A junior high school math problem to find the circumference of a right angled trapezoid

Choose B for two cases: case 1, right-angled side = 10, case 2, hypotenuse = 10 When solving the problem, for convenience, divide the trapezoid into a rectangle and a triangle, and one angle of the triangle is 45°, so the two right-angled sides are equal, so the rectangle is a square.

Case 1: When the waist length of the right-angled side is 10:

Top bottom + bottom bottom = 30

Right angle waist length = 10, hypotenuse waist length = 10 2

So, perimeter = 10 (4 + 2).

Case 2: When hypotenuse waist length = 10.

In the same way, circumference = 10 (2 2 + 1).

So choose B

Go over the vertex of the diagonal waist and make a perpendicular line at the bottom to get a right triangle and a rectangle.

The length of the upper bottom is 5 2, the lower bottom is 10 2, and the height is 5 2, so the circumference is 10 (2 2 + 1) cm

Option D, is it a mistake?

Choose B, do the height of the trapezoid, because the bottom angle is 45, so the triangle is an equiangular right triangle.

By waist is 10 cm, the side of the triangle is 5 2. And the side of the triangle plus the bottom is equal to the bottom bottom.

The lower bottom is 10 2, and the upper bottom and high are both 5 2. The waist is the sum of 10 to get the result

There is a trapezoidal shape with an upper base of 5 cm, a height of 5 cm, and a bottom that is twice as large as the upper bottom. Q: How many centimeters is its circumference?

Split into a rectangle and triangle.

You can draw your own picture to celebrate hail.

First of all, make an auxiliary line, and pass a as an ae reputation chain sail bc. BC=AE can be obtained, because in the right-angled triangle AED AD is hypotenuse equal to 20, and AE is right-angled side equal to 10, we can get the call banquet angle D equal to 30 degrees.

And because the angle b angle c is equal to 90 degrees, it is ab cd. So the angle a should be complementary to the angle d, so the angle a is equal to 150 degrees.

Obviously, according to the meaning of the quarrel, let the diagonal line intersect the point o, then the silver can set ao=do=x, bo=co=y

So, ad = root number 2 * x and bc = root number 2 * y

Therefore, the root number 2*y-root number 2*x=4*cos60°=2, that is, y-x=root number 2 is then obtained by the equal area method, (x+y)*y=y*root number 3, that is, y+x=root number 3 is obtained by the above two formulas, y=(root number 3 + root number 2) 2x=(root number 3 - root number 2) 2

If a straight line perpendicular to the bottom is made from the two points of the upper bottom excavation group, the triangles on both sides are imitation Ming isosceles right triangles (45 degree angles).

Then the length of the bottom is 8 + 4 + 4 = 16, then the perimeter is.

Solution: Because the circumference of a trapezoid is 5a (1 3 2)a, the perimeter of two trapezoids is 8a (2 3 2)a

The perimeter of the three trapezoids is 11a (3 3 2)a

The circumference of the four trapezoids is 14a (4 3 leaky skin 2)a, so the circumference of n trapezoids is (n 3 2)a

1.Let the quadrilateral pqcd be a parallelogram after X seconds.

Because. When the quadrilateral ridge pqcd is a parallelogram, pd=cqso. 24-x=3x

The ruler field is matched.

x=6 so Ling finger.

After 6 seconds, the quadrilateral PQCD is a parallelogram.

2.Let the quadrilateral PQCD be an isosceles trapezoidal after Y seconds.

x=26-3x

x = after the elapsed seconds quadrilateral pqcd is an isosceles trapezoid.

Just the second question wrote the wrong mountain trail, it should be the state circle:

2y-26）^2+8^2=8^2+2^2

So Funny Bi y = 7

After 7 seconds, the quadrilateral PQCD is isosceles trapezoidal.

Make point d about the symmetry point m of ab, connect cm, then cm is the minimum value of pc+pd, the triangles are similar, find pb=3

Solution: Let a waist length be xcm. The other should be 1 3xcm(x+1 3x) 3=48

The solution is x=12

then the other waist is = 4 (the short waist is the right-angled waist).

The sum of the upper and lower bases is = (12 + 4) 2 = 32

s trapezoidal area = 32 4 2 = 64

I don't understand, please ask, I wish you a pleasant visit.

The circumference of a right-angled trapezoid is 48 cm, and the ratio of the sum of the upper and lower bottoms to the sum of the two waists is 2:1, so the sum of the two bottoms and the sum of the two waists are 32 cm and 16 cm respectively, and the length of one of the waists is 1 3 of the other waist length, then the two waists are 4 cm and 12 cm respectively.

That is, the height of the trapezoid is 4 centimeters, and the area of this right-angled trapezoid is 32 4 2 = 64 square centimeters.

The parallel line of AB at the point D intersects with BC at the point E, because AD is parallel to BC, AB=CD=4, so CE=4, there is because, AD is parallel to BC, AB is parallel to DE, so AD=BE=6-4=2, and the perimeter just adds those sides.

Extend DA to D'so that da=d'a, as d'E CB extension cable at point E, connected to D'c, pass ab to point p

d'a=da=4

eb=d'a=4

ec=eb+bc=4+6=10

d'e=ab=5

Then use the Pythagorean theorem to calculate the length of dc, the length of dc is the minimum value, and use cde cpb to calculate pb=3 (sorry the word is not enough).