Question

the diagonals of a rhombus are 26 and 10cm. find the length of one side of a rhombus.​

Answer 1

$\huge\underbrace\mathbf\pink{Answer}$

Let ABCD be the rhombus

Where,

• AC = 10cm

• BD = 24cm

Let AC and BD intersect each other at O.

Now, diagonals of rhombus bisect each other at right angles.

Thus, we have

${ \bf{\sf \: AO = \frac{1}{2} \times AC }}$

$\red\rightarrow{\bf {\sf \: \frac{1}{2} \times 10 = 5 \: cm}}$

${ \bf{\sf \: BO= \frac{1}{2} \times BD}}$

$\red\rightarrow{ \bf{\sf \: \frac{1}{2} \times 24 = 12 \: cm}}$

In right angled △AOB

${ \bf{\sf \:( {AB)</p><p>}^{2} =(AO)²+ (BO)²}}$

${ \bf{\sf \:( {AB)</p><p>}^{2} =(5) ²+ (12) ²}}$

${\bf{\sf \:( {AB)</p><p>}^{2} = 25 + 144}}$

$\bf{\sf \:( {AB)</p><p>}^{2} = 169}$

$\bold\red{AB = 13 \: cm}$

∴ The length of each side of rhombus is 13cm.

Answer 2

Answer:

$\sqrt{194}$

Step-by-step explanation:

We know that,

the diagonals of rhombus bisect each other at 90°.

Thus,

$\pink{ad = \frac{1}{2} \times ac} = \frac{1}{2} \times 10 = 5cm$

$\red{bd = \frac{1}{2} \times bd } = \frac{1}{2} \times 26 = 13cm$

Therefore,

AoB is a right angled triangle by thearem,we have.

AB² = AD² + BD²

AB² = 5² + 13²

AB² = 25 + 169

AB² = 194²

AB² = √194

HENCE, THE LENGTH OF ONE SIDE OF RHOMBUS IS √194