Sum of line of segments in a triangle

[Funkj]

Pls help me with this question Triangle is divided to four triangles and three quadrilaterals,total perimeter for the four triangles is 20 and 25 for the three quadrilateral, total perimeter for whole triangle is 19, what is the sum of three straight line segment

 

[Dr.Peterson]

Nice problem!

What have you tried?

I'd start by observing which of the 15 disjoint segments in the figure are included in each mentioned sum. (Perhaps mark the lines included in each sum in a different color.) You'll find that you can combine the stated sums in such a way as to obtain the sum you need to find.

 

[Funkj]

Thanks so much for your response , but I still don't know how to combine the stated sum together to get the actual value of the segments ,can you please expatiate more on the solution to the problem

 

[Dr.Peterson]

Please make an attempt. It requires some thought, not just a quick glance.

Here is a start, showing which parts are included in the sum of the perimeters of the triangles:


The sum of these 12 segments is 20.

Do the same thing (in different colors) for the quadrilaterals and the outer triangle, then think about it.

 

[Funkj]

 

[Funkj]

Thanks ,have gotten it,I will post my solution soonest, thanks all the great mathematicians here

 

[Funkj]

Please make an attempt. It requires some thought, not just a quick glance.

Here is a start, showing which parts are included in the sum of the perimeters of the triangles:
View attachment 13622

The sum of these 12 segments is 20.

Do the same thing (in different colors) for the quadrilaterals and the outer triangle, then think about it.
I did the same by denoting the outer red segments A and the inner red segments B( the sum of the segments needed to know) and the black segments C. There fore 20 - A = B .. 19 - A= C and 25 - B = C .
Hence : 19 - A=25 - B
25 - 19= B - A
6 = B - A ..... i
20=B + A ....ii
Simultaneous equation
A= 7. Recall 20 - A = B .
Therefore B ( sum of the line of segments) = 13....

 

[Dr.Peterson]

I did the same by denoting the outer red segments A and the inner red segments B( the sum of the segments needed to know) and the black segments C. There fore 20 - A = B .. 19 - A= C and 25 - B = C .
Hence : 19 - A=25 - B
25 - 19= B - A
6 = B - A ..... i
20=B + A ....ii
Simultaneous equation
A= 7. Recall 20 - A = B .
Therefore B ( sum of the line of segments) = 13....

Good work.

For completeness, here's how I did it, without equations:


The red segments are the sum of the triangles, and equal 20.

The green segments are the sum of the quadrilaterals, and equal 25.

Together, these account for all the segments we care about, taking each segment twice, plus the blue segments (the outer triangle, 19), once each.

So adding 20 + 25 and subtracting 19, we are left with twice the sum of the inner lines; so the answer is (20 + 25 - 19)/2 = 26/2 = 13.

 

[Funkj]

whao! This method is fast and well understood,and easy, thanks once again, more greese to your elbow

 

[Funkj]