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Eight mathematical problems such as judging triangle shape
Do CD⊥AB after point C, and pass AB at point D.

√C( 1/2,√3/2)

∴ The coordinate of point D is (1/2,0).

∫A(- 1,0) D( 1/2,0) C( 1/2,√3/2)

∴ad=|- 1|+ 1/2=3/2 CD =√3/2-0 =√3/2

In the right triangle ACD,

AC? =AD? +CD? =(3/2)? +(√3/2)? =3

In the same way.

BC? = 1

∫A(- 1,0) B( 1,0)

∴AB=|- 1|+ 1=2

∴AB? =4

In triangle ABC

AB? =4=3+ 1=AC? +BC?

∴ Triangle ABC Right triangle ABC

If you don't understand, please ask,

Please adopt it if it is useful! !

Some parts you think can be omitted or reduced, no problem, mine can also be used as a reference for thinking.

You can also directly calculate the square of the side length according to the square of the coordinate difference, for example:

A(- 1,0) C( 1/2,√3/2)

AC? =(- 1- 1/2)? +(0-√3/2)? =9/4+3/4=3

This is simpler. I don't know if you have learned it. If not, use the first method!