Question

find the area of Rhombus whose perimeter is 200 metre and one of the diagonal is 80 metre​

Answer 1

Answer:

2400

Step-by-step explanation:

4s=200

s=50

now diagonals bisext pependicullat each other

so by pythagoras you will find diagonals

then find area

plz try otherwise I will make it more clear to you

Answer 2

Formula that we use in the given question's solution

==>

Area of Rhombus

==>

(product of diagonals)/2

Perimeter of rhombus

==>

4 times of its side

Pythagoras theoram

==>

(perpendicular)² + (base)² = (hypotenious)²

Answer

==>

Area of given Rhombus will be 2400m²

Solution

==>

Let the sides of Rhombus as s and diagonals as d¹ , d²

Perimeter of Rhombus = 200m (given)

4s. = 200. (applying formula for perimeter of rhombus)

s = 200/4

s = 50m

So , sides of Rhombus = 50m

Now, d¹ = 80m (given)

✪we know that diagonals of Rhombus perpendicularly bisect each other.

So applying Pythagoras theoram

(D¹/2)² + (D²/2)² = s²

40² + (d²/2)² = 50²

(d/2)². = 2500 -1600

d/2. = √ 900

d/2. = 30 m

d. = 60m

From above we get

d¹ = 80m

d² = 60 m

Now applying formula for area of Rhombus

d¹×d²/2 = area of Rhombus

80×60/2 = area

2400m² = area of given Rhombus.