Elementary Geometry Solving, Elementary Geometry Problems

<> as shown in the figure above, the area of the square = 6*6 (length and width) and the area of the square can be seen as the sum of the areas of 2 right triangles.

Whereas, within the square two diagonals are perpendicular to each other and equally divided.

So the square area = 2 the area of the right triangle = 2 1 2 the bottom edge (hypotenuse) is high.

High = hypotenuse. So hypotenuse 2=36, so the hypotenuse length is 6 times the root number 2. That is, the radius of the circle where the fan is located is 6 times the root number 2

Sector area = circle area 1 8 = 2 6 times root number 2 8 = 3 root number 2 2

Solution: 6x6 2=18 square centimeters is the area of the triangle.

Let the hypotenuse of the triangle be a, then the height is 1 2a, a is also the radius of the circle.

The square of ax1 2a 2=18 ax1 2a = 36 a = the square of 72 a circle formula: the square of the radius radius = the area of the circle.

The square of the circle = the area of the circle in square centimeters.

The sector angle is 45, then it is 1 8 parts of the circle.

Cm. The sector area is a square centimeter.

Solution: Make up the other half of the symmetry with this figure, , with a radius of 6cm 1 4 circles, and its area is 1 4 36 = 9 , so the fan-shaped area is 9 2 =.

aod eob, then ao:oe=ad:be=2:1, and the ratio of the height of aod to eob is 2:1 (the height on the edge of ad and be).

S oed:S oad=oe:oa=1:2, and S oad=ad*h(ad) 2,h(ad):ab=2:3

then S oed=(1 2)*(1 2)*(2 3)*ad*ab=24cm

Connecting bc, ac and be are parallel, so s aec = s abc = 1 2 * 4 * 8 = 16 (same base and equal height).

Unless there are other conditions, the SAEF cannot be found

If the quadrilateral ABCD is rectangular, then the diagram shows that the triangle AFC is similar to the triangle BFE, then there is AC BE=af BF to push out AF BE=32-4AF....It can also be seen from the diagram that the triangle feb is similar to the triangle ced, then there is bf dc=be de to push out 4bf = be af....From the above, 32-4af=4bf can be solved; There is no single answer to this question.

If af=bf, af=bf=4 can be obtained if af=bf=4, then the area of the triangle aef is 8;

If af≠bf there are several answers.

With the special value method, assuming that F is the midpoint of AB, then af=4, bf=4, be=4, then the area of the triangle AEF is 8

The answer is different depending on the position.

You have a problem with this question!

If the side is known to be the waist of a triangle, then the base of the triangle = 20-2 8=4 (cm).

If the side is known to be the base of the triangle, then the waist of the triangle = (20-8) 2=6 (cm).

That is, the three sides of an isosceles triangle are , or . So the circumference of an isosceles triangle is 20 cm, and if one side is 8 cm long, then the other side is (4) or (6) long

Hello, the area of the shaded part is equal to, the area of the triangle minus the area of a circle and then the area of a small corner on the left, the area of this corner is the area of the square minus the area of a circle, and then divide by 4Then the area of this little corner is 1 3That is, (10*10-25) 4*1 3=

So shaded area = 1 2 * 10 * 20- (10 2).

The area of the triangle minus the area of a circle!

When ag is joined, both the rectangle and the square are the same height as the triangle ADG, so that both the rectangle and the square are twice the area of the triangle ADG.

So, de 4 4 5

There seems to be a problem with this figure, if the two sides of the semicircle, there should be only one intersection point on the hypotenuse, in that case, the area of the semicircle is 36, the area of the triangle is 36, the difference between the two is divided by 2 and multiplied by 3 is the shadow area, that is, 54 (-1).

1. The shadow area is the area of the semicircle minus the empty area below.

2. The semicircle area is simple, and the formula is OK.

3. The blank area below is the area of the triangle, minus the fan area of the upper method of 45 degrees, and the radius is 12. This is OK with a formula.

There's something wrong with this shadow, where did that intersection under the hypotenuse come about.

The height of the drink in the cylindrical shape is 20 cm when it is being put upside down, and the height of the remaining part of the drink in the cylindrical shape is 5 cm when it is reversed. Then the volume ratio of air to drink is 1 to 4, so the beverage accounts for 4 (1+4) of the volume of the bottle, that is, 4 5

Beverages are available in the bottle.

30 4 5 = 24 liters 24,000 cubic centimeters.

So the drink in the bottle has a chain of 24,000 cubic centimeters.