Question

DETERMINE A ÁREA DO:

A) QUADRADO INSCRITO EM UM CIRCULO DE R= 5M.
B) HEXÁGONO INSCRITO EM UM CÍRCULO DE R= 6M.
C) TRIANGULO EQUILÁTERO CIRCUNSCRITO A UM CIRCULO DE R = 5M.

HELP ME PLEASE!!!

Answer 1

Resolvendo determinar as areas :

A) Solucao
......~~~~~~~~
diametro da circunferencia
#####################
D = 2 .x R

D = 2 x 5 m

D = 10m

diagonal do quadrado
#################
D = L x \/2

10 = L x \/2

L x \/2 = 10

L = 10
.......__
.......\/2

L = 10 x \/2
.......___....___
........\/2.......\/2

L = 10.\/2
......_____
...........2

L = 5.\/2 m

Area do quadrado
##############
A = L^2

A = (5.\/2)^2

A = (25 x 2)

A = 50m^2


B) Solucao :
.....~~~~~~~~~
1er paso calculandocom a formula da altura para encontrar a aresta :

h = 4m

h = a.\/3.
......____
..........2

4 = a.\/3
......____
..........2

4 x 2 = a.\/3

8 = a.\/3

a.\/3 = 8

a = 8
.......__
.......\/3

a = 8 x \/3
.......__......____
.......\/3........\/3

a = 8.\/3
.......____
............3


2do paso vamos calcular a area

S = 6 x \/3 x a^2
.......__________
..................4

S = 6 x \/3 x (8.\/3)^2
......._____________
....................4 x 3

S = 6 x \/3 x ( 64 x 3 )
.......______________
......................12

S = 6 x \/3 x 192
........_________
................12

S = 1.152 x \/3
.......________
................12

S = 96.\/3m^2


C) Solucao :
.....~~~~~~~~
Lado do triangulo equilatero :
######################

L = 2 x R x \/3

L = 2 x 5 x \/3

L = 10.\/3 m

Altura do triangulo equilatero
######################

(10.\/3)^2 = h^2 + (5\/3)^2

100 x 3 = h^2 + 25 x 3

300 = h^2 + 75

h^2 + 75 = 300

h^2 = 300 - 75

h^2 = 225

h = \/250

h = 15m


Area do triangulo equilatero
######################

A = b x h /2

A = 10.\/3 x 15 / 2

A = 150.\/3 / 2

A = 75.\/3m^2