Current location - Training Enrollment Network - Mathematics courses - Multiple solutions to one problem in junior high school mathematics
Multiple solutions to one problem in junior high school mathematics
There are several geometric problems with multiple solutions.

In the triangle ABC, AB = AC, the middle line BD of AC, divide the triangle ABC into two parts with perimeters of 12cm and 15cm respectively, and find the respective lengths of the triangle ABC.

BD is the AC center line.

∴AD=AC

Let AD be X, then DC is X and AB is 2x.

Such as AB+AD= 12cm, BC+CD= 15cm.

2x+x= 12

x+y= 15

x=4 y= 1 1

∴AB=AC=8,BC= 1 1

Such as AB+AD= 15cm, BC+CD= 12cm.

2x+x= 15

x+y= 12

x=5,y=7

∴AB=AC= 10,BC=7

There are two answers, and the process is clear. BD is not included in O (∩ _ ∩) O.