Question

Finding Hypotenuse Lengths. Find the length of the hypotenuse.

Answer 1

The length of the hypotenuse can be found using the Pythagorean theorem. In this case, the hypotenuse has a length of approximately 10.3 blocks.

Step-by-step explanation:

The hypotenuse of a right triangle can be found using the Pythagorean theorem. The Pythagorean theorem states that in a right triangle, the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse.

In this case, the lengths of the legs are 9 blocks and 5 blocks.

1. Square the lengths of the legs: 9² = 81 and 5² = 25.
2. Add the squared values: 81 + 25 = 106.
3. Take the square root of the sum: √106 ≈ 10.3.

Therefore, the length of the hypotenuse is approximately 10.3 blocks.

Answer 2

11)
`(3√(5))/(2)`

12)
`4√(5)`

13)
`22`

14)
`18`

15)
`2`

16)
`24√(3)`

Step-by-step explanation:

For these problems I used the pythagorean theorem:
`a^(2)+ b^(2)= c^(2)` and SOHCAHTOA

Sin =
`(opposite)/(hypotenuse)`

Cos =
`(adjacent)/(hypotenuse)`

Tan =
`(opposite)/(hypotenuse)`

11)

First find the length of the bottom side by using Cos

`cos(60)=(x)/(3)`

`3(cos(60))=x`

`1.5=x`

Then plug it into the formula for the pythagorean theorem to find the hypotenuse

`3^(2)+ 1.5^(2)= c^(2)`

`9+2.25=c^(2)`

`√(11.25)=\sqrt{c^(2)}`

`(3√(5))/(2)`

12)

Find the length of the bottom side using Cos

`cos(60)=(x)/(8)`

`8(cos(60))=x`

`4=x`

Then plug it into the formula for the pythagorean theorem to find the hypotenuse

`8^(2)+ 4^(2)= c^(2)`

`64+16=c^(2)`

`√(80) =\sqrt{c^(2) }`

`4√(5)`

13)

Find the length of the other side by using Tan

`tan(30)=(x)/(11√(3) )`

`11√(3)* (tan(30)=x`

`11=x`

Then plug it into the formula for the pythagorean theorem to find the hypotenuse

`(11√(3)) ^(2)+ 11^(2)= c^(2)`

`363+121=c^(2)`

`√(484) =\sqrt{c^(2)}`

`22`

14)

(This is probably an easier way to do these problems)

Find the hypotenuse by using Cos (
`(adjacent)/(hypotenuse)`)

`cos(60)=(9)/(x)`

`cos(60)x=9`

`x=(9)/(cos(60))`

`x=18`

15)

Find the hypotenuse using Sin (
`(opposite)/(hypotenuse)`)

`sin(30)=(1)/(x)`

`sin(30)x=1\\x=(1)/(sin(x))`

`x=2`

16)

Find the hypotenuse using Cos (
`(adjacent)/(hypotenuse)`)

`cos(60)=(12√(3) )/(x)`

`cos(60)x=12√(3)`

`x=(12√(3) )/(cos(60))`

`x=24√(3)`