Question

Each side of a rhombus is 20 cm long and one of its diagonals is 24 cm in length, find the area of ​​this rhombus?

Answer 1

Given :

• Side of rhombus = 20 cm

• Diagonal_1 of rhombus = 24 cm

To find :

• Diagonal_2 of rhombus

Note :

Kindly keep in touch with picture

Solution:

We know:

$\bigstar \boxed{ \rm Area = \dfrac{Diagonal_1 \times Diagonal_2 }{2}}$

As we can clearly see we need both diagonals, but in question only one diagonal is given.

So first let's find other diagonal.

$\red{\textsf{ \textbf{To find Diagonal}}} \red{\sf\pmb{_2}} \leadsto$

In △ AOB :

AB - Hypotenuse

AO = Perpendicular

BO = Base

Base² = Hypotenuse ² - Perpendicular²

∴ BO² = AB² - AO²

• BO² = 20² - 12²
• BO² = 400 - 12²
• BO² = 400 - 144
• BO² = 256
• BO = √(256)
• BO = √(16 × 16)
• BO = 16 cm²

• Diagonal_2 = 2BO
• Diagonal_2 = 2 × 16
• Diagonal_2 = 32 cm

$\red{\textsf{ \textbf{To find Area}}} \leadsto$

As we already know Area of rhombus so :

$\twoheadrightarrow\sf Area = \dfrac{Diagonal_1 \times Diagonal_2 }{2}$

$\twoheadrightarrow\sf Area = \dfrac{32 \times 24}{2}$

$\twoheadrightarrow\sf Area = \dfrac{768}{2}$

$\twoheadrightarrow\bf Area = \red {384 {cm}^{2} }$

Answer 2

Answer:

AB = 20 CM

OA = 1

__ × 24 cm

2

= 12 cm

AOB = 90 °

OB² == AB² - OA ²

$( {20}^{2} ) - ( {12}^{2} ) \\ = 400 - 144 \\ = 256$

$ob \: = \sqrt{256} = 16 \\ bd \: = 2 \times ob \\ = 2 \times 16 \\ = 32cm \:$

AREA OF ABCD IS

$\frac{1}{2 } \times ac \times bd \\ = \frac{1}{2} \times 24 \times 32 \\ 384cm$

$hope \: its \: helpful \: \: to \: you \:$