| 3 | ||
kąta ostrego w trapezie ABCD jest równy | . oblicz długości jego podstaw | |
| 4 |
a + b + 20 = 40
| x | |
= ctg45o | |
| 10 |
Obw.=40
40=20+a+b
a+b=20
| 1 | 1 | |||
x= | a− | b | ||
| 2 | 2 |
| h | ||
tgα= | ||
| x |
| 3 | h | ||
= | |||
| 4 | x |
| 4 | ||
x= | h | |
| 3 |
| 4 | ||
h2=100−( | h)2 | |
| 3 |
| 16 | ||
h2=100− | h2 | |
| 9 |
| 16 | ||
h2+ | h2=100 | |
| 9 |
| 4 | 4 | |||
x= | h= | *6=8 | ||
| 3 | 3 |
| 1 | 1 | |||
x= | a− | b | ||
| 2 | 2 |
| 1 | 1 | |||
8= | a− | b | ||
| 2 | 2 |
| ⎧ | a+b=20 | |
| ⎩ | a−b=16 |
| ⎧ | a=18 | |
| ⎩ | b=2 |
| 3 | 3k | |||
Ob=40 ⇒ a+b=20 i tgα= | = | , k>0 | ||
| 4 | 4k |