Question

Triangle ABL is an isosceles triangle in circle A with a radius of 11, PL = 16, and ∠PAL = 93°. Find the area of the circle enclosed by line PL and arc PL. Show all work and round your answer to two decimal places.

Answer 1

Answer:

The area of the circle enclosed by line PL and arc PL is approximately 37.62 square units

Step-by-step explanation:

The given parameters in the question are;

The radius of the circle, r = 11

The length of the chord PL = 16

The measure of angle ∠PAL = 93°

The segment of the circle for which the area is required = Minor segment PL

The shaded area of the given circle is the minor segment of the circle enclosed by line PL and arc PL

The area of a segment of a circle is given by the following formula;

Area of segment = Area of the sector - Area of the triangle

In detail, we have;

Area of segment = Area of the sector of the circle that contains the segment) - (Area of the isosceles triangle in the sector)

Area of a sector = (θ/360)×π·r²

Where;

r = The radius of the circle

θ = The angle of the sector of the circle

Plugging in the the values of r and θ, we get;

The area of the sector enclosed by arc PL and radii AP and AL = (93°/360°) × π × 11² ≈ 98.2 square units

Area of a triangle = (1/2) × Base length × Height

Therefore;

The area of ΔAPL = (1/2) × 16 × 11 × cos(93°/2) ≈ 60.58 square units

∴ The area of the segment PL ≈ (98.2 - 60.58) square units = 37.62 square units

Therefore, the area of the shaded segment PL ≈ 37.62 square units

More examples on area of a shaded segment can be found here:

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Answer 2

The area of the circle enclosed by line segment PL and circle arc PL is 37.80 square units.

The calculation of the area between line segment PL and circle arc PL is described below:

1) Calculation of the area of the circle arc.

2) Calculation of the area of the triangle.

3) Subtracting the area found in 2) from the area found in 1).

Step 1:

The area of a circle arc is determined by the following formula:

$A_{ca} = \frac{\alpha\cdot \pi\cdot r^{2}}{360}$ (1)

Where:

$A_{ca}$ - Area of the circle arc.

$\alpha$ - Arc angle, in sexagesimal degrees.

$r$ - Radius.

If we know that $\alpha = 93^{\circ}$ and $r = 11$, then the area of the circle arc is:

$A_{ca} = \frac{93\cdot \pi\cdot 11^{2}}{360}$

$A_{ca} \approx 98.201$

Step 2:

The area of the triangle is determined by Heron's formula:

$A_{t} = \sqrt{s\cdot (s-l)\cdot (s-r)^{2}}$ (2)

$s = \frac{l + 2\cdot r}{2}$

Where:

$A_{t}$ - Area of the triangle.

$r$ - Radius.

$l$ - Length of the line segment PL.

If we know that $l = 16$ and $r = 11$, then the area of the triangle is:

$s = \frac{16+2\cdot (11)}{2}$

$s = 19$

$A_{t} = \sqrt{19\cdot (19-16)\cdot (19-11)^{2}}$

$A_{t} \approx 60.399$

Step 3:

And the area between the line segment PL and the circle arc PL is:

$A_{s} = A_{ca}-A_{t}$

$A_{s} = 98.201 - 60.399$

$A_{s} = 37.802$

The area of the circle enclosed by line segment PL and circle arc PL is 37.80 square units.