Question

Given that CE is a perpendicular bisector find the length of

Answer 1

Given:

AE=EB=12.

AC=3x-12 and BC=2x+4.

By perpendicular bisector property, we get

`(AC)/(AE)=(BC)/(EB)`

Substitute AE=EB=12, AC=3x-12 and BC=2x+4, we get

`(3x-12)/(12)=(2x+4)/(12)`

Cancel out the common term, we get

`3x-12=2x+4`

Adding 12 to both sides of the equation, we get

`3x-12+12=2x+4+12`

`3x=2x+16`

Subtracting 2x from both sides of the equation, we get

`3x-2x=2x+16-2x`

`x=16`

Substitute x=16 in AC=3x-12 , we get

`AC=3(16)-12`

`AC=36`

We get AC=36 units.

Given that AD=y+16 and DB=3y+22 and AE=EB=12.

By perpendicular bisector property, we get

`(AD)/(AE)=(DB)/(EB)`

Substitute AD=y+16 and DB=3y+22 and AE=EB=12 in the equation, we get

`(y+16)/(12)=(3y+22)/(12)`

Cancel out the common terms, we get

`y+16=3y+22`

Subtracting 22 from both sides, we get

`y+16-22=3y+22-22`

`y-6=3y`

Subtracting y from both sides of the equation, we get

`y-6-y=3y-y`
`-6=2y`

Dividing both sides by 2, we get

`-(6)/(2)=(2y)/(2)`
`y=-3`

Substitute y=-3 in DB=3y+22, we get

`DB=3(-3)+22`

`DB=-9+22`
`DB=13`

We get DB=13.

Use Pythagorean theorem to find DE.

`DB^2=DE^2+EB^2`

Substitute DB=13 and EB=12 in the equation, we get

`13^2=DE^2+12^2`

`DE^2=13^2-12^2`

`DE^2=25`

Taking square root on both sides, we get

`DE=\sqrt[]{25}`

`DE=\sqrt[]{5^2}`
`DE=5`

We get DE=5 units.

Hence the answers are

`\bar{AC}=36\text{units}`
`\bar{DB}=13units`
`\bar{DE}=5units`