Question

If the diagonals of a rhombus are 24 cm, 10 cm. Find the length of its side.​

Answer 1

Answer:

Let ABCD be the rhombus where, AC=10cm and BD=24cm

Let AC and BD intersect each other at O.Now, diagonals of rhombus bisect each other at right angles.

Thus, we have

AO=21×AC=21×10=5cm and

BO=21×BD=21×24=12cm

In right angled △AOB,

⇒ (AB)2=(AO)2+(BO)2

⇒ (AB)2=(5)2+(12)2

⇒ (AB)2=25+144

⇒ (AB)2=169

∴ AB=13cm

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

Answer 2

Answer:

13cm.

Explanation:

1st diagonal of rhombus = 24cm.

2nd diagonal of rhombus = 10cm.

Length of its side = ?

Let the 1st diagonal be x and the 2nd diagonal be y.

By using formula of side,

$★\:\:{\boxed{\boxed{{\bf{Side= \dfrac{ \sqrt{x^{2} + {y}^{2} } }{2}}}}}}\:\:★ \\ \\ \\ \rm:\dashrightarrow{Side= \frac{ \sqrt{(24)^{2} +({10})^{2} } }{2} } \\ \\ \\\rm:\dashrightarrow{Side= \frac{ \sqrt{576 + 100} }{2} \: \: \: \: \: \: \: } \\ \\ \\\rm:\dashrightarrow{Side= \frac{ \sqrt{676} }{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\\rm:\dashrightarrow{Side= \frac{26}{2}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\:\\ \\ \\\bf:\dashrightarrow \boxed{\bf{Side=13cm.}} \: \: \: \: \: \: \: \: \: \:$

Therefore, length of its side is 13cm.

Additional Information

Area of rhombus = ½ × x × y.

Perimeter of rhombus = 4 × side.

Area of triangle = ½ × b × h.

Area of rectangle = l × b.

Area of square = (side)².

Area of parallelogram = b × h.

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