The area of a trapezoid is 56 square centimeters. The sum of its two bases is 16 centimeters

The area of a trapezoid is 56 square centimeters. The sum of its two bases is 16 centimeters

56×2÷16＝7

The top of the trapezoid is 6 cm, the bottom is 10 cm, the trapezoid area is 56 square cm, how many square cm is the shadow area?

The top of the trapezoid is 6 cm, the bottom is 10 cm, and the trapezoid area is 56 square cm
According to the trapezoidal formula, the height is 7 cm

1.56 square meters = () square meters () square decimeters

1 56

(x-3) / 2 - (4x-9) / 5 = 6

(x-3)/2-(4x-9)/5=6
Multiply both sides of the equation by 10
5（x-3)-2(4x-9)=60
5x-15-8x+18=60
-3x=60+15-18
-3x=57
x=-19

(- 81) △ 2 and a quarter * (- four ninths) △ 16

(- 81) △ 2 and a quarter * (- four ninths) △ 16
=(-81)*(4/9)*(-4/9)*(1/16)
=1

2. Remove the brackets, 3. Move the items, 4. Merge the similar items, 4. Change the coefficient to 1, 5 (x + 8) - 5 = 0, 2 2 - 3 (x + 3) = 24

①5(x+8)-5=0
Get rid of the brackets
5x+40-5=0
Transfer of items
5x=5-40
Merge similar items
5x=-35
The coefficient is reduced to 1
x=-7
②-3（x+3）=24
Get rid of the brackets
-3x-9=24
Transfer of items
-3x=24+9
Merge similar items
-3x=33
The coefficient is reduced to 1
x=-11

1 / 2 1 / 4 1 / 8 1 / 16 1 / 64 1 / 128 1 / 250 1 / 512 how to operate easily

It is wrong to write 1 / 2 + 1 / 4 + 1 / 8 + 1 / 16 + 1 / 32 + 1 / 64 + 1 / 128 + 1 / 256 + 1 / 512 so that a = 1 / 2 + 1 / 4 + 1 / 8 + 1 / 16 + 1 / 32 + 1 / 64 + 1 / 128 + 1 / 256 + 1 / 5122a = 1 + 1 / 2 + 1 / 4 + 1 / 8 + 1 / 16 + 1 / 32 + 1 / 64 + 1 / 128 + 1 / 256 minus the same offset 2a-a = 1-1 / 512, so 1 / 2 + 1 / 4 + 1 / 8 + 1 / 16 + 1 / 32 + 1

AB is the diameter of circle O, the extension lines of chord BD and Ca intersect at point E, and the extension line of EF perpendicular to Ba intersects at point F. it is proved that ab · AF = AE point AC
Point means multiply

It seems that the intersecting string theorem can't be used in big problems?!
Proof: connect BC
∵ AB is the diameter of ⊙ o
∴∠ACB=90°
∵EF⊥BF
∴∠F=90°
∴∠ACB=∠F=90°
And ∠ EAF = ∠ BAC
∴△FEA∽△CBA
∴AB/AE=AC/AF
∴AB·AF=AE·AC

Simple algorithm of 28 × 236-28 × 136
Calculation by the formula of "want to take off"

28*（236-136）=2800

In the triangle ABC, angle c = 90, bisectors of angles a and B intersect at P, and PE is perpendicular to AB and E. if BC = 2 and AC = 3, what is AE multiplied by EB

Note that P is the inner part of the triangle as PD, PF is perpendicular to BC, AC is perpendicular to D, F, then PD = PE = PF (inner property). In right triangle, PD = CF = CD = 0.5 (CF + CD) = 0.5 (BC + ca-af-db) note that AF = AE, BD = be (can be proved by using congruence and tangent), so PD = 0.5 (BC + ca-ae-eb) = 0.5 (BC + ca-ab)