Mathematics, 06.08.2019 01:20 MrSavannahCat

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.)

$In rectangle $abcd$, $p$ is a point on $bc$ so that $\angle apd=90^{\circ}$. $ts$ is perpendicular t$

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Answer from: potato3999

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 from: willveloz4

$\frac{28}{3}$

Step-by-step 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.

$\frac{PT}{QT}=\frac{SD}{SQ}$

$\frac{12}{9}=\frac{SD}{7}$

On cross multiplication we get

$9SD=12\times 7$

9SD=84

Divide both sides by 9.

$SD=\frac{84}{9}$

$SD=\frac{28}{3}$

Therefore, $SD=\frac{28}{3}$ .

Answer from: Quest

step-by-step explanation:

Answer from: Quest

simplifying

5(x + -6) = 2(x + 3)

reorder the terms:

5(-6 + x) = 2(x + 3)

(-6 * 5 + x * 5) = 2(x + 3)

(-30 + 5x) = 2(x + 3)

reorder the terms:

-30 + 5x = 2(3 + x)

-30 + 5x = (3 * 2 + x * 2)

-30 + 5x = (6 + 2x)

solving

-30 + 5x = 6 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-30 + 5x + -2x = 6 + 2x + -2x

combine like terms: 5x + -2x = 3x

-30 + 3x = 6 + 2x + -2x

combine like terms: 2x + -2x = 0

-30 + 3x = 6 + 0

-30 + 3x = 6

add '30' to each side of the equation.

-30 + 30 + 3x = 6 + 30

combine like terms: -30 + 30 = 0

0 + 3x = 6 + 30

3x = 6 + 30

combine like terms: 6 + 30 = 36

3x = 36

divide each side by '3'.

x = 12

simplifying

x = 12

this should .

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