#### QUESTION

Lengths of diagonals of a rhombus ABCD are 16 cm and 12 cm. Find the side and perimeter of the rhombus.

#### SOLUTION

ABCD is a rhombus.

Here, segment AC and segment BD are the diagonals of the rhombus ABCD.

l(AC) = 16 cm and l(BD) = 12 cm.

Let the diagonals of rhombus ABCD intersect at point O.

l(AO) = 12 l(AC)

…[Diagonals of a rhombus bisect each other]

∴l(AO) = 12 × 16

∴l(AO) = 8 cm

Also, l(DO) = 12 l(BD)

…[Diagonals of a rhombus bisect each other]

∴l(DO) = 12 × 12

∴l(DO) = 6 cm

In ∆DOA,

m∠DOA = 90°

..[Diagonals of a rhombus are perpendicular to each other]

[l(AD)]² = [l(AO)]² + [l(DO)]²

…[Pythagoras theorem]

= (8)² + (6)²

= 64 + 36

∴[l(AD)]² = 100

∴l(AD) = √100

… [Taking square root of both sides]

∴l(AD) = 10 cm

∴l(AB) = l(BC) = l(CD) = l(AD) = 10 cm

…[Sides of a rhombus are congruent]

Perimeter of rhombus ABCD

= l(AB) + l(BC) + l(CD) + l(AD)

= 10+10+10+10

= 40 cm

∴The side and perimeter of the rhombus are 10 cm and 40 cm respectively.

ABCD is a rhombus.

Here, segment AC and segment BD are the diagonals of the rhombus ABCD.

l(AC) = 16 cm and l(BD) = 12 cm.

Let the diagonals of rhombus ABCD intersect at point O.

l(AO) =

…[Diagonals of a rhombus bisect each other]

∴l(AO) =

∴l(AO) = 8 cm

Also, l(DO) =

…[Diagonals of a rhombus bisect each other]

∴l(DO) =

∴l(DO) = 6 cm

In ∆DOA,

m∠DOA = 90°

..[Diagonals of a rhombus are perpendicular to each other]

[l(AD)]² = [l(AO)]² + [l(DO)]²

…[Pythagoras theorem]

= (8)² + (6)²

= 64 + 36

∴[l(AD)]² = 100

∴l(AD) = √100

… [Taking square root of both sides]

∴l(AD) = 10 cm

∴l(AB) = l(BC) = l(CD) = l(AD) = 10 cm

…[Sides of a rhombus are congruent]

Perimeter of rhombus ABCD

= l(AB) + l(BC) + l(CD) + l(AD)

= 10+10+10+10

= 40 cm

∴The side and perimeter of the rhombus are 10 cm and 40 cm respectively.