Question

The diagonal pf the rhombus are 48cm and 20cm long. Find area and perimeter of the rhombus.​

Answer 1

Answer:

ABCD is the rhombus and AC and BD are the diagonals. The diagonals intersect at point O.

We know that,

<DOC = 90

= DO = OB = 1/2 DB = 1/2 x 48 = 24 cm

Similarly, AO = OC = 1/2 AC = 1/2 x 20 = 10 cm

∴ In ΔDOC, we get

= DC^2 = √DO^2 + OC^2

= √24^2 + 10^2

= √576 + 100

= √676 (Approx 26cm)

∴ <DC the side of the Rhombus

∴ We know that in Rhomus, all sides are equal.

=> Perimeter of ABCD = 26 x 4 = 104 cm

Answer 2

★ See in attachment

GIVEn

The diagonal of the rhombus are 48cm and 20cm long.

TO FINd

Find area and perimeter of the rhombus

SOLUTIOn

Diagonals of rhombus = 48cm and 20cm

As we know that

✞ Area of rhombus

½ × product of diagonals

So, area of given rhombus

➠ ½ × AC × BD

➠ ½ × 48 × 20

➠ 24 × 20

➠ 480 cm²

NOTE : -

• Diagonals of rhombus bisect each other at 90°
• Adjacent sides are equal in rhombus

So,

• AO = OC = 48/2 = 24 cm
• BO = OD = 20/2 = 10 cm

Now , In ∆ AOB

✞ By Pythagoras theorem

➠ (AB)² = (BO)² + (AO)²

➠ AB² = (10)² + (24)²

➠ AB² = 100 + 576

➠ AB = √676 = 26 cm

AB = BC = CD = DA = 26 cm

✞ Perimeter of rhombus = 4a

where a = side

→ 4 × 26

→ 104cm

Hence,

Area of Rhombus = 480 cm²

Perimeter of Rhombus = 104 cm