Question

Pls solve this problem, step by step. Find the area of rhombus where each side measures 20 cm and of its diagonal is 32 cm. ​

Answer 1

Answer:

Area of rombus = diagonal×diagonal /2

diagonal = 32 ( given)

32×32 /2

= 512 cm^2

Answer 2

Solution:-

Given:-

• It is a rhombus.
• BD = 32 cm
• AB=BC=CD=DA= 20 cm

To find:-

• Area.

Formula used here will be:-

1. All sides of Rhombus are equal.
2. Diagonals of a Rhombus bisect each other.
3. Diagonals of a Rhombus bisect each other at right angles.
4. $area \: =( \: \frac{1}{2} \times diagonal_{1}\times diagonal_{2}$)

By the Problem:-

$BD = 32 cm= > \frac{1}{2} BD \: = \frac{32}{2}cm - - - (divide \: both \: side \: by \: 2)$

=>OD = 16 cm ----(OD+OB=BD&OD = OB)----(Rule ---3)

It is given that diagonals of a Rhombus bisect each other at right angles. Means we can use Pythagoras formula in ∆ AOD to find the length of the other diagonal.

So,

${(AD)}^{2} = {(AO)}^{2} + {(OD)}^{2} \\ = > {(2 0)}^{2} = {16}^{2} + {(AO)}^{2} \\ = > {(AO)}^{2} = {(2 0)}^{2} - {16}^{2} \\ = > {(AO)}^{2} = 400 - 256 \\ = > {(AO)}^{2} = 144 \\ = >AO = \sqrt{144} \\ = > AO = 12cm$

We know that AO = OC (Diagonals of a Rhombus bisect each other)

So, AO=OC= 12 cm

So, AC = (AO+OC)

=>AC =( 12+ 12) cm

=>AC =24 cm

$area \: =( \: \frac{1}{2} \times diagonal_{1}\times diagonal_{2}) \\ area \: = (\: \frac{1}{2} \times \: 32 \times 24) \:{cm}^{2} \\ = > 384 \: {cm}^{2}$

Answer:-

$384 \: {cm}^{2}$

See this attachment.

Hope you are looking for this answer.