+0

0
27
9

I have been trying this problem for the past 2 days. What I did so far was calcualte that 27/x=1/4. But when I simplfiy and put in the answer it is wrong.

Each triangle shown is equilateral. What is the probability that a point chosen at random within the largest triangle will lie within the unshaded area? Express the result as a common fraction.

Jun 6, 2023

#1
0

The probability is 3/8.

Jun 6, 2023
#2
0

That's wrong, I would also like an explantion for how to get the answer so I could improve.

Jun 6, 2023
#3
0

If I had a printer, what I would do is print out that figure. Then, using a straight edge, I'd fill in all the diagonal lines and all the horizontal lines.  That would divide up the figure into a whole bunch of little triangles, all the same size.  Then count the number of unshaded triangles and divide that by the total of all the unshaded triangle plus all the shaded triangles.

.

Jun 6, 2023
#4
0

Shouldn't you:

1 - count the number of white triangles =27

2 - count the number of shaded triangles =19

3 - Total number of triangles: 27 + 19 =46

4 - Probability of random point landing in the unshaded area: 27 / 46

Jun 6, 2023
#5
0

That's wrong.

Guest Jun 7, 2023
#6
0

I couldn't print out the figure and actually draw the lines,

to divide the bigger shaded triangles into little triangles,

but I think I got an accurate count by imagining the lines.

How many little triangles would be contained

in the three medium sized shaded triangles         12

How many little triangles would be contained

in the single large shaded triangle                      15

Total equivalency of little shaded triangles          36

Total white triangles                                             27

Total of shaded and white triangles                      63

27                          3

Probability of landing in a unshaded area          ——   reduces to   ——

63                           7

.

Jun 7, 2023
edited by Guest  Jun 7, 2023
edited by Guest  Jun 7, 2023
#7
0

Sorry, that's wrong.

Guest Jun 7, 2023
#8
0

I must have miscounted the possible little triangles in the big shaded areas.

My original suggestion was to print out the figure, and actually draw the lines.

That will give you an accurate count, and then apply the method that I used.

.

Guest Jun 7, 2023