A chord with a length of 3√ (2 + √2) cm contracts an arc whose degree measure is 135.

A chord with a length of 3√ (2 + √2) cm contracts an arc whose degree measure is 135. Find the area of a circular sector corresponding to this arc?

Let’s draw the radii OA and OB from the center of the circle O. The AOB triangle is isosceles, since ОА = ОВ = R.

Let’s use the cosine theorem for a triangle.

AB2 = R ^ 2 + R ^ 2 – 2 * R * R * CosAOB.

(3 * √ (2 + √2) ^ 2 = 2 * R ^ 2 – 2 * R ^ 2 * (-√2 / 2).

9 * (2 + √2) = 2 * R ^ 2 + R ^ 2 * √2 = R ^ 2 * (2 + √2).

9 = R ^ 2.

R = 3 cm.

Let’s define the area of the OAB sector.

S = (n * R² * α) / 360 = (n * 9 * 135) / 360 = n * 27/8 cm2.

Answer: The area of the sector is n * 27/8 cm2.