In rectangle $abcd$, $p$ is a point on $bc$ so that $\angle apd=90^{\circ}$. $ts$ is perpendicular to $bc$ with $bp=pt$, as shown. $pd$ intersects $ts$ at $q$. point $r$ is on $cd$ such that $ra$ passes through $q$. in $\triangle pqa$, $pa=20$, $aq=25$ and $qp=15$. find $sd$. (express your answer as a common fraction.)

28/3

Step-by-step explanation:

AB = 16

AP = 20

Triangle ABP is a right triangle.

(AB)^2 + (BP)^2 = (AP)^2

16^2 + (BP)^2 = 20^2

BP^2 = 144

BP = 12

BP = PT = 12

AS = BP + PT = 12 + 12 = 24

Triangle AQS is a right triangle.

(AS)^2 + (SQ)^2 = (AQ)^2

AQ = 25

24^2 + (SQ)^2 = 25^2

SQ = 7

Triangle PQT is a right triangle.

(PT)^2 + (QT)^2 = (PQ)^2

12^2 + (QT)^2 = 15^2

QT = 9

Triangle PTQ is similar to triangle DSQ.

PT/QT = SD/SQ

12/9 = SD/7

9SD = 12 * 7

SD = (12 * 7)/9

SD = 28/3

Step-by-step explanation:

Given information: ABCD is a rectangle, , PA=20, AQ=25 and QP=15.

In a right angled triangle

In triangle ABP, AB = 16 and AP = 20. Using Pythagoras theorem we get

Since BP = PT, therefore PT = 12.

AQS is a right angled triangle and AQ = 25. Use Pythagoras theorem in triangle AQS.

Triangle PQT is a right triangle. Use Pythagoras theorem in triangle PQT.

In triangle  PTQ and DSQ,

            (Vertical angles)

            (Right angles)

Triangle PTQ is similar to triangle DSQ by AA property of similarity.

Corresponding parts of similar triangles are proportional.

On cross multiplication we get

Divide both sides by 9.

Therefore, .

d. 8 that's the correct answer