Question

Please help me
I need it as soon as possible ​

Answer 1

Given:

• Angle JKL= 112 units
• JK= 7 units

To find:

• The area of sector JKL to the nearest hundredth.

Solution:

JKL occupies
`(112)/(360)` of the circle.

Radius of the circle is 7 units.

First, we'll have to find the area of the circle.

`\large\boxed{Formula: a= \pi {r}^(2)}`

Let's solve!

Let's substitute the values according to the formula.

We are finding it in terms of pi.

`a= 7×7`

`a= 49 \pi`

Now, we can find the area of sector JKL.

`= (112)/(360)× 49 \pi`

`= 47.89183467 \: {units}^(2)`

Now, we'll have to round off to the nearest hundredth.

The value in thousandths place is less than 5 so we won't have to round up.

Final answer:

`\large\boxed{a= 47.89 \: {units}^(2)}`

Hence, the area of sector JKL is 47.89 square units.