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 $\frac{112}{360}$ 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.

$= \frac{112}{360}× 49 \pi$

$= 47.89183467 \: {units}^{2}$

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.