Area and Perimeter of Compound Shapes with Circles

Add the rectangle area and half the area of the matching circle.

1. The rectangle area is 12\times8=96\text{ cm}^2.
2. The semicircle radius is 4\text{ cm}, so its area is \frac12\pi\times4^2=8\pi\text{ cm}^2.
3. The total area is (96+8\pi)\text{ cm}^2.

The shared diameter is inside the shape and is not part of its outside boundary.

1. The straight boundary consists of two 10\text{ cm} sides and one 6\text{ cm} side, totalling 26\text{ cm}.
2. The semicircular arc is half a circumference: \frac12\pi\times6=3\pi\text{ cm}.
3. The perimeter is (26+3\pi)\text{ cm}.

Subtract the smaller circle area from the larger circle area.

1. The ring is the large circle with the small circle taken away. Area of a circle =\pi r^2.
2. The ring's area is \pi\times7^2-\pi\times3^2=49\pi-9\pi.
3. This is 40\pi\text{ cm}^2.

The square side is the circle diameter.

1. The square area is 10^2=100\text{ cm}^2.
2. The circle radius is 5\text{ cm}, giving area 25\pi\text{ cm}^2.
3. The required area is (100-25\pi)\text{ cm}^2.

Find the area of each semicircle, then subtract.

1. Half the outer circle has area \frac12\pi\times6^2=18\pi\text{ cm}^2.
2. Half the inner circle has area \frac12\pi\times4^2=8\pi\text{ cm}^2.
3. Subtracting gives 10\pi\text{ cm}^2.

Subtract only one quarter of the circle area.

1. The rectangle area is 9\times7=63\text{ cm}^2.
2. The quarter-circle area is \frac14\pi\times3^2=\frac{9\pi}{4}\text{ cm}^2.
3. The remaining area is (63-\frac{9\pi}{4})\text{ cm}^2.

The two semicircular arcs together make one full circumference.

1. The straight sides total 2\times12=24\text{ cm}.
2. The two arcs total \pi\times5=5\pi\text{ cm}.
3. The perimeter is (24+5\pi)\text{ cm}.

Include both semicircular arcs and both short straight edges.

1. The arc of a semicircle is half a circumference: \frac12\times2\pi r=\pi r.
2. The outer arc is 9\pi\text{ cm} and the inner arc is 5\pi\text{ cm}, so the arcs total 14\pi\text{ cm}.
3. Each straight edge runs from the inner semicircle to the outer one, so it is 9-5=4\text{ cm}. The two edges total 8\text{ cm}.
4. The perimeter is (8+14\pi)\text{ cm}.

Use half the full-circle area to find the radius first.

1. The radius satisfies \frac12\pi r^2=18\pi.
2. Thus r^2=36 and the positive radius is 6\text{ cm}.
3. The diameter is 2\times6=12\text{ cm}.

Subtract the area of the ring from the area of the outer circle.

1. The outer circle has area \pi\times10^2=100\pi\text{ cm}^2.
2. The ring is the outer circle with the inner circle removed, so the inner circle's area is 100\pi-64\pi=36\pi\text{ cm}^2.
3. So \pi r^2=36\pi, which gives r^2=36.
4. A radius is positive, so r=6\text{ cm}.

Follow the actual boundary through each notch rather than counting the removed sides.

1. Only the two 14\text{ cm} straight sides remain, totalling 28\text{ cm}.
2. The two notch arcs together have the circumference of a diameter-6\text{ cm} circle: 6\pi\text{ cm}.
3. The perimeter is (28+6\pi)\text{ cm}.

Find the large diameter by adding the two smaller diameters.

1. The large diameter is 6+14=20\text{ cm}, so its radius is 10\text{ cm}.
2. The three semicircle areas are 50\pi, \frac92\pi and \frac{49}{2}\pi\text{ cm}^2.
3. The shaded area is 50\pi-\frac92\pi-\frac{49}{2}\pi=21\pi\text{ cm}^2.

The four removed quarter circles have the area of one full circle.

1. The square area is 8^2=64\text{ cm}^2.
2. The four quarter circles total \pi\times4^2=16\pi\text{ cm}^2.
3. The remaining area is (64-16\pi)\text{ cm}^2.

Find the area of the backdrop, double it for two coats, then work out how many whole tins are needed.

1. The rectangle has area 5\times3=15\text{ m}^2.
2. The semicircle has radius 5\div2=2.5\text{ m}, so its area is \frac12\pi\times2.5^2=9.817\ldots\text{ m}^2.
3. The backdrop's area is 15+9.817\ldots=24.817\ldots\text{ m}^2. Two coats need enough paint for 2\times24.817\ldots=49.634\ldots\text{ m}^2.
4. Tins needed: 49.634\ldots\div6=8.27\ldots. Only whole tins can be bought, and 8 tins would not be enough, so round up to 9 tins.
5. The cost is 9\times12.50=112.50, so the tins cost £112.50.

Divide the remaining area by the original square area before converting to a percentage.

1. The square area is 144\text{ cm}^2 and the hole area is 16\pi\text{ cm}^2.
2. The percentage remaining is \frac{144-16\pi}{144}\times100\%=65.0934\ldots\%.
3. To one decimal place, 65.1\% remains.

Call the radius r and write the perimeter in terms of r: half a circumference plus a diameter.

1. Let the radius be r\text{ cm}. The arc is half a circumference, \frac12\times2\pi r=\pi r, and the diameter is 2r.
2. So the perimeter gives the equation \pi r+2r=36.
3. Factorise: r(\pi+2)=36, so r=\frac{36}{\pi+2}=7.0017\ldots\text{ cm}.
4. The area of the semicircle is \frac12\pi r^2=\frac12\pi\times7.0017\ldots^2=77.006\ldots\text{ cm}^2.
5. To 3 significant figures, the area is 77.0\text{ cm}^2.

The diagonal of the smaller square is a diameter of the circle.

1. The circle touches all four sides of the large square, so its diameter is 10\text{ cm} and its radius is 5\text{ cm}.
2. The circle's area is \pi\times5^2=25\pi\text{ cm}^2.
3. The corners of the smaller square are on the circle, so its diagonal is a diameter: 10\text{ cm}.
4. Let the smaller square have side s. Its diagonal is the hypotenuse of a right-angled triangle with two sides s, so by Pythagoras s^2+s^2=10^2, giving s^2=50.
5. The smaller square's area is s^2=50\text{ cm}^2, so the shaded area is (25\pi-50)\text{ cm}^2.

The outer semicircle radius increases by the path width, while the straight sections keep their length.

1. The inner area is 40\times20+\pi\times10^2=800+100\pi\text{ m}^2.
2. The outer width is 24\text{ m} and the outer semicircle radius is 12\text{ m}.
3. The outer area is 40\times24+\pi\times12^2=960+144\pi\text{ m}^2.
4. The path area is (960+144\pi)-(800+100\pi)=160+44\pi\text{ m}^2.

Set up the difference of the two circle areas; the squared terms will cancel.

1. The area equation is \pi(r+2)^2-\pi r^2=28\pi.
2. Divide by \pi and expand: r^2+4r+4-r^2=28.
3. Thus 4r=24, giving r=6\text{ cm}.

Use Pythagoras to find the semicircle diameter before calculating its area.

1. The hypotenuse is \sqrt{6^2+8^2}=10\text{ cm}, so the semicircle radius is 5\text{ cm}.
2. The triangle area is \frac12\times6\times8=24\text{ cm}^2.
3. The semicircle area is \frac12\pi\times5^2=\frac{25\pi}{2}\text{ cm}^2.
4. The total is (24+\frac{25\pi}{2})\text{ cm}^2.