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Combined figures

The most distinctive class of board-exam problems in this chapter shows a complex diagram with shaded regions and asks for an area. The recipe is the same every time: decompose the shaded region into known pieces, then add (or subtract) their areas. The pieces are sectors, segments, triangles, rectangles, and circles.

The general strategy

  1. Identify all the basic shapes present in the figure: full circles, semicircles, sectors, triangles, rectangles, squares.
  2. Identify the shaded region in terms of these basic shapes , what to add and what to subtract.
  3. Compute each basic shape's area using its formula.
  4. Sum or subtract as the decomposition demands.
  5. Check the answer's units and approximate magnitude.

Common configurations

Configuration A: Three semicircles on the sides of a right triangle. The classic lune of Hippocrates. Total shaded area = sum of semicircle areas on the legs minus the segments of the largest semicircle covered by the triangle.

Configuration B: Circle inscribed in a square. Shaded region (square minus circle) = s2π(s/2)2=s2(1π/4)s^2 - \pi(s/2)^2 = s^2(1 - \pi/4), where ss is the side.

Configuration C: Square inscribed in a circle. Shaded region (circle minus square) = πr22r2\pi r^2 - 2r^2 (because the inscribed square has diagonal =2r= 2r, side =r2= r\sqrt 2, area =2r2= 2r^2).

Configuration D: Four quarter-circles at the corners of a square. The shaded centre region equals s2π(s/2)2s^2 - \pi(s/2)^2 if the four arcs are quarter-circles of radius s/2s/2 from each corner.

Configuration E: Two overlapping circles (Venn-style lune). Compute via inclusion-exclusion: shaded = circle1_1 + circle2_2 – (intersection), where the intersection is two segments combined.

Examples and applications

Example 1. A square of side 1414 cm has a circle of diameter 1414 cm inscribed in it. Find the area of the region outside the circle but inside the square.

Square area =196= 196. Circle area =22/749=154= 22/7 \cdot 49 = 154. Shaded =196154=42= 196 - 154 = 42 cm2^2.

Example 2. A square of side 1414 cm has four quadrants of radius 77 cm at its four corners. Find the area of the region inside the square but outside the quadrants.

Total quadrant area =4(1/4)π(7)2=π49=154= 4 \cdot (1/4)\pi(7)^2 = \pi \cdot 49 = 154. Shaded =196154=42= 196 - 154 = 42 cm2^2. (Same answer as Example 1 , the four quadrants tile up to a full circle of radius 77.)

Example 3. A square has side 44 cm. A semicircle is drawn on each side as a diameter, all four bulging outward. Find the total area enclosed.

The four semicircles have radii 22, so each has area (1/2)π(4)=2π(1/2)\pi(4) = 2\pi. Total =8π= 8\pi. Square area =16= 16. Combined area =16+8π16+25.13=41.13= 16 + 8\pi \approx 16 + 25.13 = 41.13 cm2^2.

Example 4. A flower bed is in the shape of a circle of diameter 1414 m, with a square garden of side 1414 m surrounding it. Find the area of the garden (not the flower bed).

Square area =196= 196, circle area =154= 154 (as Example 1). Garden =196154=42= 196 - 154 = 42 m2^2.

Example 5. A piece of land is in the shape of an equilateral triangle of side 1414 m. A goat is tied at one vertex with a rope of length 77 m. Find the area the goat can graze.

At the corner of an equilateral triangle, the interior angle is 6060^\circ. The goat can graze a 6060^\circ sector of radius 77. Area =(60/360)(22/7)(49)=(1/6)(22)(7)=154/625.67= (60/360)(22/7)(49) = (1/6)(22)(7) = 154/6 \approx 25.67 m2^2.

Example 6 (compound sector). A pendant is in the shape of a circle of diameter 3232 mm hanging from a chain. There is a hole of diameter 66 mm at the centre. Find the area of the pendant (not the hole).

Outer area =π(16)2=256π= \pi(16)^2 = 256\pi. Hole =π(3)2=9π= \pi(3)^2 = 9\pi. Pendant =247π776.0= 247\pi \approx 776.0 mm2^2 (with π=22/7\pi = 22/7, =776.86= 776.86 mm2^2).

Worked examples (with diagrams to imagine)

Example A. In a circle of radius 2121, the perimeter of a sector is 5454. Find the sector's area.

Perimeter: 2r+=54=5442=122r + \ell = 54 \Rightarrow \ell = 54 - 42 = 12. Sector area =(1/2)r=(1/2)(21)(12)=126= (1/2) r \ell = (1/2)(21)(12) = 126 cm2^2.

Example B. A horse is tethered at the corner of a 1515 m by 88 m rectangular grass plot by a 77-m rope. Find the area of the plot the horse can graze.

The corner of a rectangle has interior angle 9090^\circ, so the horse grazes a 9090^\circ sector of radius 77: area =(1/4)(22/7)(49)=38.5= (1/4)(22/7)(49) = 38.5 m2^2.

Example C. A square is inscribed in a quarter circle of radius 55 cm such that one of its corners is at the centre of the quarter circle. Find the largest possible area of the square. (Concept: s2=5s=5/2s2=25/2=12.5s\sqrt 2 = 5 \Rightarrow s = 5/\sqrt 2 \Rightarrow s^2 = 25/2 = 12.5 cm2^2.)

Example D. A circular park of radius 1414 m has a square flower bed of side 1414 m inscribed in it. Find the area of the park outside the bed.

Square diagonal =14= 14 if inscribed (impossible since the side equals the diameter , re-read). If the square has side 1414, the circumscribed circle has diameter 14214\sqrt 2. If instead the square is inscribed in the circle of radius 1414, then diagonal =28= 28, side =142= 14\sqrt 2, square area =392= 392. Park area =22/7196=616= 22/7 \cdot 196 = 616. Outside-square area =616392=224= 616 - 392 = 224 m2^2.

Try it yourself

  1. A square of side 2020 cm has a circle inscribed in it. Find the area of the square minus the circle.
  2. A circular table of radius 77 cm has a square inscribed in it. Find the area of the circle minus the square.
  3. Four quadrants of radius 44 are cut from the corners of a 1414-cm square. Find the remaining area.
  4. A goat is tied to one corner of a 2020-m square plot by an 88-m rope. Find the grazing area.
  5. A semicircle of radius 1414 cm has a triangle inscribed with base == diameter. Find the area of the triangle if its third vertex is on the arc.
  6. A pizza of radius 2121 cm is divided into 88 equal slices. Find the area and arc-length of one slice.
  7. Two circles of radii 1010 and 66 touch externally. Find the area of the region enclosed by both circles' tangent lines and arcs (you'll need to sketch).
  8. A track is a rectangle of 100100 m by 5050 m with semicircular ends of diameter 5050 m. Find the total track area.
  9. A flower bed is in the shape of a sector of a circle, radius 1010, central angle 9090^\circ. Find its perimeter and area.
  10. A square has side 1414 cm. Four semicircles are drawn on its sides bulging outward. Find the total combined area.
  11. A circle of radius 77 has three equal sectors removed (each 6060^\circ). Find the remaining area.
  12. A swimming pool of dimensions 2020 m ×\times 1010 m has semicircular ends. Find the area of the pool.

Pitfalls / Insight

(1) Always decompose the figure before computing , never try to compute a complex shape's area in one stroke.

(2) Common trick: four quadrants of radius rr at the corners of a square tile up to one full circle of radius rr. Use this to short-circuit calculations.

(3) Be careful about "inscribed" vs "circumscribed". A square inscribed in a circle has diagonal == diameter. A circle inscribed in a square has diameter == side.

(4) For "goat grazing" problems, the available angle depends on the geometry of the boundary. At an interior corner of a square, the angle is 9090^\circ; at an equilateral triangle corner, 6060^\circ; in an open field, 360360^\circ.

(5) Always state your final answer in proper units (cm2^2, m2^2, etc.) and use the value of π\pi specified.

Practice quiz

Quick check on this topic.

Quiz
Quick check : Combined figures
6 questions · pick the best answer
Q1

Square side 77 cm with circle inscribed. Shaded (outside circle):

Q2

Four quadrants of radius rr at corners of square:

Q3

Square inscribed in circle of radius 77. Square side:

Q4

Square inscribed in circle of radius rr. Area of circle minus square:

Q5

A semicircle has radius 77. Area:

Q6

Track: rectangle 30×1430 \times 14 m with semicircular ends of diameter 1414 m. Track area: