The One Big Idea: Add and Subtract

"Shaded region" problems look scary but use a single strategy: build the shaded area out of simple pieces you already know — circles, sectors, semicircles, squares, rectangles and triangles — and add or subtract them.

Shaded area=(areas you include)−(areas you remove).\text{Shaded area} = (\text{areas you include}) - (\text{areas you remove}).

The whole skill is seeing the figure as a combination. Label each simple piece, write its area, then combine.

Circle Inside a Square (and Square Inside a Circle)

Largest circle in a square of side aa: the circle's diameter equals the side, so r=a2r = \dfrac{a}{2}. The leftover (corners) area is a2−π(a2)2=a2−πa24.a^2 - \pi\left(\dfrac{a}{2}\right)^2 = a^2 - \dfrac{\pi a^2}{4}.

Largest square in a circle of radius rr: the square's diagonal is the diameter 2r2r, so its area is (2r)22=2r2\dfrac{(2r)^2}{2} = 2r^2. The leftover (four circular gaps) is πr2−2r2\pi r^2 - 2r^2.

Key Point: Inscribed circle in a square: diameter == side. Inscribed square in a circle: diagonal == diameter.

Quarter Circles at the Corners

A favourite design: a square of side aa with a quarter circle drawn at each of the four corners, radius rr. The four quarter circles together make one full circle of radius rr. So Shaded (square outside the quarters)=a2−πr2,\text{Shaded (square outside the quarters)} = a^2 - \pi r^2, provided the quarters do not overlap (r≤a2r \le \tfrac{a}{2}).

Similarly, semicircles drawn on the sides of a square, or a design of overlapping circles, are handled by counting how many full circles/semicircles the pieces add up to.

Key Point: Four quarter circles of equal radius == one full circle. Two semicircles of equal radius == one full circle.

A square ABCD with a quarter circle drawn at each of its four corners; the four quarter circles together make one full circle, so the shaded region in the middle equals the square area minus the circle area.

A Working Checklist

For any shaded-region problem:

  1. Identify the outer boundary (square, rectangle, circle, triangle).
  2. Spot the circular pieces (full circles, sectors, semicircles, quadrants) inside.
  3. Decide which areas are added and which are removed to leave the shaded part.
  4. Substitute π=227\pi = \tfrac{22}{7} or 3.143.14 as told, and keep units (cm2^2, m2^2) consistent.

Key Point: Write the plan in words first ("big square −- circle ++ small triangles"), then compute. Most mistakes are planning mistakes, not arithmetic.

Solved Examples

Example 1: Circle in a square

A circle is inscribed in a square of side 14 cm. Find the area between the square and the circle. (π=227)\left(\pi=\dfrac{22}{7}\right)

Solution:

  1. Circle diameter == side =14⇒r=7=14 \Rightarrow r=7; circle area =227×49=154= \dfrac{22}{7}\times 49 = 154 cm2^2.
  2. Square area =142=196= 14^2 = 196 cm2^2.
  3. Shaded =196−154=42= 196 - 154 = 42 cm2^2.

Final Answer: 42 cm2^2.

Takeaway: Inscribed circle ⇒r=side2\Rightarrow r=\tfrac{\text{side}}{2}.

Example 2: Quarter circles at corners

From each corner of a square of side 14 cm, a quarter circle of radius 7 cm is removed. Find the area of the remaining (shaded) region. (π=227)\left(\pi=\dfrac{22}{7}\right)

Solution:

  1. Four quarter circles of radius 7 == one full circle =227×49=154= \dfrac{22}{7}\times 49 = 154 cm2^2.
  2. Square =196= 196 cm2^2.
  3. Shaded =196−154=42= 196 - 154 = 42 cm2^2.

Final Answer: 42 cm2^2.

Takeaway: Four equal quarter circles combine into one circle.

Example 3: Square inside a circle

Find the area of the region between a circle of radius 7 cm and the largest square inscribed in it. (π=227)\left(\pi=\dfrac{22}{7}\right)

Solution:

  1. Square's diagonal =2r=14= 2r = 14, so square area =1422=98= \dfrac{14^2}{2} = 98 cm2^2.
  2. Circle area =154= 154 cm2^2.
  3. Shaded =154−98=56= 154 - 98 = 56 cm2^2.

Final Answer: 56 cm2^2.

Takeaway: Inscribed square area =12d2=2r2= \tfrac12 d^2 = 2r^2.

Example 4: Two semicircles design

A rectangle is 14 cm by 7 cm with a semicircle drawn on each of the two shorter sides (diameter 7 cm), pointing outwards. Find the total area of the figure. (π=227)\left(\pi=\dfrac{22}{7}\right)

Solution:

  1. Rectangle =14×7=98= 14\times 7 = 98 cm2^2.
  2. Two semicircles of diameter 7 (radius 3.5) make one full circle: 227×3.52=227×12.25=38.5\dfrac{22}{7}\times 3.5^2 = \dfrac{22}{7}\times 12.25 = 38.5 cm2^2.
  3. Total =98+38.5=136.5= 98 + 38.5 = 136.5 cm2^2.

Final Answer: 136.5 cm2^2.

Takeaway: Two equal semicircles == one circle; here they are added to the rectangle.