Question

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.)

Answer 1

`(28)/(3)`

Explanation:

Given information: ABCD is a rectangle,
`\angle APD=90^(\circ)`, PA=20, AQ=25 and QP=15.

In a right angled triangle

`hypotenuse^2=base^2+perpendicular^2`

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

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

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

`(BP)^2 = 20^2-16^2`

`BP^2 = 144`

`BP = 12`

Since BP = PT, therefore PT = 12.

`AS = BP + PT = 12 + 12 = 24`

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

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

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

`SQ = 7`

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

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

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

`QT = 9`

In triangle PTQ and DSQ,

`\angle TQP=\angle SQD` (Vertical angles)

`\angle PTQ=\angle DSQ` (Right angles)

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

Corresponding parts of similar triangles are proportional.

`(PT)/(QT)=(SD)/(SQ)`

`(12)/(9)=(SD)/(7)`

On cross multiplication we get

`9SD=12* 7`

`9SD=84`

Divide both sides by 9.

`SD=(84)/(9)`

`SD=(28)/(3)`

Therefore,
`SD=(28)/(3)`.