Find the area of the shaded region.

Question

d

c

Answer 1

D)

angle MNO is 45 degrees;

MO is perpendicular radius to LN;

angle LMO is 45 degrees;

Triangle LMO is an isoceles right with base angle 45.

So LO and MO are 4/sqrt(2) = 4*Sqrt(2)/2 = 2*sqrt(2);

Note that by pythagorean: LO^2 + MO^2 = (2*sqrt(2))^2 + (2*sqrt(2))^2 =

4*2 + 4*2 = 16 = 4^2 = LM^2

So the radius is exactly 2*sqrt(2).

The area of the half circle is then pi * (2*sqrt(2))^2 =

pi * 4 * 2 = 8 * pi

The area of the triangle LMN is 1/2* 4*Sqrt(2) * 2*Sqrt(2) = 8

Therefore the area of the shaded region is 8 * pi - 8 = 8(pi - 1)

C)

The area of the entire circle is pi * 64;

The area of the unshaded circle is pi * 16;

Therefore the area of the shaded region is 64 * pi - 16 * pi = 48 * pi

Answer 2

The area of the shaded region is equal to the area of the entire circle minus the area of the semicircle and the triangle

The triangle is a 45-45-90 triangle so LM=MN=4 so the area of the triangle is (1/2) 4 x 4=8

The hypotenuse is 4sqrt(2)This is also the diameter of the circle so radius is 2sqrt (2)

The area is pi r square or 8pi and the area of the semicircle is 4pi

Area of shaded region is then 8pi - 4pi - 8 =4pi - 8

Answer 3

d.

LN = Diameter

Pitagoras:

LN² = LM² + MN²

LN² = (4cm)² + (4cm)²

LN = (16 + 16) ^1/2

LN = 5.66 cm = Diameter

Radio = 2.83cm

Area triangle = 4 x 4 / 2 = 8 cm2

Area semi-circle = Pi x (radio)^2 / 2

Area semi-circle= 3.1416 x (2.83)² / 2

Area semi-circle= 12.57cm²

Shaded Area = (Area semi-circle) - (Area triangle)

Shaded Area = 12.57 cm² - 8 cm ²

Shaded Area = 4.57cm²

c

OP = Radio₁ = R₁

OA = Radio₂ = R₂

AB = Radio₃ = R₃

Area ₁ = Pi x (R1)²

Area ₁ = 64 x Pi

Area ₂ = Pi x (R₂)²

Area ₂ = 16 x Pi

Area ₃= Pi x (R₃)²

Area ₃ = 4 x Pi

Shaded Area = Area ₁ - Area ₂ + Area ₃

Shaded Area = 64 Pi - 16 Pi + 4 Pi = 52 Pi

Shaded Area = 52 x 3.1416 = 163.36 cm²