Finding an isosceles trapezoidal area, elementary school math Olympiad problem

See Figure: <>

Let the length of the upper and bottom edge be A, and the height should be H1

According to the characteristics of the isosceles trapezoid, it can be obtained.

abe∽△cde

In ABE, the area S1 is 15

s1=a·h1=15

ab cd = 3 5

It is obtained that in the CDE, the lower bottom edge cd=5a3, the height is 5h1 3, and finally the isosceles trapezoidal area formula:

s=(ab+cd)·h/2

Substituting ab=a, cd=5a3, h=8h1 3s=64a·h1 9=64 15 9

The triangle with a known area is similar to the triangle at the top, so the ratio of the two bases of the two triangles is equal to the ratio of the high. Okay, you draw my following description on paper, the top bottom is A, the bottom bottom is B, the height of the triangle corresponding to the top bottom is H, the bottom bottom corresponds to the height of the triangle is M, and the total height of the trapezoid is H, then there is the following relationship:

b = a5 3, m = h5 3, h + m = h, the trapezoidal area is s = (a + b) * h 2, start substitution, a + b = a + a5 3, h = h + h5 3, substitute the area formula, s = (a + a5 3) * (h + h5 3) 2 = ah32 9

According to "the area of the triangle formed by the intersection of the diagonal line as the vertex and the upper base is 15", that is, ab 2 = 15

AB=30, substitute the above formula, s=30*32 9 haha done!! This is definitely the standard answer! I'm not embarrassed to teach you!

Let the upper bottom be a, the lower bottom be 5 3a, and the height is h, then.

Known triangle area: a*(h2) 2=15

Get; ah=15*2*2=60

Then the trapezoidal area: (a+5 3a)*h 2

8/3a*h/2

Answer: 302 3

Solution: If the length of the bottom is x, then the bottom is 3x 5;Then let the height of the triangle opposite by the bottom be y, because the triangle corresponding to the bottom is similar to the triangle corresponding to the upper bottom, so the height of the triangle corresponding to the top bottom is 3y 5.

From the meaning of the title, it can be obtained that (3x 5*3y 5) 2=15, so x*y=250 3 The area of the trapezoid is: s = (upper bottom + lower bottom) * high 2

So: s=(x+3x 5)*(y+3y 5) 232x*y 25

Hello! It can be calculated that the area of the trapezoid is 40Thank you!

If you use the similarity of triangles, it is easy to calculate the area of the isosceles trapezoid, which is 320 3, but I wonder if I learned the similarity of triangles in elementary school?!I remember when I was in junior high school.

106 and 2 3

It is mainly solved by using several properties that are similar to triangles.

The area of the four parts is 15,25,25,125 3

I'm only in sixth grade, and it's too hard.

The figure on the right is made up of four identical isosceles trapezoids, and it is known that the upper bottom of the trapezoidal is 8 decimeters and the lower bottom is 14 decimeters, and the area of one of the trapezoids is found.

There is a square in the middle, and there are 4 trapezoids next to the 4 sides of the small square (the bottom), and they combine to form a large square. Combined area of the large square: 142=196m2 The area of the small square in the middle is the area of the square in the middle:

82 = 64 m24 trapezoidal total area: 142- 82 = 196-64 = 132m2a isosceles trapezoidal area 132 + 4 = 33m2 The height of the isosceles trapezoid is 33x2 + (8 + 14) = 3m trapezoidal carrying area formula: s = (upper bottom + lower bottom) * height 2 substitution can obtain the height of the isosceles trapezoid.

Through the two ends of the upper bottom to make the perpendicular line of the lower bottom swift imitation, the congruent three hidden angles can be obtained.

And the right-angled side of one acre of fiber strip is 5-3 2=1cm

So the height is 1cm

So s=(3+5)*1 2=4cm2

308 (square centimeters).

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The circumference of each isosceles trapezoid is 18cm, i.e., a+b+2c=18.........

Assembled into a flat quadrilateral, its circumference is 30cm

That is, 2(a+b+c)=30

a+b+c=15………

Get; c=3 waist is 3 cm.

6cm^2

Take the bottom edge of the midpoint e, set the upper left corner point as a, the upper right corner point as b, the lower right corner point as c, and the lower left corner point as d

Even AE, then the triangle Ade, AEC, and ABC have equal areas, and the area of the blank space is S,2(S+8)=20+8S=6

If 3 is the waist, then the waist length and the trapezoidal height and the lower bottom part of the part of the right triangle, the right angle side hypotenuse, does not conform to the reality, so give up;If 4 is the waist, then the waist length and the trapezoidal height and part of the lower bottom part of the right-angled triangle, the right-angled side = hypotenuse, does not conform to the reality, so it is discarded;If 11 is the waist, in line with reality, then the circumference = 11 2 + 3 + 4 = 29;So the circumference of this isosceles trapezoidal is: 29

The area of the trapezoid = (upper bottom + lower bottom) height 2;

This should be a square (as mentioned in the title), then the trapezoidal height is 6dm, and the area is (8+14)x6 2=66dm

Trapezoidal height:

14-8) Divide by 2

3 trapezoidal area:

8+14) multiplied by 3 divided by 2=33