Question

The diagonal of a rhombus are 26cm and 12cm.find its area and perimeter.

Answer 1

$\huge{\mathcal{Hi\: there!}}$

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$\huge{\underline{\bold{Solution :}}}$

Let the diagonals of the rhombus be P and Q.

Given ; P = 26 cm and Q = 12 cm.

Area of rhombus = $\mathsf{ \frac{pq}{2}}$

Area of rhombus = 26 × 12 / 2

Area of rhombus = 312 / 2

Area of rhombus = 156 cm².

Now,
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Let us know first, the properties of rhombus.

1. All sides of a rhombus are equal.

2. Diagonals of a rhombus bisect each other.

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According to question,

AC = 26 cm and BD = 12 cm.

And,

AE = EC = 13 cm.

BE = ED = 6 cm.

[see the above diagram]

Now,

In ΔAED, using pythagoras theorem,

We have ;

AD² = AE² + ED²

AD² = 13² + 6²

AD² = 169 + 36

AD² = 205

AD = √205

AD = 14. 31

As we know,

Each side of a rhombus are equal.

Therefore,

Perimeter = AD + DC + CB + BA

Perimeter = 14.31 × 4

Perimeter ≈ 57. 24 cm

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Thanks for the question!

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Answer 2

here is your answer ✌

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Let ABCD be the given rhombus.

We know that the Diagonals of a rhombus are perpendicular bisector of each other. Let the point of intersection of diagonals AC and BD be M.
Given AC = 26 cm and BD = 12 cm

Now, AE = EC/2 = 26/2 = 15 cm

and BE = ED /2 = 12/2 = 6 cm

Now, in right triangle AMD, by pythagoras theorem,

AD2 = AE'2 +ED'2

⇒AD2 =13'2 + 6'2

⇒AD2 =169 + 36 = 205

⇒ AD = √205 =14.31 cm

Again, all the sides of a rhombus are equal.

Therefore, AB = BC = CD = AD = 14.31 x 4 = cm

Now the perimeter of a rhombus = sum of all sides = AB + BC + CD + AD = 14.31cm = 57.24 cm OK

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