

$\large \pink{ \pmb{ \bf{\frak{Given:-}}}}$

• Area of rhombus = 24 cm²

• One of it's diagonal = 4 cm

$\large \pink{ \pmb{ \bf{ \frak{To \: Find:-}}}}$

• Perimeter of rhombus.

$\large \pink{ \pmb{ \bf{ \frak{Solution: - }}}}$

As we know that -

$: \mapsto \purple{\boxed{\bf \: Area \: of \: rhombus = \frac{1}{2} \times d_1 \times d_2}}$

So,

ACQ,

$\therefore \sf \: \frac{1}{2} \times AC \times BD = 24 \\ \\ \\ \sf➛ \: \frac{1}{2} \times BD \times 4 = 24 \\ \\ \\ ➛ \sf{2} \times BD = 24 \\ \\ \\ ➛ \large { \bf{ \boxed{ \bf\red{BD = 12 \: cm}}}}$

Thus, we have AC = 4 cm and BD = 12 cm

$\dashrightarrow \: \sf OA = \frac{1}{2} AC = \blue{2 \: cm} \\ \\ \\ \dashrightarrow \: \sf OB = \frac{1}{2} BD = \green{6 \: cm}$

Since the diagonals of a rhombus bisects each other at right angle. Therefore, ∆OAB is right ∆, right angled at O.

By Pythagoras theorem in ∆AOB, we have

$\sf \: AB {}^{2} = OA {}^{2} + {OB}^{2} \\ \\ \\ \implies \sf \:AB {}^{2} = {2}^{2} + {6}^{2} \\ \\ \\ \implies \sf \: AB = \sqrt{40} \\ \\ \\ \implies \bf\large \pink{AB = 2 \sqrt{10} cm}$

Hence,

$\sf \: Perimeter _{(rhombus)} = 4 \times side \\ \\ \\ ➥ \sf \:Perimeter _{(rhombus)} = 4 \times 2 \sqrt{10} \: cm \\ \\ \\ \large\pmb{ \bf{➥ \color{indigo} \:Perimeter _{(rhombus) = 8 \sqrt{10} \: cm }}}$

$\tiny \underline{\pink{\therefore\: \sf\:Perimeter _{(rhombus)} = 8 \sqrt{10} \: cm }}$

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