Question

Find the area of a rhombus each side of which measures 20 cm and one of whose diagonals is 24 cm​

Answer 1

The side of rhombus measure 20cm each and one of the diagonals is 24cm.

We can make use of the Pythagoras theorem here. The diagonals intersect at the middle, mark it as O. Mark the respective 4 vertices ad A, B, C and D.

Now here, AB^2=AO^2+BO^2

Then we calculate it to get that BO^2= 256cm

Then BO = 16cm.

Now we can extend this BO to BD so that BD is 2BO which is nothing but 32cm.

Area of rhombus = (d1 x d2)/2

Then the answer is 384cm^2.

Mark my answer as brainliest.

Answer 2

Given:

The area of a rhombus each side of which measures 20 cm and one of whose diagonals is 24 cm

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To find?

Find the area of a rhombus

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Explanation:-

• All sides of rhombus are equal

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Solution:

$\sf \pink \implies$ AB=BC=CD=DA=20cm

Let BD be the diagonal of 24 cm

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Let's consider △ABD,

s=$\dfrac{20+20+24}{2}$ =32

A(△ABD)=

$\sqrt{s(s -a)(s -b)(s -c)}$

$= \sqrt{32 \times 12 \times 12 \times 8} = \sqrt{8 \times 4 \times {12}^{2} \times 8 }$

$= \sqrt{ {8}^{2} \times {2}^{2} \times {12}^{2} } = 8 \times 2 \times 12 = 192 {cm}^{2}$

A(□ABCD) =(△ABD) +(△BCD)

=192+192

=384$\sf{cm^2}$

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