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# Perimeter and Area of Composite Figures > help?

An archery target has a diameter of 80 cm. It contains a circle in the centre with a radius of 8cm and four additional concentric rings each 8 cm wide.

a) Find the area of the outer ring, to the nearest square centimetre.

b) What percent of the total area is the outer ring?

This is a question from the perimeter and area of composite figures unit that I'm working on independently at the moment. So yeah if you have the answer to this question, I would really appreciate it. Please help me, asap. Oh & this is grade 9 academic math, hope that helps?

Relevance

Inner radius of outer ring = 8 + 3(8) = 32 cm

outer radius of outer ring = 8 + 4(8) = 40 cm

a) Area of outer ring = π(40)² - π(32)² = 576π = 1809.56 cm²

b) Percent of total area = [π(40)² - π(32)²] / π(40)² = 0.36 = 36 %

• 4 years ago

Let L = length; w=width Perimeter = 2L + 2w = 44 Or L + W = 22 (dividing perimeter by 2) Let W = 22 - L Substituting for W and L in the area function (L)(22 - L) = 40 22 L - L^2 = 40 Rewrite into standard quadratic form L^2 - 22L + 40 = 0 Factoring by using FOIL method (L - 20)(L - 2) = 0 So, L = 20 or 2 If L = 20, W = 22 - L = 2 So, W = 2, L = 20 If we tried L = 2 W = 22 - L W = 2 So W = 20; L = 2 By convention, the longer side is denoted by L so L = 20, W = 2 Either way, the dimensions of rectangle are 2 yards x 20 yards

the target has an area of pi (40 cm)^2 = 5026.548246, the radii are at 8 cm, 16 cm, 24 cm, and 32 cm. The area inside the last ring is pi (40 cm)^2 - pi (32 cm)^2 = 1809.5574. The ratio of this area to the entire target area is 0.36, or 36%.

center r = 8

ring 1 r = 16

ring 2 r = 24

ring 3 r = 32

ring 4 r = 40

area ring 4 = pi 40^2 - pi 32^2 = pi(1600 - 1024) = 576 pi = 1810 sq cm

% of total area = 576pi/ 1600 pi = 576 /1600 * 100 = 36 %

Area = pi r^2

Whole circle is 3.14 40 40

5024 sq cm

Area of inner circles 3.14 32 32

3215.36

5024 total minus

3215.36 inner

1808.64

I say 1809 sq cm in the outside

1809/5024= 0.36 or 36 percent