Answer:
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
Answer:
[tex]\frac{28}{3}[/tex]
Step-by-step explanation:
Given information: ABCD is a rectangle, [tex]\angle APD=90^{\circ}[/tex], PA=20, AQ=25 and QP=15.
In a right angled triangle
[tex]hypotenuse^2=base^2+perpendicular^2[/tex]
In triangle ABP, AB = 16 and AP = 20. Using Pythagoras theorem we get
[tex](AB)^2 + (BP)^2 = (AP)^2[/tex]
[tex]16^2 + (BP)^2 = 20^2[/tex]
[tex](BP)^2 = 20^2-16^2[/tex]
[tex]BP^2 = 144[/tex]
[tex]BP = 12[/tex]
Since BP = PT, therefore PT = 12.
[tex]AS = BP + PT = 12 + 12 = 24[/tex]
AQS is a right angled triangle and AQ = 25. Use Pythagoras theorem in triangle AQS.
[tex](AS)^2 + (SQ)^2 = (AQ)^2[/tex]
[tex]24^2 + (SQ)^2 = 25^2[/tex]
[tex]SQ = 7[/tex]
Triangle PQT is a right triangle. Use Pythagoras theorem in triangle PQT.
[tex](PT)^2 + (QT)^2 = (PQ)^2[/tex]
[tex]12^2 + (QT)^2 = 15^2[/tex]
[tex]QT = 9[/tex]
In triangle PTQ and DSQ,
[tex]\angle TQP=\angle SQD[/tex] (Vertical angles)
[tex]\angle PTQ=\angle DSQ[/tex] (Right angles)
Triangle PTQ is similar to triangle DSQ by AA property of similarity.
Corresponding parts of similar triangles are proportional.
[tex]\frac{PT}{QT}=\frac{SD}{SQ}[/tex]
[tex]\frac{12}{9}=\frac{SD}{7}[/tex]
On cross multiplication we get
[tex]9SD=12\times 7[/tex]
[tex]9SD=84[/tex]
Divide both sides by 9.
[tex]SD=\frac{84}{9}[/tex]
[tex]SD=\frac{28}{3}[/tex]
Therefore, [tex]SD=\frac{28}{3}[/tex].
