Question

Find the area of a rhombus, one side of which measures 20 cm.and one diagonal is 24 cm.

Answer 1

The area of the rhombus is $\bold{384\mathrm{cm}^{2}}$

Given:

One side of a rhombus = 20cm and one diagonal = 24 cm

To Find:

The area of a rhombus

Solution:

In the rhombus ABCD,  

Let, AB = 20 cm  

And AC = 24 cm

From the diagram,

AO + CO = 24 cm  

AO = 12 cm  [∵ AO = CO]      

In $\Delta \mathrm{AOB},$  

AB = 20 cm and AO= 12 cm

Using Pythagoras Theorem,

$\mathrm{BO}=\sqrt{20^{2}-12^{2}}\ \mathrm{cm}$

$=\sqrt{400-144}\ \mathrm{cm}$

$=\sqrt{256}\ \mathrm{cm}$

$=16\ \mathrm{cm}$  

Hence, BO = DO = 16 cm

∴ BD = 16 cm + 16 cm = 32 cm.

Area of a Rhombus $=\frac{1}{2} \times \text { Diagonal } 1 \times \text { Diagonal } 2$

$=\frac{1}{2} \times A C \times B D$

$=\frac{1}{2} \times 24\ \mathrm{cm} \times 32\ \mathrm{cm}$  

$=384\ \mathrm{cm}^{2}$

Answer 2

Answer:

The area of the rhombus is \bold{384\mathrm{cm}^{2}}384cm

2

Given:

One side of a rhombus = 20cm and one diagonal = 24 cm

To Find:

The area of a rhombus

Solution:

In the rhombus ABCD,

Let, AB = 20 cm

And AC = 24 cm

From the diagram,

AO + CO = 24 cm

AO = 12 cm [∵ AO = CO]

In \Delta \mathrm{AOB},ΔAOB,

AB = 20 cm and AO= 12 cm

Using Pythagoras Theorem,

\mathrm{BO}=\sqrt{20^{2}-12^{2}}\ \mathrm{cm}BO=

20

2

−12

2

cm

=\sqrt{400-144}\ \mathrm{cm}=

400−144

cm

=\sqrt{256}\ \mathrm{cm}=

256

cm

=16\ \mathrm{cm}=16 cm

Hence, BO = DO = 16 cm

∴ BD = 16 cm + 16 cm = 32 cm.

Area of a Rhombus =\frac{1}{2} \times \text { Diagonal } 1 \times \text { Diagonal } 2=

2

1

× Diagonal 1× Diagonal 2

=\frac{1}{2} \times A C \times B D=

2

1

×AC×BD

=\frac{1}{2} \times 24\ \mathrm{cm} \times 32\ \mathrm{cm}=

2

1

×24 cm×32 cm

=384\ \mathrm{cm}^{2}=384 cm

2