Question

the diagonal of romhous are 48cm , 20cm.
find...
i) Area of romhous.
ii) perimeter of romhous ​

Answer 1

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Area of the Rhombus = 1/2 X d1 X d2 ( d1 and d2 are the two diagonals of the Rhombus)

so ,

i) Diagonal 1 = 48 cm

Diagonal 2 = 20 cm

Area of the Rhombus = 1/2 X 48 X 20

= 960/2

= 480 cm^2

so the area of the Rhombus = 480 cm^2

ii) Here to find the perimeter we need to know the measurement of atleast 1 side

(Diagram of the Rhombus is in the attachment)

so let's consider ∆AOB

In ∆AOB

side AO = 1/2 of side AC.

AO = 1/2 X 48

= 24 cm

side BO = 1/2 of side BD

BO = 1/2 X 20

= 10 cm

Now we have to find the measurement of side AB

so,

By Pythagorean theorem

AO^2 + BO^2 = AB^2

(24)^2 + (10)^2 = AB^2

576+100 = AB^2

676 = AB^2

AB^2 = 676

AB = ✓676

AB = 26 cm

so ,

The side of the Rhombus = 26 cm.

Now,

perimeter of the Rhombus = 4 X side

= 4 X 26

= 104 cm

Therefore,

Perimeter of the Rhombus = 104 cm.

Note :

diagonal 1 = AC

diagonal 2 = BD

Hope this is useful....

Answer 2

Answer:

Area of the Rhombus = 1/2 X d1 X d2 ( d1 and d2 are the two diagonals of the Rhombus)

so ,

i) Diagonal 1 = 48 cm

Diagonal 2 = 20 cm

Area of the Rhombus = 1/2 X 48 X 20

= 960/2

= 480 cm^2

so the area of the Rhombus = 480 cm^2

ii) Here to find the perimeter we need to know the measurement of atleast 1 side

(Diagram of the Rhombus is in the attachment)

so let's consider ∆AOB

In ∆AOB

side AO = 1/2 of side AC.

AO = 1/2 X 48

= 24 cm

side BO = 1/2 of side BD

BO = 1/2 X 20

= 10 cm

Now we have to find the measurement of side AB

so,

By Pythagorean theorem

AO^2 + BO^2 = AB^2

(24)^2 + (10)^2 = AB^2

576+100 = AB^2

676 = AB^2

AB^2 = 676

AB = ✓676

AB = 26 cm

so ,

The side of the Rhombus = 26 cm.

Now,

perimeter of the Rhombus = 4 X side

= 4 X 26

= 104 cm

Therefore,

Perimeter of the Rhombus = 104 cm.

Note :

diagonal 1 = AC

diagonal 2 = BD

Hope this is useful....