Question

ABCD is a rhombus with AC=16 cm and AB=10 cm.what is the area of the rhombus ABCD?

Answer 1

Given:

• A rhombus ABCD
• AC = 16 cm
• AB = 10 cm

To be found:

Area of the rhombus ABCD

Note:  An image of rhombus representing the given dimensions is attached.

Formula to be used:

Area of a rhombus = $(1/2) . d_1 . d_2$..............(1)

where, d₁ and d₂ are the lengths of the diagonals of rhombus.

Rhombus: Rhombus is a closed polygon with four equal sides, but unlike square, it does not have equal angles and the sides do not meet perpendicularly.

Properties of rhombus:

• The diagonals of rhombus meet each other perpendicularly.
• The diagonals of rhombus bisect each other, i.e., a diagonal divides other diagonal into two equal parts.

Solution:

Let d₁ be the length of the diagonal AC.

Let d₂ be the length of diagonal BD.

From the image, one side of the rhombus is given as AB = 10cm

And, a diagonal is given as AC = 16cm

i.e., $d_1 = 16cm$

To find the area of the given rhombus, the length of other diagonal (d₂) is required.

From the image, consider the intersection point of the diagonals AC and BD as "O".

Now, from the properties of the rhombus,

AO = 8cm

Other diagonal's length "d₂" can be calculated as:

$d_2 = BD = BO + OD$........ (2)

Since, $BO = OD$ (from the properties), $BD = 2(BO)$....... (3)

BO can be calculated using the pythagoras theorem, as AOB forms a triangle with right angle at "O".

Now, from Pythagoras theorem:

$AB^{2} = AO ^{2} + BO ^{2}\\10^{2} = 8^{2} + BO ^{2}\\100 = 64 + BO^{2}\\BO^{2} = 100 - 64\\BO^{2} = 36\\BO = \sqrt{36} \\BO = 6 cm$

Now, using equation (2), the length of the diagonal BD = $d_2 = 2(BO) = 2(6)$

⇒$d_2 = 2(BO) = 2(6) = 12 cm$

The values of d₁ and d₂ are available as 16cm and 12cm.

Now, from equation (1), the area of the rhombus ABCD is given by:

$A = (1/2) * 16 * 12\\A= 8 * 12\\A = 96 cm^{2}$

∴ The required area of rhombus ABCD is 96cm².

Hence, found.

Answer 2

Answer:

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Step-by-step explanation:

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