The formula for the area of a sector with radius r and central angle θ (in degrees) is given by (θ)/(360^°) × π r². We will substitute the values of r and θ into this formula.

Substitute θ = 90^° and r = 6 cm into the formula. Simplify the fraction (90)/(360) to (1)/(4) and calculate 6² = 36.

Multiply (1)/(4) by 36π to get the final area of the sector OEF.

The diagonals of a rhombus ABCD intersect at O. Taking 'O' as the centre, an arc of radius 6 cm is drawn intersecting OA and OD at E and F respectively. The area of the sector OEF is :

Answer: $9\pi\text{ cm}^2$

Step-by-step solution

Step 1: Identify properties of a rhombus

In a rhombus, the diagonals intersect at right angles. Therefore, the angle formed by the intersection of diagonals OA and OD, which is $\angle AOD$ , is $90^\circ$ . This angle is also the central angle of the sector OEF.

Step 2: Determine the radius and central angle of the sector

The problem states that an arc of radius 6 cm is drawn with O as the center. This means the radius of the sector OEF is $r = 6\text{ cm}$ . From the previous step, we know the central angle $\theta = \angle AOD = 90^\circ$ .

Step 3: Apply the formula for the area of a sector

The formula for the area of a sector with radius $r$ and central angle $\theta$ (in degrees) is given by $\frac{\theta}{360^\circ} \times \pi r^2$ . We will substitute the values of $r$ and $\theta$ into this formula.

Step 4: Calculate the area of the sector

Substitute $\theta = 90^\circ$ and $r = 6\text{ cm}$ into the formula. Simplify the fraction $\frac{90}{360}$ to $\frac{1}{4}$ and calculate $6^2 = 36$ .

Step 5: Final calculation

Multiply $\frac{1}{4}$ by $36\pi$ to get the final area of the sector OEF.

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