10 min readPathway sheet3-mark shaded-region question; 5-mark long answer; case-based question

Problems on areas and perimeters of these figures

Break a design into circles, sectors, segments, triangles and squares. Add or subtract their areas. For a perimeter, add only the outer boundary lengths.

Shaded areaArea of the whole figure − area of the unshaded parts

PerimeterSum of the arcs and straight edges on the outer boundary only

Four corner quadrants of radius rTogether make one full circle, area πr²

Perimeter of a quadrant2r + πr/2

Method for shaded-region problems

1
Name each shape in the figure: square, triangle, sector, segment, semicircle.
2
Write the area formula for each shape.
3★
Decide whether to add or subtract each area to get the region asked for.
4
Substitute, using the value of π the question gives.
5
For a perimeter, list every boundary piece (arcs and straight lines) and add them.

★ key step

How to draw the figure

Copy the figure neatly and mark every centre and radius.
Shade only the region the question asks for.
Write the name of each part on it: quadrant, semicircle, segment.
Write each given length on the figure, with units.

Square tile with corner quadrants

Problem

A square tile of side 14 cm has a quadrant of radius 7 cm drawn at each of its four corners. Find the area of the tile outside the quadrants. (Use π = 22/7)

Working

Area of the square = 14 × 14 = 196 cm².
The four quadrants together make one circle of radius 7 cm.
Area of the quadrants = (22/7) × 7 × 7 = 154 cm².
Area outside = 196 − 154 = 42 cm².

Answer

42 cm²

Word problem: an arched window

Problem

A window is a rectangle 1.4 m wide and 2 m high, with a semicircle on top whose diameter is the width. Find (i) the area of the window and (ii) the length of the wooden frame along its outer edge. (Use π = 22/7)

Working

Area of the rectangle = 1.4 × 2 = 2.8 m².
Radius of the semicircle = 1.4 ÷ 2 = 0.7 m.
Area of the semicircle = (1/2) × (22/7) × 0.7 × 0.7 = 0.77 m².
Area of the window = 2.8 + 0.77 = 3.57 m².
Curved edge = (1/2) × 2πr = (22/7) × 0.7 = 2.2 m.
Outer frame = two sides + bottom + curved edge = 2 + 2 + 1.4 + 2.2 = 7.6 m. The top of the rectangle is inside, so it is not counted.

Answer

Area = 3.57 m²; frame = 7.6 m

A leaf design made of two segments

Problem

In a square ABCD of side 14 cm, two quadrants of radius 14 cm are drawn with centres A and C. Find the area of the leaf-shaped region where they overlap. (Use π = 22/7)

Working

Each quadrant has area (1/4) × (22/7) × 14 × 14 = 154 cm².
The diagonal BD cuts the leaf into two segments, each of central angle 90°.
Area of ΔABD = (1/2) × 14 × 14 = 98 cm².
One segment = 154 − 98 = 56 cm².
Leaf = 2 × 56 = 112 cm².

Answer

112 cm²

In real life

A kolam or rangoli made of circles and petals, an arched temple window or a clock face design is built from these shapes. Artists and carpenters estimate paint, glass and wood from their areas and perimeters.

Exam angle

2 marks: “Find the perimeter of a quadrant of a circle of radius 14 cm. (Use π = 22/7)” Arc = (1/4) × 2 × (22/7) × 14 = 22 cm. Perimeter = 14 + 14 + 22 = 50 cm.

3 marks: “A square of side 28 cm has a quadrant of radius 14 cm drawn at each corner. Find the area of the square outside the quadrants. (Use π = 22/7)” Square = 784 cm². Four quadrants = one circle = (22/7) × 196 = 616 cm². Area outside = 784 − 616 = 168 cm².

5 marks: “A circular garden has radius 14 m. A chord subtends 60° at its centre, and the minor segment it cuts off is a flower bed. Find (i) the area of the bed, (ii) the area of the rest of the garden and (iii) the length of the fence around the bed. (Use π = 22/7 and √3 = 1.73)” Sector = 102.67 m²; equilateral triangle = 49√3 = 84.77 m². Bed = 17.90 m²; rest = 616 − 17.90 = 598.10 m². Fence = arc 14.67 m + chord 14 m = 28.67 m.

Remember it

Area: add or subtract shapes. Perimeter: walk the outer edge only.