Question

z.5 str.206

Przyjmując oznaczenia jak na rysunku, oblicz:

a)|AB|, jeśli |BC| = 7 cm, |AC| = 24 cm.

b) |AC|, jeśli |BC| = 13 i jedna trzecia cm, |AB| = 13 i dwie trzecie cm.

c) |BC|, jeśli |AC| = pierwiastek z 2 cm, |AB|= pierwiastek z 3 cm.

Answer 1

a)|AB|=c

|BC|=b=7cm

|AC|=a=24cm

 

a²+b²=c²

c²=24²+7²

c²=576+49

c²=625/√

c=25 cm

 

b)|AC|=a

|BC|=b=13⅓cm

|AB|=c=13⅔cm

 

a²=c²-b²

a²=(13⅔)²-(13⅓)²

a²=(⁴¹/₃)²-(⁴⁰/₃)²

a²=¹⁶⁸¹/₉-¹⁶⁰⁰/₉

a²=⁸¹/₉

a²=9 /√

a=3

 

c)

|BC|=b

|AC|=a=√2

|AB|=c=√3

 

 

b²=c²-a²

b²=(√3)²-(√2)²

b²=3-2

b²=1 /√

b=1cm

 

Answer 2

^2- do kwadratu

a) |BC||^2 + |AC|^2 =|AB|^2

    7^2+24^2=|AB|^2

    49+576= |AB|^2

   625= |AB|^2

   25 = |AB|

b) |BC||^2 + |AC|^2 =|AB|^2

    ( 13 i 1/3)^2 + |AC|^2 = (13 i 2/3)^2

    (40/3)^2 + |AC|^2 = (41/3)^2

    1600/9 + |AC|^2 = 1681/9

    |AC|^2 =1681/9 - 1600/9

    |AC|^2 = 81/9

     |AC|= 9/3 = 3

c) |BC||^2 + |AC|^2 =|AB|^2

     |BC||^2 + (pier z 2)^2 =(pier z 3)^2

     |BC||^2  +2 = 3

     |BC||^2  =1

     |BC|= 1