1. oblicz pole trojkata rownobocznego, ktorego wysokosc jest rowna 3\sqrt3[/tex]3
2. dany jest trojkat o bokach dlugosc 2,3 i 4. Oblicz pole tego trojkata i jego wyskosci.
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z.1
h = 3 p(3)
zatem
a p(3)/2 = 3 p(3)
a/2 = 3
a = 6
====
Pole
P = (1/2) a*h = (1/2) *6 * 3 p(3) = 9 p(3)
=====================================
z.2
h - wysokośc tego trójkąta
Mamy
h^2 + x^2 = 2^2
h^2 + ( 4 - x)^2 = 3^2
----------------------------
h^2 + x^2 = 4
h^2 + 16 - 8 x + x^2 = 9
-------------------------------- odejmujemy stronami
( h^2 + 16 - 8 x + x^2) - ( h^2 + x^2) = 9 - 4
16 - 8 x = 5
8 x = 11
x = 11/8
========
h^2 = 4 - x^2 = 4 - ( 11/8)^2 = 4 - 121/64 = 256 /64 - 121/64 = 135/64
h = p( 135/64 ) = ( 3/8) p(15)
-------------------------------------
Pole
P = 0,5 a*h = 0,5 *4*(3/8) p(15) = ( 3/4) p(15)
=======================================
p(15) - pierwiastek kwadratowy z 15