Question

find the area of a rhombus if the measure of each side is 26 CM and one of its diagonals is 48 CM.​

Answer 1

$\large{\rm{\underline{\underline{Given:-}}}}$

• One of the diagonal of the rhombus = $\sf{d_1}$ = 48 cm

• Measure of each side of the rhombus = 26 cm.

$\large{\rm{\underline{\underline{To\:Find:-}}}}$

• Area of the rhombus.

$\large{\rm{\underline{\underline{Solution:-}}}}$

We know that,

$\underline{\boxed{\mathtt{Area \:of\: the \:rhombus= \frac{1}{2} \times d_1 \times d_2}}}$

Let ABCD be the rhombus.

Each side of a rhombus,

$\dashrightarrow\:\:\:\rm{AB = BC = CD = DA = 26\:cm }$

✯ Diagonals of a rhombus bisect each other at right angles.

$\therefore\rm{ \angle AOB = \angle BOC = \angle COD = \angle DOA = 90^{\circ} }$

Now,

$\dashrightarrow\:\:\:\rm{ OA = OC }$

$\dashrightarrow\:\:\:\rm{ \dfrac{1}{2} \times AC }$

$\dashrightarrow\:\:\:\rm{ \dfrac{1}{2} \times 48 }$

$\dashrightarrow\:\:\:\rm{ 24 \:cm }$

According to Pythagoras theorem :-

$:\implies\;\;\:\rm{AB^2 = OB^2 + OA^2}$

$:\implies\;\;\:\rm{OB = \sqrt{(AB^2 - OA^2})}$

$:\implies\;\;\:\rm{OB = \sqrt{(26^2 - 24^2})}$

$:\implies\;\;\:\rm{OB = \sqrt{(100})}$

$:\implies\;\;\:\rm{OB = 10\:cm}$

∴ OB = OD = 10 cm

Other diagonal of the rhombus :-

$\dashrightarrow\:\:\:\rm{BD = 2 \times OB }$

$\dashrightarrow\:\:\:\rm{BD = 2 \times 10 }$

$\dashrightarrow\:\:\:\rm{BD = 20\:cm }$

∴ Other diagonal of the rhombus = $\sf{d_2}$ = 20 cm.

Now,

Area of the rhombus :-

$:\implies\:\:\:\rm{Area =\dfrac{1}{2}\times d_1 \times d_2 }$

$:\implies\:\:\:\rm{Area =\dfrac{1}{2}\times 48 \times 20}$

$:\implies\:\:\:\rm{Area = 10 \times 48}$

$:\implies\:\:\:\rm{Area = 480\:cm^2}$

∴ Area of the rhombus = 480 cm²

Answer 2

The area of the rhombus is 480 cm².

Step-by-step explanation:

Referring to the figure attached below, we have

ABCD is a rhombus

Each side of a rhombus, AB = BC = CD = DA = 26cm

One of the diagonal, AC = 48 cm

We know that the diagonals of a rhombus bisect each other at right angles.

∴ ∠AOB = ∠BOC = ∠COD = ∠DOA = 90°

and

∴ OA = OC = ½ * AC = ½ * 48 = 24 cm

Now, considering ∆ AOB and applying Pythagoras theorem, we get

AB² = OB² + OA²

⇒ OB = √[AB² – OA²]

⇒ OB = √[26² – 24²]

⇒ OB = √[100]

⇒ OB = 10 cm

∴ OB = OD = 10 cm

∴ Other diagonal of the rhombus, BD = 2 * OB = 10 * 2 = 20 cm

Thus,

The area of the rhombus ABCD,

= ½ * AC * BD

= ½ * 48 * 20

= 10 * 48

= 480 cm²

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