Question

One of its diagonals of a rhombus is 6cm. If its area is 24 sq cm. Then find the length of each side.​

Answer 1

Answer:

5

Step-by-step explanation:

One of its diagonals of a rhombus is 6cm and

its area is 24 sq cm.

Then another diagonal of the Rhombus is,

\begin{gathered}area = \frac{1}{2} \times d1 \times d2 \\ 24 = \frac{1}{2} \times 6 \times d2 \\ d2 = 8 \: cm\end{gathered}

area=

2

1

×d1×d2

24=

2

1

×6×d2

d2=8cm

The diagonal of Rhombus bisect each other at perpendicular.

So,

\frac{1}{2} \times d1 = \frac{1}{2} \times 6 = 3 \: cm

2

1

×d1=

2

1

×6=3cm

and,

\frac{1}{2} \times d2 = \frac{1}{2} \times 8 = 4 \: cm

2

1

×d2=

2

1

×8=4cm

Applying the property of Right Angle Theorem:

side \: a = \sqrt{ {3}^{2} + {4}^{2} } = 5 \: cmsidea=

3

2

+4

2

=5cm

The length of each side is 5 cm.

Answer 2

Answer:

One of its diagonals of a rhombus is 6cm and the area of Rhombus is 24 sq cm.

Then another diagonal of Rhombus is,

$area = \frac{1}{2} \times d1 \times d2 \\ 24 = \frac{1}{2} \times 6 \times d2 \\ d2 = 8 \: cm$

The diagonal of Rhombus bisect each other at perpendicular,

so,

$\frac{1}{2} \times d1 = \frac{1}{2} \times 6 = 3 \: cm$

and

$\frac{1}{2} \times d2 = \frac{1}{2} \times 8 = 4 \: cm$

Applying the property of the Right Angle Theorem:

Side of Rhombus is,

$side \: a = \sqrt{ {3}^{2} + {4}^{2} } = 5 \: cm$

The length of each side is 5 cm.