QuestionFind $D E$ given that $B$ is the midpoint of $A C$ , $A C=C D$ , $A B=3 x+4$ , $A C=11 x-17$ , and $C E=49$ .

Studdy Solution

STEP 1

Assumptions1. $B$ is the midpoint of $AC$ . $AC = CD$
3. $AB =3x +4$
4. $AC =11x -17$
5. $CE =49$

STEP 2

Since $B$ is the midpoint of $AC$ , $AB = BC$ . So we can write the equation $AB = BC$

STEP 3

Substitute the given value of $AB$ into the equation.
$3x + = BC$

STEP 4

Since $AC = CD$ , we can write the equation $AC = CD$

STEP 5

Substitute the given value of $AC$ into the equation.
$11x -17 = CD$

STEP 6

Since $BC = CD$ , we can equate the expressions we found for $BC$ and $CD$ .
$3x +4 =11x -17$

STEP 7

olve the equation for $x$ .
$3x -11x = -17 -4$

STEP 8

implify the equation.
$-8x = -21$

STEP 9

Divide both sides by -8 to solve for $x$ .
$x = \frac{-21}{-8} = \frac{21}{8}$

STEP 10

Substitute $x = \frac{21}{8}$ into the equation $AC =x -17$ to find the length of $AC$ .
$AC = \times \frac{21}{8} -17$

STEP 11

implify the equation to find the length of $AC$ .
$AC = \frac{231}{8} -17 = \frac{231 -136}{8} = \frac{95}{8}$

STEP 12

Since $AC = CD$ , the length of $CD$ is also $\frac{95}{8}$ .

STEP 13

The length of $DE$ is the difference between the lengths of $CE$ and $CD$ .
$DE = CE - CD$

STEP 14

Substitute the given value of $CE$ and the calculated value of $CD$ into the equation.
$DE =49 - \frac{95}{8}$

STEP 15

Convert $49$ to a fraction with $8$ as the denominator to simplify the subtraction.
$DE = \frac{392}{8} - \frac{95}{8}$

STEP 16

Subtract the fractions to find the length of $DE$ .
$DE = \frac{392 -95}{8} = \frac{297}{8} =37.125$ So, $DE =37.125$ units.

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QuestionFind $D E$ given that $B$ is the midpoint of $A C$ , $A C=C D$ , $A B=3 x+4$ , $A C=11 x-17$ , and $C E=49$ .

Studdy Solution

STEP 1

Assumptions1. $B$ is the midpoint of $AC$ . $AC = CD$
3. $AB =3x +4$
4. $AC =11x -17$
5. $CE =49$

STEP 2

Since $B$ is the midpoint of $AC$ , $AB = BC$ . So we can write the equation $AB = BC$

STEP 3

Substitute the given value of $AB$ into the equation.
$3x + = BC$

STEP 4

Since $AC = CD$ , we can write the equation $AC = CD$

STEP 5

Substitute the given value of $AC$ into the equation.
$11x -17 = CD$

STEP 6

Since $BC = CD$ , we can equate the expressions we found for $BC$ and $CD$ .
$3x +4 =11x -17$

STEP 7

olve the equation for $x$ .
$3x -11x = -17 -4$

STEP 8

implify the equation.
$-8x = -21$

STEP 9

Divide both sides by -8 to solve for $x$ .
$x = \frac{-21}{-8} = \frac{21}{8}$

STEP 10

Substitute $x = \frac{21}{8}$ into the equation $AC =x -17$ to find the length of $AC$ .
$AC = \times \frac{21}{8} -17$

STEP 11

implify the equation to find the length of $AC$ .
$AC = \frac{231}{8} -17 = \frac{231 -136}{8} = \frac{95}{8}$

STEP 12

Since $AC = CD$ , the length of $CD$ is also $\frac{95}{8}$ .

STEP 13

The length of $DE$ is the difference between the lengths of $CE$ and $CD$ .
$DE = CE - CD$

STEP 14

Substitute the given value of $CE$ and the calculated value of $CD$ into the equation.
$DE =49 - \frac{95}{8}$

STEP 15

Convert $49$ to a fraction with $8$ as the denominator to simplify the subtraction.
$DE = \frac{392}{8} - \frac{95}{8}$

STEP 16

Subtract the fractions to find the length of $DE$ .
$DE = \frac{392 -95}{8} = \frac{297}{8} =37.125$ So, $DE =37.125$ units.

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QuestionFind DED EDE given that BBB is the midpoint of ACA CAC, AC=CDA C=C DAC=CD, AB=3x+4A B=3 x+4AB=3x+4, AC=11x−17A C=11 x-17AC=11x−17, and CE=49C E=49CE=49.

Studdy Solution

STEP 1

Assumptions1. BBB is the midpoint of ACACAC . AC=CDAC = CDAC=CD3. AB=3x+4AB =3x +4AB=3x+44. AC=11x−17AC =11x -17AC=11x−175. CE=49CE =49CE=49

STEP 2

Since BBB is the midpoint of ACACAC, AB=BCAB = BCAB=BC. So we can write the equationAB=BCAB = BCAB=BC

STEP 3

Substitute the given value of ABABAB into the equation.3x+=BC3x + = BC3x+=BC

STEP 4

Since AC=CDAC = CDAC=CD, we can write the equationAC=CDAC = CDAC=CD

STEP 5

Substitute the given value of ACACAC into the equation.11x−17=CD11x -17 = CD11x−17=CD

STEP 6

Since BC=CDBC = CDBC=CD, we can equate the expressions we found for BCBCBC and CDCDCD.3x+4=11x−173x +4 =11x -173x+4=11x−17

STEP 7

olve the equation for xxx.3x−11x=−17−43x -11x = -17 -43x−11x=−17−4

STEP 8

implify the equation.−8x=−21-8x = -21−8x=−21

STEP 9

Divide both sides by -8 to solve for xxx.x=−21−8=218x = \frac{-21}{-8} = \frac{21}{8}x=−8−21​=821​

STEP 10

Substitute x=218x = \frac{21}{8}x=821​ into the equation AC=x−17AC =x -17AC=x−17 to find the length of ACACAC.AC=×218−17AC = \times \frac{21}{8} -17AC=×821​−17

STEP 11

implify the equation to find the length of ACACAC.AC=2318−17=231−1368=958AC = \frac{231}{8} -17 = \frac{231 -136}{8} = \frac{95}{8}AC=8231​−17=8231−136​=895​

STEP 12

Since AC=CDAC = CDAC=CD, the length of CDCDCD is also 958\frac{95}{8}895​.

STEP 13

The length of DEDEDE is the difference between the lengths of CECECE and CDCDCD.DE=CE−CDDE = CE - CDDE=CE−CD

STEP 14

Substitute the given value of CECECE and the calculated value of CDCDCD into the equation.DE=49−958DE =49 - \frac{95}{8}DE=49−895​

STEP 15

Convert 494949 to a fraction with 888 as the denominator to simplify the subtraction.DE=3928−958DE = \frac{392}{8} - \frac{95}{8}DE=8392​−895​

STEP 16

Subtract the fractions to find the length of DEDEDE.DE=392−958=2978=37.125DE = \frac{392 -95}{8} = \frac{297}{8} =37.125DE=8392−95​=8297​=37.125So, DE=37.125DE =37.125DE=37.125 units.

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QuestionFind $D E$ given that $B$ is the midpoint of $A C$ , $A C=C D$ , $A B=3 x+4$ , $A C=11 x-17$ , and $C E=49$ .

Assumptions1. $B$ is the midpoint of $AC$ . $AC = CD$
3. $AB =3x +4$
4. $AC =11x -17$
5. $CE =49$

Since $B$ is the midpoint of $AC$ , $AB = BC$ . So we can write the equation $AB = BC$

Substitute the given value of $AB$ into the equation.
$3x + = BC$

Since $AC = CD$ , we can write the equation $AC = CD$

Substitute the given value of $AC$ into the equation.
$11x -17 = CD$

Since $BC = CD$ , we can equate the expressions we found for $BC$ and $CD$ .
$3x +4 =11x -17$

olve the equation for $x$ .
$3x -11x = -17 -4$

implify the equation.
$-8x = -21$

Divide both sides by -8 to solve for $x$ .
$x = \frac{-21}{-8} = \frac{21}{8}$

Substitute $x = \frac{21}{8}$ into the equation $AC =x -17$ to find the length of $AC$ .
$AC = \times \frac{21}{8} -17$

implify the equation to find the length of $AC$ .
$AC = \frac{231}{8} -17 = \frac{231 -136}{8} = \frac{95}{8}$

Since $AC = CD$ , the length of $CD$ is also $\frac{95}{8}$ .

The length of $DE$ is the difference between the lengths of $CE$ and $CD$ .
$DE = CE - CD$

Substitute the given value of $CE$ and the calculated value of $CD$ into the equation.
$DE =49 - \frac{95}{8}$

Convert $49$ to a fraction with $8$ as the denominator to simplify the subtraction.
$DE = \frac{392}{8} - \frac{95}{8}$

Subtract the fractions to find the length of $DE$ .
$DE = \frac{392 -95}{8} = \frac{297}{8} =37.125$ So, $DE =37.125$ units.