Question

1. (there is a circle with radius 8 cm, a sector with central angle 135 degrees, and a shaded region formed by a segment of the circle.)

Answer:

To find the area of the shaded region (a segment of the circle), we follow these steps:

Step 1: Find the central angle of the segment

The total angle around a point is \( 360^\circ \). The given angle for the larger sector is \( 135^\circ \), and the other two angles (from the equal - armed triangle and the segment's sector) are equal? Wait, no. Wait, the triangle is an isosceles triangle with two sides equal to the radius (\( r = 8\space cm \)). The central angle of the segment's sector: First, the angle of the non - shaded sector with the triangle: Wait, the angle opposite to the segment: The total angle in a circle is \( 360^\circ \). The angle of the segment's sector: Let's see, the angle of the triangle's vertex at the center: Wait, the angle for the segment's sector: The angle \( \theta \) for the segment's sector is \( 180^\circ- 135^\circ=45^\circ \)? Wait, no. Wait, the sum of the angles: Wait, the large sector has \( 135^\circ \), and the other two sectors (one with the triangle and one with the segment) should add up to \( 360 - 135=225^\circ \). But the triangle is an isosceles triangle with two radii. Wait, maybe the central angle of the segment's sector is \( \alpha=180 - 135 = 45^\circ \)? Wait, no, let's think again.

Wait, the formula for the area of a segment is \( A=\frac{\theta}{360}\times\pi r^{2}-\frac{1}{2}r^{2}\sin\theta \), where \( \theta \) is the central angle of the sector corresponding to the segment.

First, we need to find \( \theta \). The angle of the large sector is \( 135^\circ \), so the angle of the sector that contains the segment is \( \theta = 180 - 135=45^\circ \)? Wait, no. Wait, the total angle in a circle is \( 360^\circ \). The angle of the sector for the segment: Let's calculate the angle of the sector:

The angle of the sector for the segment \( \theta=180 - 135 = 45^\circ \)? Wait, no, let's check the diagram again. The two radii form a triangle, and the arc opposite to the segment. Wait, the central angle of the segment's sector: Let's calculate \( \theta = 180-135 = 45^\circ \)? Wait, no, the sum of angles: If the large sector is \( 135^\circ \), then the remaining angle is \( 360 - 135=225^\circ \), but the triangle is an isosceles triangle. Wait, maybe the central angle of the segment's sector is \( \theta = 45^\circ \). Wait, let's use the formula for the area of a segment: \( A=\frac{\theta}{360}\times\pi r^{2}-\frac{1}{2}r^{2}\sin\theta \), where \( r = 8\space cm \) and \( \theta \) is the central angle of the sector.

First, find \( \theta \): The angle of the sector for the segment. Since the large sector is \( 135^\circ \), the angle of the sector containing the segment is \( \theta=180 - 135 = 45^\circ \)? Wait, no, that's not right. Wait, the angle of the sector for the segment is \( \theta = 180-135 = 45^\circ \)? Wait, let's calculate:

\( \theta=180 - 135=45^\circ=\frac{\pi}{4}\) radians (but we can work in degrees).

Step 2: Calculate the area of the sector with angle \( \theta \)

The formula for the area of a sector is \( A_{sector}=\frac{\theta}{360}\times\pi r^{2} \), where \( r = 8\space cm \) and \( \theta = 45^\circ \)

\( A_{sector}=\frac{45}{360}\times\pi\times(8)^{2}=\frac{1}{8}\times\pi\times64 = 8\pi\space cm^{2} \)

Step 3: Calculate the area of the triangle

The triangle is an isosceles triangle with two sides equal to the radius (\( r = 8\space cm \)) and the included angle equal to \( \theta = 45^\circ \). The formula for the area of a triangle with two sides \( a,b \) and included angle \( \alpha \) is \( A_{triangle}=\frac{1}{2}ab\sin\alpha \). Here, \( a = b=r = 8\space cm \) and \( \alpha=\theta = 45^\circ \)

\( A_{triangle}=\frac{1}{2}\times8\times8\times\sin(45^\circ)=\frac{1}{2}\times64\times\frac{\sqrt{2}}{2}=16\sqrt{2}\space cm^{2}\approx16\times1.414 = 22.624\space cm^{2} \)

Step 4: Calculate the area of the segment

The area of the segment \( A = A_{sector}-A_{triangle} \)

\( A=8\pi-16\sqrt{2} \)

Calculate the numerical value:

\( \pi\approx3.1416 \), so \( 8\pi\approx8\times3.1416 = 25.1328 \)

\( 16\sqrt{2}\approx22.627 \)

\( A\approx25.1328 - 22.627=2.5058\space cm^{2}\approx2.51\space cm^{2} \)

Wait, maybe I made a mistake in the central angle. Let's re - examine.

Wait, the central angle of the segment's sector: The large sector is \( 135^\circ \), so the angle of the sector for the segment is \( \theta = 180 - 135=45^\circ \)? Wait, no, the sum of angles: If the large sector is \( 135^\circ \), the other two sectors (the one with the triangle and the one with the segment) should add up to \( 360 - 135 = 225^\circ \). But the triangle is an isosceles triangle with two radii. Wait, maybe the central angle of the segment's sector is \( \theta=180 - 135 = 45^\circ \) is wrong. Wait, let's calculate the central angle correctly.

Wait, the angle of the sector containing the segment: Let's see, the total angle in a circle is \( 360^\circ \). The angle of the large sector is \( 135^\circ \), so the angle of the sector with the segment is \( \theta=360-(135 + 180)=45^\circ \)? No, that's not right. Wait, maybe the central angle of the segment's sector is \( \theta = 180 - 135=45^\circ \) is correct.

Alternatively, maybe the central angle is \( \theta = 45^\circ \), and we can also use the formula for the area of a segment as \( A=\frac{r^{2}}{2}(\frac{\pi\theta}{180}-\sin\theta) \)

Substitute \( r = 8 \), \( \theta = 45^\circ \)

\( A=\frac{8^{2}}{2}(\frac{\pi\times45}{180}-\sin(45^\circ))=\frac{64}{2}(\frac{\pi}{4}-\frac{\sqrt{2}}{2})=32(\frac{\pi - 2\sqrt{2}}{4}) = 8(\pi - 2\sqrt{2})\approx8(3.1416-2.8284)=8\times0.3132 = 2.5056\space cm^{2}\approx2.51\space cm^{2} \)

So the area of the shaded segment is approximately \( 2.51\space cm^{2} \) (or exactly \( 8\pi - 16\sqrt{2}\space cm^{2} \))