Question

Find the length of the second diagonal of a rhombus, whose side is 5 cm and one of the diagonals is 6 cm.

Answer 1

Plz refers to the attachments

Given:

ABCD is a rhombus, in which each side is 5 cm, AC = 6 cm.

To find out:

Find the length of the second diagonal ( BD ) of a rhombus.

Solution:

AC = 6 cm

∠1 = ∠2 = ∠3 = ∠4 = 90°

OA = OC = AC/2 = 6/2 = 3 cm

and, OB = OD = BD/2 [ Diagonals of a rhombus are perpendicular and bisects each other]

In right ∆AOB,

By pythagoras therome

OA² + OB² = AB²

⇒ 3² + OB² = 5²

⇒ 9 + OB² = 25

⇒ OB² = 25 - 9

⇒ OB² = 16

⇒OB = √16

⇒ OB = 4 cm

Therefore,

Second Diagonal, BD = 2 ( OB )

= 2 × 4 cm

= 8 cm

Answer 2

Step-by-step explanation:

$\bf \underline{Question}$

Find the length of the second diagonal of a rhombus, whose side is 5 cm and one of the diagonals is 6 cm.

_______________________________

$\bf \underline{given \to}$

• whose side is 5 cm
• one of the diagonals is 6 cm.
• AB=BC=CD=DA=5cm
• AC=6 cm

_______________________________

$\bf \underline{to \: find \to}$

• The length of the second diagonal

___________________________

$\tt\underline{according \: to \: the \: question}$

$\tt \: ao = oc \: ao = 3cm \\ \tt \: aob \: is \: right \: angel \:triangle \: as \: diagonal \: \\ \tt\: of \: rhombus \: intersect \: at \: right \: angled \\ \tt \therefore \: \: by \: Pythagoras \: theory \to\: ob = 4c \\ \tt \: since \: do = ob \: and \: bd = 8cm \\ \tt \: length \: of \: the \: other \: diagonal \: = 2(bo) \\ \tt \: putting \: all \: values \: \\ \: \tt \: hence \: bd = 2 \times 4 = 8cm$

$\tt \: the \: length \: of \: the \: second \: diagonal = 8cm$