QuestionFind the length $x$ of $\overline{TP}$ if pentagons $JKLMN$ and $PQRST$ are similar, with $QR=3.6$ , $QP=0.9$ , $RS=0.9$ , and $ST=1.8$ .

Studdy Solution

STEP 1

Assumptions1. The pentagons $J K L M N$ and $Q R$ are similar. . The sides of similar polygons are proportional.
3. The given lengths are $QR=3.6$ , $QP=0.9$ , $RS=0.9$ , and $=1.8$ .
4. We are asked to find the length $x$ of $\overline{T}$ .

STEP 2

Since the pentagons are similar, the ratio of corresponding sides is constant. We can set up a proportion to find the length $x$ of $\overline{T}$ .
Let's denote the corresponding sides of pentagon $J K L M N$ as $a$ , $b$ , $c$ , $d$ , and $e$ in the same order as $Q R$ .
We know that $aQP = bQR = cRS = d = eTP$ .

STEP 3

We can choose any two ratios to set up our proportion. Since we know the lengths of $QR$ , $QP$ , $RS$ , and $$, we can use these to set up our proportion. Let's use $aQP = d$.
$\frac{a}{QP} = \frac{d}{}$

STEP 4

Substitute the given lengths into the proportion.
$\frac{a}{0.9} = \frac{d}{1.8}$

STEP 5

We can simplify this equation by multiplying both sides by0.9.
$a = \frac{d \times0.9}{1.8}$

STEP 6

implify the right side of the equation.
$a = \frac{d}{2}$

STEP 7

Since $a$ and $d$ are corresponding sides of the same pentagon, they are equal. So, we can say that $d =2a$ .

STEP 8

Now, we can use another ratio to find $x$ , the length of $\overline{T}$ . Let's use $aQP = eTP$ .
$\frac{a}{QP} = \frac{e}{TP}$

STEP 9

Substitute $a =2d$ and $TP = x$ into the equation.
$\frac{2d}{QP} = \frac{e}{x}$

STEP 10

Substitute the given length $QP =0.9$ into the equation.
$\frac{2d}{0.9} = \frac{e}{x}$

STEP 11

We know that $d = =.8$ , substitute this into the equation.
$\frac{ \times.8}{0.9} = \frac{e}{x}$

STEP 12

implify the left side of the equation.
$4 = \frac{e}{x}$

STEP 13

olve for $x$ .
$x = \frac{e}{}$

STEP 14

Since $e$ and $d$ are corresponding sides of the same pentagon, they are equal. So, we can say that $e = d =.8$ .

STEP 15

Substitute $e =.8$ into the equation.
$x = \frac{.8}{4}$

STEP 16

Calculate the value of $x$ .
$x = \frac{.8}{4} =0.45$ The length of $\overline{T}$ is0.45.

Was this helpful?

QuestionFind the length $x$ of $\overline{TP}$ if pentagons $JKLMN$ and $PQRST$ are similar, with $QR=3.6$ , $QP=0.9$ , $RS=0.9$ , and $ST=1.8$ .

Studdy Solution

STEP 1

Assumptions1. The pentagons $J K L M N$ and $Q R$ are similar. . The sides of similar polygons are proportional.
3. The given lengths are $QR=3.6$ , $QP=0.9$ , $RS=0.9$ , and $=1.8$ .
4. We are asked to find the length $x$ of $\overline{T}$ .

STEP 2

STEP 3

We can choose any two ratios to set up our proportion. Since we know the lengths of $QR$ , $QP$ , $RS$ , and $$, we can use these to set up our proportion. Let's use $aQP = d$.
$\frac{a}{QP} = \frac{d}{}$

STEP 4

Substitute the given lengths into the proportion.
$\frac{a}{0.9} = \frac{d}{1.8}$

STEP 5

We can simplify this equation by multiplying both sides by0.9.
$a = \frac{d \times0.9}{1.8}$

STEP 6

implify the right side of the equation.
$a = \frac{d}{2}$

STEP 7

Since $a$ and $d$ are corresponding sides of the same pentagon, they are equal. So, we can say that $d =2a$ .

STEP 8

Now, we can use another ratio to find $x$ , the length of $\overline{T}$ . Let's use $aQP = eTP$ .
$\frac{a}{QP} = \frac{e}{TP}$

STEP 9

Substitute $a =2d$ and $TP = x$ into the equation.
$\frac{2d}{QP} = \frac{e}{x}$

STEP 10

Substitute the given length $QP =0.9$ into the equation.
$\frac{2d}{0.9} = \frac{e}{x}$

STEP 11

We know that $d = =.8$ , substitute this into the equation.
$\frac{ \times.8}{0.9} = \frac{e}{x}$

STEP 12

implify the left side of the equation.
$4 = \frac{e}{x}$

STEP 13

olve for $x$ .
$x = \frac{e}{}$

STEP 14

Since $e$ and $d$ are corresponding sides of the same pentagon, they are equal. So, we can say that $e = d =.8$ .

STEP 15

Substitute $e =.8$ into the equation.
$x = \frac{.8}{4}$

STEP 16

Calculate the value of $x$ .
$x = \frac{.8}{4} =0.45$ The length of $\overline{T}$ is0.45.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionFind the length xxx of TP‾\overline{TP}TP if pentagons JKLMNJKLMNJKLMN and PQRSTPQRSTPQRST are similar, with QR=3.6QR=3.6QR=3.6, QP=0.9QP=0.9QP=0.9, RS=0.9RS=0.9RS=0.9, and ST=1.8ST=1.8ST=1.8.

Studdy Solution

STEP 1

Assumptions1. The pentagons JKLMNJ K L M NJKLMN and QR Q RQR are similar. . The sides of similar polygons are proportional.3. The given lengths are QR=3.6QR=3.6QR=3.6, QP=0.9QP=0.9QP=0.9, RS=0.9RS=0.9RS=0.9, and =1.8=1.8=1.8.4. We are asked to find the length xxx of T‾\overline{T}T.

STEP 2

STEP 3

We can choose any two ratios to set up our proportion. Since we know the lengths of QRQRQR, QPQPQP, RSRSRS, and $$, we can use these to set up our proportion. Let's use $aQP = d$.aQP=d\frac{a}{QP} = \frac{d}{}QPa​=d​

STEP 4

Substitute the given lengths into the proportion.a0.9=d1.8\frac{a}{0.9} = \frac{d}{1.8}0.9a​=1.8d​

STEP 5

We can simplify this equation by multiplying both sides by0.9.a=d×0.91.8a = \frac{d \times0.9}{1.8}a=1.8d×0.9​

STEP 6

implify the right side of the equation.a=d2a = \frac{d}{2}a=2d​

STEP 7

Since aaa and ddd are corresponding sides of the same pentagon, they are equal. So, we can say that d=2ad =2ad=2a.

STEP 8

Now, we can use another ratio to find xxx, the length of T‾\overline{T}T. Let's use aQP=eTPaQP = eTPaQP=eTP.aQP=eTP\frac{a}{QP} = \frac{e}{TP}QPa​=TPe​

STEP 9

Substitute a=2da =2da=2d and TP=xTP = xTP=x into the equation.2dQP=ex\frac{2d}{QP} = \frac{e}{x}QP2d​=xe​

STEP 10

Substitute the given length QP=0.9QP =0.9QP=0.9 into the equation.2d0.9=ex\frac{2d}{0.9} = \frac{e}{x}0.92d​=xe​

STEP 11

We know that d==.8d = =.8d==.8, substitute this into the equation.×.80.9=ex\frac{ \times.8}{0.9} = \frac{e}{x}0.9×.8​=xe​

STEP 12

implify the left side of the equation.4=ex4 = \frac{e}{x}4=xe​

STEP 13

olve for xxx.x=ex = \frac{e}{}x=e​

STEP 14

Since eee and ddd are corresponding sides of the same pentagon, they are equal. So, we can say that e=d=.8e = d =.8e=d=.8.

STEP 15

Substitute e=.8e =.8e=.8 into the equation.x=.84x = \frac{.8}{4}x=4.8​

STEP 16

Calculate the value of xxx.x=.84=0.45x = \frac{.8}{4} =0.45x=4.8​=0.45The length of T‾\overline{T}T is0.45.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionFind the length $x$ of $\overline{TP}$ if pentagons $JKLMN$ and $PQRST$ are similar, with $QR=3.6$ , $QP=0.9$ , $RS=0.9$ , and $ST=1.8$ .

Assumptions1. The pentagons $J K L M N$ and $Q R$ are similar. . The sides of similar polygons are proportional.
3. The given lengths are $QR=3.6$ , $QP=0.9$ , $RS=0.9$ , and $=1.8$ .
4. We are asked to find the length $x$ of $\overline{T}$ .

We can choose any two ratios to set up our proportion. Since we know the lengths of $QR$ , $QP$ , $RS$ , and $$, we can use these to set up our proportion. Let's use $aQP = d$.
$\frac{a}{QP} = \frac{d}{}$

Substitute the given lengths into the proportion.
$\frac{a}{0.9} = \frac{d}{1.8}$

We can simplify this equation by multiplying both sides by0.9.
$a = \frac{d \times0.9}{1.8}$

implify the right side of the equation.
$a = \frac{d}{2}$

Since $a$ and $d$ are corresponding sides of the same pentagon, they are equal. So, we can say that $d =2a$ .

Now, we can use another ratio to find $x$ , the length of $\overline{T}$ . Let's use $aQP = eTP$ .
$\frac{a}{QP} = \frac{e}{TP}$

Substitute $a =2d$ and $TP = x$ into the equation.
$\frac{2d}{QP} = \frac{e}{x}$

Substitute the given length $QP =0.9$ into the equation.
$\frac{2d}{0.9} = \frac{e}{x}$

We know that $d = =.8$ , substitute this into the equation.
$\frac{ \times.8}{0.9} = \frac{e}{x}$

implify the left side of the equation.
$4 = \frac{e}{x}$

olve for $x$ .
$x = \frac{e}{}$

Since $e$ and $d$ are corresponding sides of the same pentagon, they are equal. So, we can say that $e = d =.8$ .

Substitute $e =.8$ into the equation.
$x = \frac{.8}{4}$

Calculate the value of $x$ .
$x = \frac{.8}{4} =0.45$ The length of $\overline{T}$ is0.45.