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Dahasolnce [82]
3 years ago
15

Please help, thank you!

Mathematics
1 answer:
Nadya [2.5K]3 years ago
4 0

Answer:

37 and 26

Step-by-step explanation:

Let the bigger number be x and the smaller number be y.

Then the two numbers total 63 will give us:

y+x=63------>eqn1

The two numbers have a difference of 11.

y-x=11------->eqn2

We add equations 1 and 2 to obtain:

2y=74

Divide both sides to get:

y=37

Put y=37 into eqn2 to get:

37-x=11

x=37-11

x=26

This larger number is 37 and the smaller number is 26

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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 encl
katovenus [111]

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 <em>r</em> and <em>θ</em>, 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:

brainly.com/question/22599425

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