The intersection o is the height of △BOC and intersects with BC at point D. This height is actually the height of △ABC (intersection a is the height of BC because BC//AO is equal). Moreover, because △BOC and △ABC have the same base and BC, the areas of △BOC and △ABC are equal, so finding the area of the shaded part is equivalent to finding the area of the sector OBC.

S=3. 14x3x3x 1/6

=4.7 1

7. If you look at the picture carefully, you will find that a radius perpendicular to the diameter below is made in the middle, and then the shadow part on the right side of the radius is placed in the blank part on the lower left, which just gives you half a semicircle, that is, the area of the shadow part. That is to find the circular area of 1/4, so

3. 14x4x4x 1/4 = 12.56

8. Have a good look. Yin 1+ Yin 3= semicircle = 3.14x10x/0x0.5 =157 Yin 1= 157- Yin 3.

According to the known conditions, Yin 1- Yin 2= 18 replaces Yin 1, 157- Yin 3- Yin 2= 18.

It can be concluded that Yin 2+ Yin 3 =157-18 =139 (that is, the area of triangle ABC). It is known that a cardinal number is 20, which is the diameter. Now find another right-angled edge = 139x2 divided by 20= 13.9.