A chord of 9cm is constructed in a circle with a radius of 7cm.What is the area of the minor segment cut off by the chord?Leave your answer to 1 decimal place.

To find the area of the minor segment cut off by the chord, we first need to find the area of the sector and then subtract the area of the triangle.

Step 1: Find the length of the perpendicular from the center to the chord. The radius of the circle is 7cm and half the length of the chord is 4.5cm. We can use the Pythagorean theorem to find the length of the perpendicular (h) from the center to the chord.

h = √(r² - (chord/2)²) = √(7² - 4.5²) = √(49 - 20.25) = √28.75 ≈ 5.4cm

Step 2: Find the angle of the sector. We can use the cosine rule to find the angle (θ) of the sector.

cos(θ/2) = (Perpendicular/Radius) = 5.4/7 θ/2 = cos⁻¹(5.4/7) θ = 2cos⁻¹(5.4/7) ≈ 1.4 rad

Step 3: Find the area of the sector. The area (A) of the sector can be found using the formula A = 0.5r²θ.

A = 0.5 * 7² * 1.4 ≈ 34.3cm²

Step 4: Find the area of the triangle. The area of the triangle can be found using the formula A = 0.5 * base * height.

A = 0.5 * 9 * 5.4 = 24.3cm²

Step 5: Find the area of the minor segment. The area of the minor segment is the area of the sector minus the area of the triangle.

Area of minor segment = 34.3 - 24.3 = 10cm²

So, the area of the minor segment cut off by the chord is approximately 10.0cm².