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Explanation of math problems in the second day of junior high school
1, the intersection a is AE⊥BC, and the intersection d is DF⊥BC. Because ∠ b = 45, then △ABE is an isosceles right triangle. From AB=8, BE=AE=4, the radical 2 is obtained.

Because the trapezoid ABCD is an isosceles trapezoid, CD=AB=8, CF=DF=AE=4, and the root number is 2.

So perimeter = ad+ab+BC+CD = ad+2ab+(be+ef+cf) = 4+2 * 8+2 * 4 root number 2+4=24+8 root number 2.

Area =(AD+BC)*AE/2=(4+4+8 roots 2)*4 roots 2/2=32+ 16 roots 2.

2. The cut area is 6 square centimeters, which is the area of two triangles, so the area of a triangle is 3 square centimeters = 1/2*3*h, and h=2, so the hypotenuse of this triangle is the root number (9+4)= 13, which is the waist length of an isosceles trapezoid.

When H is longer than the original rectangle, the perimeter of the trapezoid after opening is: 4+8+2* root number 13= 12+2 root number 13.

When h is the width of the original rectangle, the perimeter of the trapezoid after opening is: 2+4+2 root 13=6+2 root 13.