Question

12. A piece of land is in the shape of a rhombus whose perimeter is 400 m and one of its
diagonals is 160 m. find the area of the land.​

Answer 1

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Answer 2

Answer:

• Perimeter (Rhombus) = 400m
• Diagonal = 160 m

To Find:

• Area of the Rhombus?

We can derive side from perimeter by dividing by 4,

= 400/4

= 100 m (side)

Sides of ∆ABD,

• 100m
• 100m
• 160m

Sides of ∆BCD,

• 100m
• 100m
• ?

We can find the third side using Pythagorean theorem;

(h)² = (b)² + (p)²

(h)² = (100)² + (100)²

(h)² = 20000

h = √20000

h = 100√2 m (2nd Diagonal)

Formula Used:

• Area of Rhombus = ½ × D1 × D2

Substitute the values, we get;

➻ ½ × 160 × 100√2

➻ 80 × 100√2

➻ 8000√2 m²

So,

the area of Rhombus is 8000√2 m².