Math problems, one problem in Grade 6 Geometry

One acute angle of a right triangle is 45°, and the other angle must be (180°-90°-45°=)45°.

That is, the two equal angles correspond to the equal edges, and the diameter of the circle r=8

The area of the right semicircle = r = 8

The shadow inside the triangle should be a fan with a radius of 4 and a central angle of 45 °.

Then the area is: (45° 360°) r =1 8 r =2 The total shadow area is 10.

Drawings are not allowed to ......

According to my thoughts.

It should be 10

The shadow on the left is 2 and 8 on the right

It's so difficult, I can't be a freshman in junior high school, it's hard for you, we didn't have this question at the time. o(╯□o

Count the semicircular ones first, with a diameter of 8

Diameter 2) square * 360 * 180 = 8

Then count the shadow in the triangle, and follow the arc from o.

Diameter 2) squared * 360*45=2 The two are added.

As for the substitution number, you can do it yourself.

In terms of the known conditions and graphs in the question, the answer is 8 on the right plus the answer on the left, depending on the radius of the shadow on the left.

First of all, the diagram is not accurate, this triangle should be an isosceles right triangle, and the circle should be a semicircle.

The area of the semicircle is multiplied by (8 divided by 2) multiplied by (8 divided by 2) divided by 2 = small sector area multiplied by (8 divided by 2) multiplied by (8 divided by 2) multiplied by 45 360 = total is.

The picture is too broken, it is 10 factions.

Question 1: Primary 6 Math Geometry Problem Look at the diagram Shape the area of the large trapezoid+ B into a triangle:

10 + 6) x 10 2 = 80 square centimeters.

Large body shape area:

10x10 = 100 square centimeters.

Area A is larger than B:

100-80 = 20 square centimeters.

Question 2: Primary 6 Plane Geometry Olympiad Problem 1It is known that the sum of two small triangles with an area is a large triangle, and the area is 2+6=8, so the ratio of its height to the height of a triangle with an area of 6 is 8:

6 = 4:3 (its bottom edge is the same), so the ratio of the height of the upper triangle to the lower triangle is 1:3, because the two are similar triangles, so the ratio of the upper bottom edge and the lower bottom edge of Qing Chao Jian is 1:

3 Suppose the bottom edge is x and the bottom triangle is y, thus.

xy=2*6=12

Trapezoidal area (upper bottom + lower bottom * height 2) = (x 3 + x) * (4 * y) 3 2 = (1 + 1 3) 4xy 3 2 = 4 3 * 4 * 12 3 2 = 10 and 2 3

2.It is impossible to distinguish c and d, and the total median area is 2*(a+b)=120, and the lower triangle is lower. Similar to question 1, the ratio to the area of a is 1:

9 (1:3 on the similar side of the trousers), so the area is 4, and the area of the figure on the left is 60-4=56

Question 3: Is it really that hard to love someone It's hard to fall in love with security because there are viruses.

It's also hard to fall in love with the virus, because it's illegal.

Security and virus boards, that's all it takes, hehe.

Question 4: Primary 6 Mathematics Geometry Problem Diagram.

Question 5: Grade 6 Geometry Proof Olympiad Questions: Due to the incomplete conditions of the question stem, the conditions cannot be answered normally.

Question 6: Primary 6 Mathematics Olympiad in Plane Geometry 100 points (Note: The result is 7/30).

<> use the bucket or compare, Li Pin is high. Which vertical.

This is not easy: because ab is 20 cm, the area of the semicircle is rr 2 = multiplied by 10 times 10 2 = 157 square centimeters, so the triangle area is 157-48 = 109 square centimeters, so the triangle area formula base multiplied by height divided by 2 obtains: bc multiplied by ab divided by 2 = 109, bringing ab into ob gets:

BC is equal to 109 divided by 10 = centimeters.

s -s = (s + white) - (s + white) = s semicircle - s triangle abc = vulture * (10 square) 2-bc*20 2 = 48

The solution is bc=5 vultures.

White set x

then the triangle is the area of the semicircle.

BC 5 Vultures.

120 degrees, i.e. 1 3 of the circle;

Calculate the area of the ring by 3!

30*30* -10*10*) 3= 800 3 cm choose a

The height of the parallelogram ABCD = 90 10 = 9, the triangle BIF is similar to the triangle DAF, [AAA] the area of the triangle BEF: the area of the triangle DAF = (5:10) = 1 4 The area of the triangle DAF = 4 * the area of the triangle BEF, the area of the triangle BAD = 10 * 9 2 = 90 2, that is, the area of the triangle ABF + the area of the triangle DAF = 90 2, the area of the triangle ABF + 4 * the area of the triangle BEF = 90 2,..

1) Area of triangle ABE = 5 * 9 2 = 45 2, area of triangle ABF + area of triangle BEF = 45 2, area of triangle ABF = 45 2 - area of triangle BEF ,..2) 2) Substitution 1):

Area of triangle BEF = (90 2-45 2) 3 = 15 2 area of triangle def = area of triangle bed - area of triangle BEF = 5 * 9 2-15 2 = 15

The triangle BEF is equal to the triangle DFA

So ad:be=fm:fn=2:1

So nf length = 3

Area of the triangle EFD = area of the triangle BCD - area of the triangle def - area of the triangle BEF.

s def=s deb-s bef (s denotes area).

BEF is similar to DAF, with high values of 6 and 3, respectively, and S BEF=5*3 2=

s△deb=5*9/2=，s△def=