A math Olympiad problem about area .

The quadrilateral AEFD area is 11. See the picture for a detailed explanation.

The most basic similar triangle may be a bit difficult to learn at first, but it is not at the level of the Olympiad.

Let's not talk about the analysis process, it's cumbersome, just look at the answer.

The key to finding gf:fe is obviously cg:ce=dg:

ae=dg:be=3:5=gf:

fe then ef:fc=5 ((5+3)*3 2+3)=1:3 is a simple equiquantitative substitution and parallel line property in the middle.

sade=2 5 *saec=1 5*sabcsdef=1 4*scde=3 40sabc. saefd=11

I don't know if I'm wrong, but check it out for myself, and that's how it works.

Solution: By the condition, AEF area = BFE area (equal height at the base) is the same reason, AEC area = BCE area = 20, because the two triangles are of equal height, and the bottom edge is equal in proportion.

So: ADF area = 2 3cdf area.

In the same way, ADB area = 16, CDB area = 24

cdf-efb = ace-adb = 20-16 = 4 and because aef = efb

So cdf=AEF+4 equation (1).

And because ADF area = 2 3cdf area Equation (2) AEC = CDF + ADF + AEF = 20 Equation (3) is joined by equations (1) (2) (3), and three of the triangle areas can be solved.

aefd=adf+aef

The problem is solved.

Add 2 cm and 5 cm respectively to the two adjacent sides of the resulting square. The increased area is two rectangles with the length of the sides of the square and the width of . The sum of the areas is the length of the sides of the square multiplied by (5+2).

At this time, a small rectangle 5 cm long and 2 cm wide was missing from the original rectangle.

Thus, the reduction of 31 square centimeters includes a square side length of 7 and a rectangle of 10 square centimeters, so the side length is 7 = 31 - 10 = 21 and the square side length is 3 centimeters.

Originally, the rectangle was 8 centimeters long and 5 centimeters wide, so the area was 40 square centimeters.

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[Part 1].

Analysis: From the diagram, it can be seen that the shaded part is a triangle, and its base and height are equal to the side length of the small square, so the area of the triangle s=(1 2)ah, substitute the data to solve the solution: solution: (1 2) 3 3=;

A: The area of the shaded part is.

Therefore, the answer is: Comment: The key to answering this question is to understand that the shaded part is a triangle, and its base and height are equal to the side length of the small square, so it can be solved by using the area formula of the triangle

[Part 2].Analysis: If the width of the chicken farm is x meters, the length is (60-2x) meters, and then the area can be found by enumerating the area formula of the rectangle s=ab

Answer: Solution: If the width of the chicken farm is x meters, the length is (60-2x) meters, according to the topic.

When the width is 1 meter, the length is 58 meters, the area is 58 1 = 58 (square meters), when the width is 2 meters, the length is 56 meters, the area is 56 2 = 112 (square meters), when the width is 3 meters, the length is 54 meters, the area is 54 3 = 162 (square meters), when the width is 4 meters, the length is 52 meters, the area is 52 4 = 208 (square meters), when the width is 5 meters, the length is 50 meters, the area is 50 5 = 250 (square meters), when the width is 6 meters, the length is 48 meters, and the area is 48 6 =288 (square meters), when the width is 7 meters, the length is 46 meters, the area is 46 7 = 322 (square meters), when the width is 8 meters, the length is 44 meters, the area is 44 8 = 352 (square meters), when the width is 9 meters, the length is 42 meters, the area is 42 9 = 378 (square meters), when the width is 10 meters, the length is 40 meters, the area is 40 10 = 400 (square meters), when the width is 11 meters, the length is 38 meters, the area is 38 11 = 418 (square meters), the width is 12 meters, the length is 36 meters, The area is 36 12 = 432 (square meters), when the width is 13 meters, the length is 34 meters, the area is 34 13 = 442 (square meters), when the width is 14 meters, the length is 32 meters, the area is 32 14 = 448 (square meters), when the width is 15 meters, the length is 30 meters, the area is 30 15 = 450 (square meters), when the width is 16 meters, the length is 28 meters, and the area is 28 16 = 448 (square meters), so it can be seen that when the width is 15 meters, the length is 30 meters, and the area is 30 15 =450 (square meters), A: The area of this chicken farm is 450 square meters

So the answer is: 450 square meters

Comments: According to the area formula of the rectangle, the enumeration method is used to derive how to enclose the area

[Part 3].

Let the area of the semicircle with BC as the diameter be .

s1, the area of the circle with ab as the radius is s2

So: s shadow = s1

s△abcs2） πbc／2）²

ab ) Because the triangle abc is an isosceles right triangle and has an area of 12cm, ab=bc=root number 12

Reduced to s shadow.

5πab²／4-12

5 root number 12 4-12

The area of the shaded part is 12cm larger than the area of the triangle adh, then the area of the parallelogram is 12cm larger than the area of the triangle abc, so the area of the parallelogram is: 8 7 2 + 12 = 40 square centimeters.

So hc = 40 8 = 5 cm.

Answer: Let ch=h parallelogram, right-angled trapezoidal height, dh=a, shadow area=8h a 8 h, adh area= a 7 h, 8h a 8 h a 7 h =12, merge and simplify to:

h=31／8，∴hc=31／8㎝

Move CE to BF and set ah=7x, so ADN area = 28x and shadow area is 1 2(8-8x)(7-7x)

Let ch=xdh=((7-x)*8) 7

s( adh) = (7-x)7) squared * 28

s(trapezoidal) = 28 - ((7-x) 7) squared * 28s (shaded part) = 8x-28 + ((7-x) 7) square * 28s (shaded part) - s (adh) = 12

The solution is x=5, and the length and area are found using the proportional relationship in similar triangles.

Let the HC length be X, the quadrilateral BCHD area is S1, the triangle ADH area is S2, and the shaded area is S3. You can list the following formula: s1+s2=1 2*8*7

s3-s2=12

s1+s3=8x

The comprehensive three-formula can be solved to x=5cm

ch=5 process:

The area of the shaded part is 12cm larger than the area of the triangle ADH, the area of the triangle ABC is 12cm smaller than the area of the quadrilateral BCEF, and the area of the triangle is (8 7) 2=28cm

The area of the quadrilateral is 28 12 = 40cm

Quadrilateral area = bc ch, bc = 8cm

ch=40÷8=5cm

ch=5cm

Again because 2r 2=1

The area sought = r 2 4 = 8 square meters.

The area swept by the AB edge when rotated is square meters.

Triangle hypotenuse = root number 2=

Each side sweeps the area of a circle of radius - the area of two triangles.

Square metre.

It's his own area

(6 2 6 4 2 4) 2 = 88 (square centimeters).

2 4 4 = 32 (square centimeters).

88 32 = 120 (square centimeters).

The shape is a square with 5 sides and a semicircle on the top and left, right?

To find the area of the shaded part, we draw two diagonal lines and divide the original square into 4 pieces, so that we can see that the shadow area is actually half the area of the square.

Square area = 5 x 5 = 25

Shaded area = 25 2 =

Split, move, or rotate to get the shaded part exactly halfway through the square. So yes:

Separate the shadow in the upper left corner and fill it on top of another shape, just half a square.

s=5×5÷2=