Answer :
Let's solve the given problems step-by-step.
Find DE: AB:
- We have the line segment [tex]AB[/tex] such that [tex]AB: DB = 4:1[/tex].
- This means if [tex]AB = 4x[/tex], then [tex]DB = 1x[/tex].
- We also have [tex]AE: EB = 3:1[/tex].
- Since [tex]AE + EB = AB[/tex], and [tex]AB = 4x[/tex], let's find [tex]AE[/tex] and [tex]EB[/tex]:
- [tex]\frac{AE}{EB} = \frac{3}{1}[/tex] implies [tex]AE = 3k[/tex] and [tex]EB = 1k[/tex].
- As their sum is equal to [tex]AB[/tex], we have [tex]3k + 1k = 4k = 4x[/tex], which gives [tex]k = x[/tex].
- Now let's find [tex]DE[/tex]:
- [tex]D[/tex] divides [tex]DB[/tex] from [tex]AB[/tex] such that [tex]AB: DB = 4:1[/tex].
- [tex]E[/tex] divides [tex]EB[/tex] from [tex]AE[/tex] such that [tex]AE: EB = 3:1[/tex].
- To find [tex]DE[/tex], we need to consider the portion [tex]DE = AE - AD[/tex], where [tex]AD = AB - DB = 4x - 1x = 3x[/tex], and [tex]AE = 3k = 3x[/tex]. Hence, [tex]DE = 3x - 3x = 0[/tex].
- Since [tex]DE = 0[/tex], the ratio [tex]DE: AB = 0:4x = 0[/tex].
Find AP: TB:
- We have [tex]AP: PB = 2:5[/tex].
- Let [tex]AP = 2y[/tex] and [tex]PB = 5y[/tex], then [tex]AB = AP + PB = 7y[/tex].
- We also have [tex]AP: PT = 3:4[/tex].
- Let [tex]PT = 4z[/tex], so [tex]AP = 3z[/tex].
- From the first equation we know [tex]AP = 2y = 3z[/tex], implying [tex]y = \frac{3z}{2}[/tex].
- We need to find [tex]TB = AB - AP[/tex]:
- [tex]AB = 7y[/tex] and [tex]AP = 2y[/tex], so [tex]TB = 7y - 2y = 5y[/tex] or [tex]5(\frac{3z}{2}) = \frac{15z}{2}[/tex].
- The ratio [tex]AP: TB[/tex] is [tex]2y:5y = 2:5[/tex].
Thus, the solution to each ratio problem is given as follows:
- DE:AB is [tex]0[/tex].
- AP:TB is [tex]2:5[/tex].
