Question

2. Find the value of x to the nearest tenth.

a. 4.5
b. 5.4
c. 6.3
d. 7.2

3. Find the value of x.
a. 7
b. 7.5
c. 8
d. 8.5


4. FG ⊥ OP, RS ⊥ OQ. FG=40, RS=40, OP=15. Find x.
a. 15
b. 17
c. 20
d. 21


5. Find the value of x to the nearest tenth.
a. 7.5
b. 7.9
c. 8.1
d. 8.9

Answer 1

Part 2) Option b. 5.4

Part 3) Option c. 8

Part 4) Option a. 15

Part 5) Option d. 8.9

Explanation:

Part 2) Find the value of x to the nearest tenth

we know that

x is the radius of the circle

Applying the Pythagoras Theorem

`x^(2)=3.6^(2)+(8/2)^(2)`

`x^(2)=28.96`

`x=5.4\ units`

Part 3) Find the value of x

In this problem

x=8

Verify

step 1

Find the radius of the circle

Let

r -----> the radius of the circle

Applying the Pythagoras Theorem

`r^(2)=8^(2)+(15/2)^(2)`

`r^(2)=120.25`

`r=√(120.25)`

step 2

Find the value of x

Applying the Pythagoras Theorem

`r^(2)=x^(2)+(15/2)^(2)`

substitute

`120.25=x^(2)+56.25`

`x^(2)=120.25-56.25`

`x^(2)=64`

`x=8\ units`

Part 4) Find the value of x

In this problem

x=OP=15

Verify

step 1

Find the radius of the circle

Let

r -----> the radius of the circle

In the right triangle FPO

Applying the Pythagoras Theorem

`r^(2)=15^(2)+(40/2)^(2)`

`r^(2)=625`

`r=25`

step 2

Find the value of x

In the right triangle RQO

Applying the Pythagoras Theorem

`25^(2)=x^(2)+(40/2)^(2)`

`625=x^(2)+400`

`x^(2)=625-400`

`x^(2)=225`

`x=15\ units`

Part 5) Find the value of x

Applying the Pythagoras Theorem

`6^(2)=4^(2)+(x/2)^(2)`

`36=16+(x/2)^(2)`

`(x/2)^(2)=36-16`

`(x/2)^(2)=20`

`(x/2)=4.47`

`x=8.9`