Question

the area of a rhombus whose perimeter is 80 m and one of its diagonal is 24 m is?

plz plz answer this...it will be a great help for me​

Answer 1

Given :

• Perimeter of rhombus = 80m
• One of it's diagonals = 24m

Aim :

• To find the area of the rhombus

Solution :

Properties of Rhombus :

• All sides of the rhombus are equal (opposite sides are equal)
• Opposite angles are equal.
• Adjacent angles add up to 180°.
• Diagonals bisect each other at 90°
• Rhombus is a parallelogram

Answer :

$\boxed{ \sf Area \: of \: a \: rhombus = \dfrac{1}{2} \times diagonal_{1} \: \times diagonal_{2}}$

Perimeter of rhombus = side + side + side + side → 4 × side

→ 80m

Let the side be x.

$\implies \sf 4 \times x = 80$

$\implies \sf x = \dfrac{80}{4}$

→ x = 20m

Hence, side of the rhombus = 20m

• We know that, the diagonals bisect each other at 90°.
• A right angled triangle is formed

By using the Pythagoras theorem let us find the other Diagonal.

☞ ½ × 24 = 12m

Area of ∆AOD :

(base)² + (height)² = (hypotenuse)²

Here,

• Base = unknown ½ of Diagonal 2.
• Height = ½ of Diagonal 1 (12m)
• Hypotenuse = side of the rhombus (20m)

Substituting the values,

➜ y² + 12² = 20²

➜ y² = 20² - 12²

Using identity a² - b² = (a + b)(a - b)

➜ y² = (20 - 12)(20 + 12)

➜ y² = (8)(32)

➜ y² = 256 m

Transposing the powers,

➜ y = √256

➜ y = 16m

½ of Diagonal 2 = 16m

Hence,

Diagonal 2 = 16 × 2 → 32m

Now that we have the values of both the diagonals,

Area of Rhombus :

$\implies \sf \dfrac{1}{2} \times 32 \times 24$

$\implies \sf 32 \times 12$

$\implies \sf 384 {m}^{2}$

Therefore, the Area of the Rhombus is 384m²

More Formulas :

• Area of triangle = ½ × base × height
• Area of square = side × side → (side)²
• Area of rectangle = Length × breadth
• Area of parallelogram = base × height
• Area of trapezium = ½ × height × (sum of parallel sides)