Question

using Pythagoras Theorem, find the length of the second diagonal of a rhombus of side 5 cm and having one side of its diagonal as 8cm.​

Answer 1

Solution:-

• Side of Rhombus = 5 cm

• Diagonal of Rhombus = 8 cm

In Rhombus all side are equal and

diagonal bisect each other at 90°.

Diagonal of Rhombus = 8cm

When it is bisected by another diagonal at 90°

= 8/2

= 4 cm

Therefore, other diagonal of Rhombus will be 5 cm

Now, Considered triangle OCD

DC = 5 cm ( Hypotenuse)

OD = 4 cm (base)

So Using the Pythagoras Theorem

h² = p² + b²

here,

P is the length of Perpendicular.

h is the length of hypotenuse.

b is the length of base .

OC² + OD² = DC²

Substitute the value we get

→ 0C² + 4² = 5²

→ OC² + 16 = 25

→ 0C² = 25-16

→ OC² = 9

→ OC = √9

→ OC = 3 cm

Diagonal AC = AO + OC

Diagonal AC = 3+3 = 6 cm

Therefore, the other diagonal of Rhombus is 6 cm.

Answer 2

Hope it Helps !!!!!!!!!!!!!!!!