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Dot R version of the eighth grade math answer.
①∵CD⊥AB

△ CDB and△ ACD are right triangles.

∴CD? =BC? -DB? =AC? -Advertising? (Pythagorean theorem)

And ∵AD=AB-DB AB=5, BC=4, AC=6.

∴4? -DB? =6? -(AB-DB)?

4? -DB? =6? -(5 decibels)?

16-DB? =36-(5? - 10DB+DB? )

16-DB? =36-25+ 10DB-DB?

16-DB? = 1 1+ 10DB-DB?

Add DB on both sides?

16 = 1 1+ 10DB

5= 10DB

DB= 1/2=0.5

∫CE is the center line of AB side.

∴E is the midpoint of AB

∴AE=EB

AB = 5

∴AB=2AE=2EB

∴5=2AE=2EB

AE=EB=5/2

∫EB = 5/2 DB = 1/2 = 0.5。

∴EB-DB=DE

∴DE=5/2- 1/2=4/2=2

②∵DE⊥AB in E,DF⊥AC in F.

BD? +CD?

= (Yes? +DE? )+(DF? +CF? )

∫∠BAC = 90 DE ⊥ AB in E,DF⊥AC in F.

∴ af ‖ deaedfadf is a parallelogram.

∵∠ BAC = 90 AEDF is a rectangle.

De? +DF? =EF? =AD? (Diagonal lines of rectangles are equal)

AB = AC,∠BAC=90

∴∠B=∠C=45

∵DE⊥AB is in E,DF⊥AC is in F.

∴∠EDB=∠FDC=45

∴△EBD,△FCD is an isosceles right triangle

∴BE=ED DF=CF

∵EB? +CF? =DE? +DF? =EF? =AD?

=2DE? +2DF?

=2AD?

③BC⊥AD

△ ACD and△ ACD are right triangles.

Ba? -DB? =AC? -CD? =AD?

∫BC = 14,AC= 15,AB= 13

Ba? -BD? =AC? -CD?

∴ 13? -BD? = 15? -CD?

∫BC = BD+CD

BD=BC-CD

BD= 14-CD

∴ 13? -( 14-CD)? = 15? -CD?

CD=9

△ ACD is a right triangle AC? -CD? =AD?

15? -9? =AD?

AD= 12 or AD=- 12 (omitted)

Give it to me, thank you!